General, Organic, and Biological Chemistry: A Guided Inquiry [1 ed.] 0471763594, 9780471763598

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GENERAL, ORGANIC, AND BIOLOGICAL CHEMISTRY A GUIDED INQUIRY MICHAEL P. GAROUTTE Missouri Southern State University

JOHN WILEY & SONS, INC

Copyright ¤ 2007 John Wiley & Sons, Inc. All rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning or otherwise, except as permitted under Sections 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc. 222 Rosewood Drive, Danvers, MA 01923, website www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030-5774, (201) 748-6011, fax (201) 7486008, Web site http://www.wiley.com/go/permissions. To order books or for customer service please, call 1-800-CALL WILEY (225-5945). ISBN 0-471-76359-4 Printed in the United States of America 10 9 8 7 6 5 4 3 2 1

To Susan, who put up wit h m e while I was writ ing t his book, and t o m y daught ers Audrey and Madeleine.

To the Instructor The act ivit ies present ed in General, Organic, and Biological Chem ist ry: A Guided I nquiry have been writ t en or adapt ed t o support process- orient ed guided- inquiry learning ( POGI L) in an allied healt h or GOB ( General- Organic- Biological) chem ist ry course. Current educat ional research shows t hat m ost st udent s experience im proved learning when t hey are act ively engaged, and when t hey are given t he opport unit y t o const ruct t heir own knowledge. POGI L act ivit ies are designed for use wit h self- m anaged t eam s t hat em ploy t he inst ruct or as a facilit at or of learning rat her t han as a source of inform at ion. Each act ivit y in t his set m ay be used independent ly, but som e quest ions m ay depend upon t opics exam ined in earlier act ivit ies. These act ivit ies do not replace a t radit ional t ext book, but rat her enhance it s use. Any st andard t ext m ay be used, and you are encouraged t o correlat e reading and/ or hom ework assignm ent s from t he t ext wit h t he Chem Act ivit ies in t his book. Many inst ruct ors will choose t o t each using POGI L during every class m eet ing. I f you are new t o guided inquiry, you m ight select one act ivit y per week t o int roduce m ore st udent part icipat ion int o your class. Chem Act ivit ies 3, 7, 10, 13, 15, 16, 19, and 21 are adapt ed from act ivit ies published in Moog, R.S. and Farrell, J.J. Chem ist ry: A Guided I nquiry, 2 nd and 3 r d edit ions, Wiley, 2002 and 2006. An I nst ruct or’s Guide, available as a PDF download from Wiley.com , cont ains inst ruct or not es for each act ivit y, exam ple quizzes, suggest ed answers for t he Crit ical Thinking Quest ions and solut ions t o t he Exercises. Anyone who t eaches a GOB nonm aj ors course will know t hat , from t he point of view of a chem ist , t he course is all about com prom ises. This set of act ivit ies is no except ion. I t ell m y st udent s in t his course t hat it is five sem est ers of chem ist ry com pressed int o one. Som e of you m ay have a t wo- sem est er GOB sequence. Regardless, choices m ust be m ade about which t opics are and are not im port ant , and about t he dept h t o which select ed t opics are invest igat ed. I have at t em pt ed t o focus here on a core set of skills t hat will help st udent s succeed in t heir fut ure courses and careers. This has m eant , for exam ple, m uch less em phasis on t radit ional organic chem ist ry react ions t han is present in m any current t ext books. For m ore inform at ion about t he POGI L proj ect and guided- inquiry learning, including a list of available t eacher resources, m at erials t o use in t he classroom , and references t o relevant research, please visit t he POGI L Web sit e at ht t p: / / www.pogil.org. Feedback regarding t he effect iveness of t he m at erials and suggest ions for im provem ent s would be appreciat ed. Send t his assessm ent inform at ion t o t he aut hor by em ail ( address available from t he Missouri Sout hern St at e Universit y Web sit e) .

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To the Student Many of you m ay see t his chem ist ry class as sim ply a requirem ent t o be m et so t hat you can get on wit h t he educat ion you “ really need” t o begin your career. And m any of you have been very successful in your academ ic careers by list ening t o lect ures and perform ing well on exam s. When you found out ( or perhaps are finding out right now) t hat in t his course, you will be spending m uch class t im e doing “ group work,” you m ay have felt t hat som eone placed a hurdle in your pat h. This class is different . This class is not sim ply about “ how t o do t he calculat ion,” nor is a m aj or goal of t his class sat isfied by sim ply learning what num bers t o plug int o what equat ion—t hough t hose t hings are good t o know. The m ain goal is t o underst and what t he num bers m ean. I n a class such as t his, you m ay be frust rat ed at t im es because you cannot im m ediat ely see t he “ right ” answer t o a quest ion. Act ually, it is by design t hat som e answers are not im m ediat ely obvious. Som et im es you will writ e an answer and go on t o a lat er quest ion, only t o find t hat t he lat er quest ion causes you t o reevaluat e t he earlier quest ion. This is OK! Lat er, when you have t he “ aha! ” m om ent , you will not easily forget what you have learned. The m ost com m on reason t hat st udent s feel t heir group is “ not working” is t hat one m em ber is cont ribut ing eit her t oo m uch or t oo lit t le. I t is each group m em ber’s responsibilit y t o m ake sure t hat all group m em bers cont ribut e. Do not hesit at e t o t alk t o your inst ruct or if you feel t hat your group needs help learning how t o be m ore product ive. Here is t he best single suggest ion, given by m any inst ruct ors and succeed in and enj oy t his course: find a st udy part ner or group, and you have st udy part ners, you have a reason t o be prepared ( t hey are if you can’t com e up wit h t he answer t oget her, t hen you are less inst ruct or—if none of you can figure it out , it m ust need som e t eacher

st udent s, for how t o m eet regularly! When count ing on you) , and shy about asking t he clarificat ion!

Hopefully in t his class, you will learn t hat chem ist ry is act ually relevant t o your career choice, and t hat st udying it can be fun as well! I f you have any suggest ions for im proving t his book, please em ail m e at m y universit y. Michael Garout t e Missouri Sout hern St at e Universit y July 2006

Some comments from former students in this course I t hought t hat I wouldn't like t his class but it has been really int erest ing t o com e t o class every day and act ually learn som et hing t hat pert ains t o life. I f you work well wit h ot hers and are able t o learn as a group and be challenged by your group of peers t o st rive t o really learn t he subj ect s t hen I recom m end t his class. But if you'd rat her work on your own and not get help from ot hers t hen t his class would be of no benefit t o you. This is one of t he hardest classes I have ever t aken. But was t he only one t hat t aught m e t o seek out t he answer inst ead of having it handed t o you. This class will help m e in fut ure classes, because I have gained good st udy skills. For t hat I wish t o t hank [ t he inst ruct or] : ) I t hink an open m ind and pat ience for t he first few weeks can best describe m y recom m endat ion t o ot her st udent 's want ing t o t ake t his class… I didn't like t his class at first , but it grew on m e and I learned m ore t han I t hought I would have. So, I will recom m end t his class t o ot her st udent s. When I t ook chem ist ry in High School, I did not underst and or rem em ber a t hing. Now t hat I have t aken t his class, I LOVE CHEMI STRY! I really like learning t hings on m y own and not j ust som eone t alking at m e. I feel t hat I have m ade a TREMENDOUS im provem ent in t he field of Chem ist ry, and j ust m akes m e furt her appreciat e t he subj ect . - v -

Acknowledgements Thanks t o Richard L. Schowen, Professor Em erit us, Universit y of Kansas, who was m y graduat e advisor, and who aft er reading t he first draft of m y dissert at ion, praised m y writ ing st yle and st at ed t hat I m ight want t o consider aut horing a t ext book som eday. Dick, your st at em ent gave m e t he confidence t o undert ake t his proj ect . This isn’t a t ext book, but I hope it m eet s wit h your approval. Thanks t o Andrei St raum anis, College of Charlest on, and Renée Cole, Cent ral Missouri St at e Universit y, who int roduced m e t o POGI L at a workshop in 2003, and who have provided m uch advice and encouragem ent . Thanks t o Rick Moog and Jim Spencer of Franklin & Marshall College, who didn’t laugh when I showed t hem t he first draft of t hese act ivit ies, and who apparent ly believed in t he adage t hat “ if you can’t say som et hing nice, don’t say anyt hing at all.” Special t hanks t o Rick, who has cont inued not t o laugh as I have repeat edly barraged him wit h quest ions about , well, everyt hing, and who reviewed som e of t hese act ivit ies and gave suggest ions for t heir im provem ent . Thanks also t o John Farrell, who along wit h Rick, wrot e t he book t hat inspired t his one. Thanks t o all t he people of t he POGI L proj ect , who have kindled m y ent husiasm by sharing t heir ideas and m at erials, by list ening, by allowing m e t o cont ribut e, and by cont inuing t o invit e m e t o share m y ideas. I can’t im agine a group of m ore helpful and support ive individuals. Thanks t o m y colleagues and depart m ent head at Missouri Sout hern, who have support ed m y choice t o “ t ry som et hing different ” in class. Many t eachers don’t have it so good. Thanks especially t o Mel Mosher, who gave m e advice when I was t eaching t his course for t he first t im e; and t o Marsi Archer who graciously agreed t o use an earlier version of t hese m at erials in a course in 2004, and who provided valuable feedback. Thanks t o Deborah Edson and t he folks at Wiley who agreed t o publish t his book, and who have been pat ient wit h all m y quest ions. Thanks t o all m y st udent s at Missouri Sout hern St at e Universit y, who have helped m e “ bet a- t est ” various versions of t hese m at erials. And finally, m any t hanks t o m y fam ily—parent s, in- laws, wife, and children—who have all cont ribut ed in som e way by m aking it possible for m e t o spend t im e working on t his book. I want all of you t o know t hat I haven’t forgot t en you, and t hat your sacrifices are appreciat ed.

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Ge n e r a l, Or ga n ic, a nd Biologica l Che m ist r y: A Gu ide d I nqu ir y

Table of Contents To t he I nst ruct or...................................................................................................... iv To t he St udent ........................................................................................................ v Acknowledgem ent s .................................................................................................. vi Table of Cont ent s..................................................................................................... 1 Recorder’s Report A ................................................................................................. 4 Recorder’s Report B ................................................................................................. 5 Chem Act ivit y 1 ........................................................................................................ 6 Working in Groups; Est im at ion ................................................................................ 6 Chem Act ivit y 2 ........................................................................................................ 7 Types of Mat t er; Chem ical and Physical Changes ....................................................... 7 Chem Act ivit y 3 ....................................................................................................... 12 At om s and The Periodic Table ................................................................................ 12 Chem Act ivit y 4 ....................................................................................................... 16 Unit Conversions: Met ric Syst em ............................................................................ 16 Chem Act ivit y 5 ....................................................................................................... 21 Measurem ent s and Significant Figures ..................................................................... 21 Chem Act ivit y 6 ....................................................................................................... 25 Densit y and Tem perat ure ...................................................................................... 25 Chem Act ivit y 7 ....................................................................................................... 28 Elect ron Configurat ion and The Periodic Table .......................................................... 28 Chem Act ivit y 8 ....................................................................................................... 33 Nuclear Chem ist ry ................................................................................................ 33 Chem Act ivit y 9 ....................................................................................................... 37 I ons and I onic Com pounds .................................................................................... 37 Chem Act ivit y 10 ..................................................................................................... 40 Covalent and I onic Bonds ...................................................................................... 40 Chem Act ivit y 11 ..................................................................................................... 44 Elect rolyt es, Acids and Bases ................................................................................. 44 Chem Act ivit y 12 ..................................................................................................... 48 Nam ing Binary Molecules, Acids and Bases .............................................................. 48 Chem Act ivit y 13 ..................................................................................................... 52 Molecular Shapes ................................................................................................. 52 Chem Act ivit y 14 ..................................................................................................... 61 Polar and Nonpolar Covalent Bonds......................................................................... 61 Chem Act ivit y 15 ..................................................................................................... 65 The Mole Concept ................................................................................................. 65 Chem Act ivit y 16 ..................................................................................................... 69 Balancing Chem ical Equat ions ................................................................................ 69 Chem Act ivit y 17 ..................................................................................................... 72 Predict ing Binary React ions.................................................................................... 72 - 1 -

Chem Act ivit y 18 ..................................................................................................... 74 Oxidat ion- Reduct ion React ions ............................................................................... 74 Chem Act ivit y 19 ..................................................................................................... 77 Mass Relat ionships ( St oichiom et ry) ......................................................................... 77 Chem Act ivit y 20 ..................................................................................................... 82 Therm ochem ist ry ................................................................................................. 82 Chem Act ivit y 21 ..................................................................................................... 87 Equilibrium .......................................................................................................... 87 Chem Act ivit y 22 ..................................................................................................... 92 Rat es of React ions................................................................................................ 92 Chem Act ivit y 23 ..................................................................................................... 96 Gases ................................................................................................................. 96 Chem Act ivit y 24 ................................................................................................... 100 Solut ions and Molarit y......................................................................................... 100 Chem Act ivit y 25 ................................................................................................... 103 Hypot onic and Hypert onic Solut ions ...................................................................... 103 Chem Act ivit y 26 ................................................................................................... 106 Acids and Bases ................................................................................................. 106 Chem Act ivit y 27 ................................................................................................... 110 Buffers ............................................................................................................. 110 Chem Act ivit y 28 ................................................................................................... 112 Alkanes, Cycloalkanes and Alkyl Halides ................................................................ 112 Chem Act ivit y 29 ................................................................................................... 116 Conform ers ....................................................................................................... 116 Chem Act ivit y 30 ................................................................................................... 120 Const it ut ional and Geom et ric I som ers ................................................................... 120 Chem Act ivit y 31 ................................................................................................... 124 I som ers ............................................................................................................ 124 Chem Act ivit y 32 ................................................................................................... 128 Propert ies of Organic Molecules ............................................................................ 128 Chem Act ivit y 33 ................................................................................................... 132 React ions of Organic Molecules............................................................................. 132 Chem Act ivit y 34 ................................................................................................... 135 Carbohydrat es ................................................................................................... 135 Chem Act ivit y 35 ................................................................................................... 140 Lipids ............................................................................................................... 140 Chem Act ivit y 36 ................................................................................................... 145 Am ino Acids and Prot eins .................................................................................... 145 Chem Act ivit y 37 ................................................................................................... 152 Energy and Met abolism ....................................................................................... 152 Chem Act ivit y 38 ................................................................................................... 157 Enzym es ........................................................................................................... 157 Chem Act ivit y 39 ................................................................................................... 160 Nucleic Acids ..................................................................................................... 160

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Chem Act ivit y 40 ................................................................................................... 166 Glycolyis ........................................................................................................... 166 Chem Act ivit y 41 ................................................................................................... 170 Cit ric Acid Cycle ................................................................................................. 170 Chem Act ivit y 42 ................................................................................................... 174 Elect ron Transport / Oxidat ive Phosphorylat ion ......................................................... 174 Chem Act ivit y 43 ................................................................................................... 179 Fat t y Acid Oxidat ion ........................................................................................... 179 Chem Act ivit y 44 ................................................................................................... 183 Ot her Met abolic Pat hways ................................................................................... 183 Chem Worksheet 1 ................................................................................................. 189 St oichiom et ry ( Mole Relat ionships) : Pract ice Worksheet 1 ........................................ 189 Chem Worksheet 2 ................................................................................................. 194 Gases: Pract ice Worksheet .................................................................................. 194 Chem Worksheet 3 ................................................................................................. 196 St oichiom et ry ( Mole Relat ionships) : Pract ice Worksheet 2 ........................................ 196 Chem Worksheet 4 ................................................................................................. 198 Funct ional Groups .............................................................................................. 198

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Recorder’s Report A Dat e

Chem Act ivit y

Manager

Recorder

Present er

Reflect or

Group Assessm ent :

/ 10

One t hing we learned t oday is:

One concept we need t o work on is:

Ot her Com m ent s:

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Recorder’s Report B Dat e

Chem Act ivit y

Manager

Recorder

Present er

Reflect or

Group Assessm ent :

/ 10

Com m ent s:

I m port ant ideas, concept s, result s, et c.

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Ch e m Act ivit y 1

Working in Groups; Estimation Information: Brief description of roles Much of t he class t im e in t his course will be spent working in groups of t hree or four. Each m em ber of t he group will be assigned t o a part icular role. Som e t ypical roles ( and t heir descript ions) are list ed below. Your inst ruct or will let you know how t he roles will funct ion in your course. M a n a ge r : Manages t he group. Ensures t hat m em bers are fulfilling t heir roles, t hat t he assigned t asks are being accom plished on t im e, and t hat a ll m e m be r s of t h e gr oup pa r t icipa t e t oge t he r in act ivit ies and un de r st a n d t h e con ce pt s. Pr e se n t e r : Present s t he work of t he group t o t he ent ire class when called upon. The present er m ay be called t o t he board t o writ e out and explain t he group's solut ion t o a problem . Frequent ly t he inst ruct or will ask what t he group responded t o a part icular quest ion or whet her t he group agrees wit h anot her group's response. I t is t he present er's role t o reply t o t hese quest ions. Re cor de r : Records ( on report form ) t he nam es of each of t he group m em bers at t he beginning of each day. Keeps t rack of t he group answers and explanat ions, along wit h any ot her im port ant observat ions, insight s, et c. The com plet ed report wit h answers t o any quest ions asked m ay be subm it t ed t o t he inst ruct or at t he end of t he class m eet ing. Re fle ct or or St r a t e gy An a lyst : Observes and com m ent s on group dynam ics and behavior wit h respect t o t he learning process. For exam ple, t he Reflect or m ight com m ent t hat a part icular group m em ber is dom inat ing t he discussion. I f a group m em ber is absent , t hen one m em ber m ay have t o fulfill m ore t han one role.

Model 1: A centimeter ruler

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Critical Thinking Question: 1. Est im at e t he num ber of ping- pong ( t able t ennis) balls t hat would com plet ely fill t his classroom . First , decide upon a “ plan of at t ack” as a group. You m ay or m ay not choose t o use t he cent im et er ruler. For t he purposes of t his exercise, you m ay assum e t hat t he room is rect angular in shape and t hat it is com plet ely em pt y of desks, people, et c. You m ay get up and m ove around t he room . When your group has an answer, t he recorder m ay be asked t o writ e it on t he board.

Exercise: 1. Read t he assigned pages in t he t ext , and work t he assigned problem s. CA01

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Ch e m Act ivit y 2

Types of Matter; Chemical and Physical Changes ( How can we classify m at t er?)

Model 1: Examples of various types of matter at room temperature I t em

Classificat ion

St at e ( or st at es)

Form ula ( or form ulas)

alum inum

elem ent

solid

Al( s)

hydrogen

elem ent

gas

H2 ( g)

wat er

com pound

liquid

H2 O( l)

t able salt

com pound

solid

NaCl( s)

salt wat er

hom ogeneous m ixt ure

aqueous solut ion

H2 O( l) and NaCl( aq)

coffee ( “ black” )

hom ogeneous m ixt ure

aqueous solut ion

H2 O( l) and m any ot hers

m uddy wat er

het erogeneous m ixt ure

liquid + solid

H2 O( l) and ot her st uff

Critical Thinking Questions: 1. Consider Model 1. How does t he form ula of an elem ent differ from t hat of a com pound?

2. How can you dist inguish elem ent s from com pounds based on t heir chem ical form ulas?

3. How does a pure subst ance ( i. e., elem ent or com pound) differ from a m ixt ure? Describe.

4. Hypot hesize on t he m eaning of t he labels ( s) , ( l) , ( g) , and ( aq) on t he form ulas.

Information: States of matter Mat t er can be classified by it s physical st a t e ( or pha se ) : solid, liqu id, or ga s. Most subst ances can exist in each of t he t hree st at es depending on t em perat ure and pressure. For exam ple, H2 O is norm ally a liquid at room t em perat ure, ice at t em perat ures below 0°C, and vapor at t em perat ures above 100°C. Alt hough we t hink of iron as a solid, it can be m elt ed and even vaporized if t he t em perat ure is high enough. Changes bet ween t hese st at es are usually considered physical—not chem ical—changes, since t he chem ical form ula is t he sam e.

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CA02

The pha se la be ls ( s) , ( l) , or ( g) can be writ t en aft er a form ula t o signify t he physical st at e. So, H2 O( g) would m ean gaseous wat er, i. e., wat er vapor.

Critical Thinking Questions: 5. Describe what is happening in t his process: H2 O( l) o H2 O( s)

6. Would t he process in Crit ical Thinking Quest ion 5 ( CTQ 5 ) be considered a chem ical or a physical process? Explain.

Information: Classifications of matter Chem ist ry is t he science t hat deals wit h m at t er and t he changes m at t er undergoes. Mat t er can be divided int o t wo m ain t ypes: pure su bst a n ce s and m ix t u r e s of subst ances. A su bst a n ce cannot be separat ed int o ot her kinds of m at t er by physical processes such as filt ering or evaporat ion, and is eit her an e le m e n t ( e. g., alum inum ) or a com pou n d ( e. g., H2 O) . Com pounds are m ade of t wo or m ore elem ent s chem ically com bined. The elem ent s t hem selves cannot be separat ed int o sim pler subst ances even by a chem ical react ion. On t he ot her hand, m ix t u r e s can be separat ed by physical m eans. Mixt ures t hat have t he sam e com posit ion t hroughout are called h om oge n e ou s ( e. g., salt wat er) ; t hose t hat do not are called h e t e r oge n e ou s ( e. g., I t alian salad dressing) . At om s of a single elem ent can exist as individual at om s ( e. g., alum inum , Al) or in chem ical com binat ion t o form m olecules ( e. g., hydrogen, H2 ) . When at om s of t wo or m ore elem ent s com bine chem ically t hey form com pounds ( e. g., wat er, H2 O) . Subscript s following an elem ent in t he form ula represent t he num ber of at om s of t hat elem ent in t he form ula.

Model 2: Flow chart for classifying matter

Matter

Mixture

Pure Substance

Element

Homogeneous

Compound

Heterogeneous

Model 3: Some different representations of the water molecule H2 O

HOH

form ulas

CA02

H–O–H

O H

H

Lewis st ruct ures

ball- and- st ick

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spacefilling

Critical Thinking Questions: 7. Look at a periodic t able of t he elem ent s. About how m any elem ent s are known?

8. Approxim at ely how m any elem ent s are m et als? ( Est im at e, don’t count ! )

9. There are t h ou sa nds of organic com pounds known—com pounds form ed out of only a few different elem ent s ( carbon, hydrogen, oxygen, nit rogen) . Explain how t his can be possible.

Exercises: 1. Writ e t he form ula of each m olecule for which t he ball- and- st ick st ruct ures are shown. Key:

= carbon

= oxygen

= hydrogen

a.

b.

c.

d.

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CA02

2. Using a periodic t able, ident ify t he elem ent s represent ed in each form ula, and st at e t he num ber of at om s of each elem ent in t he form ula. The first one has been done for you. a. NH3 ( am m onia)

one nit rogen at om , t hree hydrogen at om s

b. C6 H12 O6 ( glucose) c. Mg( OH) 2 ( m ilk of m agnesia) d. H2 SO4 ( sulfuric acid, “ bat t ery acid” ) e. C17 H18 F3 NO ( fluoxet ine, Prozac) 3. Using t he flow chart in Model 2 t o help you, classify each of t he following as eit her a m ixt ure or pure subst ance. For each su bst a n ce , t ell whet her is it an e le m e n t or a com pou n d. For each m ix t u r e , t ell whet her it is hom oge ne ous or h e t e r oge n e ou s; t hen list t wo or m ore com ponent s of t he m ixt ure. a. a lead weight b. apple j uice c. baking soda ( NaHCO3 ) d. air e. a 14- karat gold ring f.

a 24- karat gold coin

g. helium in a balloon h. beach sand i.

concret e

j.

whole blood

k. carbon dioxide 4. I n t he space below, draw a pict ure of t hree wat er m olecules in t he ball- and- st ick represent at ion.

CA02

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5. Which of t he choices below ( I or I I ) would best represent t he t hree m olecules you drew in Exercise 4? Explain your choice. Choice I H 6 O3

Choice I I 3 H2 O

6. Learn t he n a m e s and sym bols of t he elem ent s your inst ruct or suggest s. A good st art ing point is t he first 30 elem ent s, plus Br, Sr, Ag, Sn, I , Ba, Pt , Au, Hg, Pb. Spelling count s! You do n ot need t o m em orize any n u m be r s, as a periodic t able will always be available for your use. 7. Read t he assigned pages in t he t ext , and work t he assigned problem s.

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CA02

Ch e m Act ivit y 3

Atoms and The Periodic Table* ( What are at om s?)

Model 1: Schematic diagrams for various atoms proton (+) neutron (no charge) electron (-) Hydrogen 1

Hydrogen 2

H

1.0078 amu

Hydrogen 3

H

2.0141 amu

Carbon

24

Mg

12 protons 12 neutrons

23.9850 amu 1

13

C

13 -

6 protons 7 neutrons

6 protons 7 neutrons

13.0034 amu Magnesium

C

13.0039 amu Magnesium 26

Mg

12 protons 13 neutrons

24.9858 amu

Magnesium ion 24

Mg

12 protons 14 neutrons

25.9826 amu

H, 2H and 3H are isotopes of hydrogen. 12

1.0073 amu Carbon ion

C

25

H

3.01605 amu

6 protons 6 neutrons

exactly 12 amu

1 +

H

Carbon 12

Magnesium

Hydrogen ion

12

Mg+2

12 protons 12 neutrons

23.9839 amu

C and 13C are isotopes of carbon.

C and 13C may also be written as "carbon-12" and "carbon-13"

The nucleus of an atom contains the protons and the neutrons (if any).

*

Adapt ed from Chem Act ivit y 1, Moog, R.S.; Farrell, J.J. Chem ist ry: A Guided I nquiry, 3 rd ed., Wiley, 2006, pp. 2- 5.

CA03

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Critical Thinking Questions: 1. Look at t he Schem at ic Diagram s for carbon. What do all t hree carbon at om s ( and ions) have in com m on?

2. What do all four hydrogen at om s and ions have in com m on?

3. What do all m agnesium at om s and ions have in com m on?

4. Look at a periodic t able ( for exam ple, in your t ext ) . Considering your answers t o CTQs 1- 3, what is t he significance of t he a t om ic num be r , above each elem ent in t he t able?

5. How m any prot ons are in all chlorine at om s?_____ wit h 18 prot ons? _____ Why or why not ?

Do you t hink chlorine at om s exist

6. What is t he significance of t he superscript ed + or – sym bols t o t he right of t he sym bols a) for t he hydrogen ion? b) for t he carbon ion? c) Writ e one com plet e sent ence t o define t he t erm ion.

d) A posit ively charged ion is called a cat ion ( pronounced “ cat ion” ) , and a negat ively charged ion is called an anion. Which t erm applies t o t he m agnesium ion?

7. I n a box in t he corner of each schem at ic in Model 1 is t he elem ent sym bol and t he m ass num ber for t he at om ( superscript on t he left side of t he elem ent sym bol) . a) How is t his m ass num ber det erm ined?

b) Why is it called a “ m ass” num ber?

- 13 -

CA03

8. What is t he m ass num ber for t he following at om s: a)

37

Cl _____

b)

238

U _____

c) carbon- 12 _____ d) carbon- 13 _____

9. What st ruct ural elem ent do all isot opes of an elem ent have in com m on? How are t heir st ruct ures different ?

10. Considering what you know about isot opes, do all at om s of an elem ent weigh t he sam e? Explain.

Model 2: The periodic table Look at a periodic t able ( e. g., in your t ext book) . For each elem ent , t he a t om ic nu m be r is shown above t he elem ent sym bol, and t he average a t om ic m a ss ( or a t om ic w e ight ) is shown underneat h. Since t he t able will always be available, t here is no reason t o m em orize eit her of t hese num bers. There are 18 colum ns, known as fa m ilie s or gr oups. The groups can be num bered in various ways. We will use t he num bers 1- 18 for group designat ions. Elem ent s in t he sam e group oft en have sim ilar propert ies. Som e groups have fam ily nam es. Your t able m ay be divided int o t he m et als, sem im et als, and nonm et als. I f not , not e t hat a “ st air- st ep” line which st art s bet ween boron ( B) and alum inum ( Al) roughly divides m et als from nonm et als. Sem im et als ( or “ m et alloids” ) lie along t he line. There are seven periods, or rows, in t he t able. The t wo rows of 14 elem ent s at t he bot t om act ually fit bet ween groups 2 and 3 on t he m ain t able. They are shown at t he bot t om because t hat way, t hey “ fit ” on t he page bet t er.

Critical Thinking Questions: 11. Are m ost elem ent s m et als or nonm et als? 12. From your experience, list som e propert ies of m et als.

13. From your experience, list som e propert ies of nonm et als.

14. Transist ors and com put er “ chips” are m ade of sem iconduct or m at erials. What kind of elem ent s would be used for t his purpose? Give an exam ple.

CA03

- 14 -

Exercises: 1. Fill in t he following t able. Sym bol 40

At om ic N u m be r

M a ss N u m be r

N um be r of Pr ot on s

N um be r of N e u t r on s

N um be r of Ele ct r on s

15

17

18

K

Zn

2+

38 35

81

36

2. Com plet e t he t able below, classifying each elem ent as a m et al, nonm et al, or sem im et al. Wat ch spelling! Sym bol

Nam e

Cla ssifica t ion

Pd Cl germ anium 3. Com plet e t he t able below, classifying each elem ent as an alkali m et al, alkaline eart h m et al, t ransit ion m et al, halogen, or noble gas. Wat ch spelling! Sym bol

Nam e

Cla ssifica t ion

F Ca rubidium Cr krypt on 4. Learn t he fam ily nam es of groups 1, 2, 17, 18. 5. Learn t o recognize which elem ent s are m et als, nonm et als, sem im et als ( m et alloids) , t ransit ion elem ent s or m ain group elem ent s. 6. Read t he assigned pages in t he t ext , and work t he assigned problem s.

- 15 -

CA03

Ch e m Act ivit y 4

Unit Conversions: Metric System ( What is a conversion fact or?)

Model 1: Fuel efficiency of a particular automobile A part icular aut om obile can t ravel 27 m iles per gallon of gasoline used. The aut om obile has a 12- gallon gasoline t ank. At a part icular locat ion, gasoline cost s $3.00 ( 3.00 USD) per gallon.

Critical Thinking Questions: 1. Three st at ist ics are given in Model 1. Circle t he t wo st at em ent s t hat give num erical r a t ios. 2. One st at em ent in Model 1 gives a m easured qu a n t it y. Writ e t he quant it y ( wit h t he associat ed u n it ) .

3. Writ e each rat io t hat you circled in Model 1 as a fr a ct ion. Your fract ion should have a n u m be r and a u n it in bot h t he num erat or and t he denom inat or of t he fract ion.

4. How m any m iles can t he aut om obile t ravel on a full t ank of gasoline? Show your work by writ ing t he qu a n t it y from CTQ 2 m ult iplied by t he appropriat e fr a ct ion from CTQ 3. Show all unit s.

5. Explain why t he a n sw e r t o CTQ 4 does not include t he unit “ gallons.”

6. Explain why t he fr a ct ion used in CTQ 4 m ay be called a con ve r sion fa ct or .

7. Do all four conversion fact ors below give equivalent inform at ion? Explain your answer. 27 m i 1 gal

CA04

27 m i gal

1

1 gal 27 m i

27

gal

mi

- 16 -

Model 2: Definitions of the inch and the foot 1 inch = 2.54 cm ( exact ly)

0

0

1

2

3

4

1

5

6

7

2

8

9

3

10

11

12

4

13

14

5

15

6

There are exact ly 12 inches in one foot .

Critical Thinking Questions: 8. How m any cent im et ers are in one inch? 9. Draw a large X t hrough each in cor r e ct conversion fact or below.

1 cm 2.54 in

2.54 cm 1 in

2.54 in 1 cm

1 in 2.54 cm

10. Suppose you want t o convert a height from inches int o cent im et ers. Circle t he conversion fact or in CTQ 9 t hat you would use. Explain your choice.

Model 3: The unit plan A unit plan begins wit h t he unit of t h e k now n qu a n t it y and shows how t he unit s will change aft er m ult iplying by each conversion fact or used, in order. The unit plan for CTQ 10 would be: in o cm Each arrow in t he unit plan represent s one conversion fact or.

Critical Thinking Questions: 11. A basket ball player is seven feet t all. a. Using Models 2 and 3 for reference, com plet e t he unit plan for convert ing t he height of t he basket ball player int o cent im et ers. feet oo b. Writ e t he t wo conversion fact ors corresponding t o each arrow in part ( a) .

c. Perform t he calculat ion by m ult iplying t he qua n t it y by each conve r sion fa ct or in order. Show all work wit h unit s.

- 17 -

CA04

Information: Metric system units and prefixes Ta ble 1 : Som e m e t r ic syst e m un it s

Ta ble 2 : Som e m e t r ic syst e m pr e fix e s

Quant it y

Unit

Sym bol

Prefix

Sym bol

Fact or

Meaning

dist ance t im e m ass volum e t em perat ure

m et er second gram lit er kelvin

m s g L K

m ega kilo deci cent i m illi m icro

M k d c m P

10 6 10 3 10 - 1 10 - 2 10 - 3 10 - 6

one m illion one t housand one t ent h one hundredt h one t housandt h one m illiont h

The I nt ernat ional ( m et ric) Syst em prefixes can be associat ed wit h any unit . So for exam ple, one can writ e m illim et ers ( m m ) , m illiseconds ( m s) , m illigram s ( m g) , or m illilit ers ( m L) . There are m ore prefixes t han t hese. For exam ple, personal com put er clock speeds are m easured in gigahert z ( 1 GHz = 10 9 cycles per second) , and t he MCV ( Mean Cell Volum e) of red blood cells is m easured in fem t olit ers ( 1 fL = 10 - 15 lit ers) . But we will st ick t o t he m ost com m on ones. You w ill n e e d t o m e m or ize t h e infor m a t ion in Ta ble s 1 a nd 2 .

Critical Thinking Question: 12. Circle t he unit t hat would be com m only used t o m easure t he indicat ed it em . a. The volum e of liquid in a can of Coca- Cola:

PL

mL

L

kL

ML

b. The m ass of a hum an being:

Pg

mg

g

kg

Mg

c. The m ass of aspirin in one t ablet :

Pg

mg

g

kg

Mg

d. The volum e of t he gasoline t ank in a Volkswagen

PL

mL

L

kL

ML

Model 4: Conversion factors from metric system prefixes The prefixes from Table 2 give you an e qua lit y t o t he base unit ( t he unit wit h no prefix) . For exam ple: 1 m m = 10 –3 m

1 ks = 10 3 s

1 cg = 10 –2 g

1 PL = 10 - 6 L

The equalit y can be used t o creat e a con ve r sion fa ct or . For m illim et ers, t he possible conversion fact ors are* : 1 mm 10 - 3 m and 1 mm 10 -3 m Ex a m ple Pr oble m : How m any m m are in 1.89 m ? The unit plan is: m o m m 1 mm

, will cause t he m et ers ( m ) t o cancel 10 -3 m out , so we should choose t hat one. Then do t he conversion: Using t he fir st conversion fact or in Model 4,

1.89 m u

1 mm 10 -3 m

1890 m m

Rem em ber, t o ent er 10 –3 int o m ost sim ple scient ific ( non- graphing) calculat ors, use t he e x pon e n t ( EE or EXP) key ( e. g. , press 1 EE 3 ± ) . Do not use t he pow e r ( y x ) key. * I f you have already m em orized equalit ies such as “ 1 km = 1000 m ” and “ 1000 m m = 1 m ,” you m ay use t hem . However, t his book will always use t he powers of 10 and prefixes shown in Table 2.

CA04

- 18 -

Critical Thinking Question: 13. How m any cent im et ers are in 2.2045× 10 - 2 kilom et ers? a. Com plet e t he unit plan. km oo b. Writ e t he equalit ies from Table 2 for each st ep in part ( a) .

c. Writ e t he conversion fact ors from t he equalit ies in part ( b) .

d. Perform t he calculat ion, showing all work.

Exercises: 1. Writ e unit plans, and t hen perform t he unit conversions below. Use only t he conversion fact ors given in t he problem or t hose given in Table 2. a. Convert 36 m g t o g. com plet e t he unit plan: m g o

com plet e t he equalit y: 1 m g =

writ e t he t wo conversion fact ors: perform t he conversion: b. Convert 5.51 m s t o Ps. unit plan: conversion fact ors: perform t he conversion:

2. A t raveler want s t o know t he cost ( in dollars) of t he gasoline t o t ravel 75 m iles in t he aut om obile in Model 1. a. Writ e a unit plan for t he conversion. b. Writ e t he t wo conversion fact ors needed for t he conversion.

c. Perform t he calculat ion, showing all work and unit s.

- 19 -

CA04

3. What is t he weight in gram s of a 7.5- karat diam ond engagem ent ring? There are exact ly 200 m g in one karat . Make a unit plan first .

4. The speed of light is 299,792,458 m / s. a. What is t his speed in km / hour? ( Com plet e t he unit plan:

m km m o o o s min

).

Writ e t he t hree equalit ies first .

b. How long does it t ake for light t o t ravel 100 m et ers? ( Unit plan: m o ?)

Information: Units of volume One m illilit er ( m L) is equal t o one cubic cent im et er ( 1 cc or 1 cm 3 ) . One lit er ( L) is equal t o one cubic decim et er ( 1 dm 3 ) . 5. Convert 1000 cm 3 t o dm 3 using only t he equalit ies given in t he inform at ion above and t he m et ric syst em prefix m illi ( 10 - 3 ) . Show work. Make a unit plan first .

6. There are 5280 feet in a m ile. Convert 161 km t o m iles, using only equalit ies t aken from t his act ivit y. Make a unit plan first .

7. The daily dose of am picillin for t he t reat m ent of an ear infect ion is 115 m g am picillin per kilogram of body weight ( 115 m g/ kg) . What is t he daily dose for a 27- lb child? Make a unit plan first .

8. Read t he assigned pages in t he t ext , and work t he assigned problem s. CA04

- 20 -

Ch e m Act ivit y 5

Measurements and Significant Figures ( How do we m easure t hings scient ifically?)

Model 1: Outlines of a quarter and a glass bottle on top of a centimeter ruler

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

Critical Thinking Questions: 1. Use t he ruler shown t o det erm ine t he diam et er of t he quart er in Model 1 in cent im et ers. Express your answer as a de cim a l n u m be r w it h 2 digit s follow in g t h e de cim a l poin t . I nclude unit s.

2. Explain how you det erm ined t he final decim al place in t he m easurem ent in CTQ 1.

3. Use t he ruler shown t o det erm ine t he diam et er of t he glass bot t le in Model 1 in cent im et ers. Express your answer as a decim al num ber wit h t wo digit s aft er t he decim al point . I nclude unit s.

4. Explain how you det erm ined t he final decim al place in t he m easurem ent in CTQ 3.

5. Writ e t he t wo conversion fact ors t hat relat e inches t o cent im et ers.

- 21 -

CA05

6. Convert t he diam et er of t he glass bot t le t o inches. Show work.

Information: Precision Generally, when m easurem ent s are report ed, t here is unce r t a int y in on ly t h e fina l digit . This leads t o a problem deciding how t o report t he answer t o CTQ 6. We knew t he diam et er of t he bot t le t o ( at m ost ) t hree digit s of pr e cision ( for exam ple, 5.70 cm ) t o begin wit h. These t hree digit s are called sign ifica n t figur e s ( “ sig figs” ) or sign ifica n t digit s. When t he conversion t o inches is perform ed, t here are m any digit s aft er t he decim al point . We cannot cr e a t e m ore precision j ust by changing t o t he English syst em , so we should r ou nd off t o t hree sig figs. I f t he digit following t he t hird significant figure is at least 5, we should round u p; ot herwise, sim ply drop t he t railing digit s.

Critical Thinking Questions: 7. 2.244094488 cm rounded t o t hree “ sig figs” becom es 8. 2.248031496 cm rounded t o t hree “ sig figs” becom es 9. According t o t he U.S. m int , quart ers are act ually 2.426 cm in diam et er. Does your m easurem ent in CTQ 1 agree wit h t his? Explain.

10. Convert t he diam et er of a quart er int o inches. Show your work. Round your answer t o t he correct num ber of significant figures.

Information: Determining which figures in reported numbers are significant Ta ble 1 : Ru le s for sign ifica n t figu r e s in m e a sur e m e n t s Rule

Exam ple ( sig figs are underlined) 2.25

1. All non- zeroes are significant 2. Leading zeroes are n ot significant

0.0054

3. Trapped zeroes a r e significant

203, 0.0302

4. Trailing zeroes are significant on ly if an explicit decim al point is present * 5. Num bers t hat are e x a ct by definit ion have an in fin it e num ber of significant figures. This includes conversions wit hin a m easurem ent syst em ( and also 1 in = 2.54 cm ) . *

See why scient ific not at ion is useful? I t elim inat es any am biguit y.

CA05

- 22 -

700, 700., 7.00 × 10 2 189 cm u

1m 100 cm ( exact )

1.89 m

Critical Thinking Questions: 11. Give t he num ber of significant figures in each m easurem ent . a. 100 m b. 0.00095 g c. 3600.0 s 12. Writ e t he following m easurem ent s in scient ific not at ion, wit hout changing t he num ber of significant digit s. a. 100 m b. 0.00095 g c. 3600.0 s 13. Round each of t he following t o t hree significant digit s. a. 0.00320700 L b. 3.265 × 10 –4 m c. 129762 s

Information: Significant figures in calculations Table 2 gives som e rules of t hum b t hat help det erm ine how m uch precision t o report in t he result of a calculat ion. These rules involve det erm ining which digit s in each num ber are sign ifica nt ( i. e., t hey indicat e t he precision) and how t o t reat t hem in calculat ions. These rules are sim ple t o apply but only provide an e st im a t e of t he correct level of precision. This est im at e is preferable t o sim ply including all possible digit s in t he result of a calculat ion, but not as good as a rigorous error analysis. Furt herm ore, since m ost calculat ions in chem ist ry involve m ult iplicat ion or division, we can oft en use t he follow ing rule of t hum b: Keep only as m any significant figures in your answer as in t he m easurem ent wit h t he fewest num ber of sig figs. Also, t o avoid com pounding errors, one should never round off int erm ediat e result s, but wait t o round unt il you have t he final answer. Ta ble 2 : Ru le s for ca lcu la t ion s w it h sign ifica n t figur e s Operat ion

Rule

Exam ple ( sig figs underlined) 250 ÷ 7.134 = 35.043 = 35

Mult iplicat ion and division

Keep t he sm allest num ber of sig figs.

Addit ion and subt ract ion ( do a ft e r m ult iplicat ion and division)

Keep t he sm allest num ber of de cim a l pla ce s.

73.147 + 52.1 + 0.05411 = 125.30111 = 1 2 5 .3

Only digit s in t he m ant issa ( aft er t he decim al) are significant

log( 2003) = 3.3016809 = 3 .3 0 1 7

Logarit hm s

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CA05

Exercises: 1. Consider t he quart er in Model 1 and it s diam et er t hat you m easured in cent im et ers. a. Writ e a unit plan t o convert t he diam et er int o m illim et ers.

b. Writ e t he t wo conversion fact ors needed.

c. Perform t he conversion, showing all work.

2. Follow t he procedure in Exercise 1 t o convert t he diam et er of t he glass bot t le int o m icr om e t e r s.

3. Perform t he calculat ions, report ing your answer wit h t he correct num ber of sig figs. a. 26.234 g – 5.6 g = cups

b. 67.6 oz ÷ 8.0 oz/ cup =

2.54 cm cm = 1 in 4. Calculat e t he circum ference of t he glass bot t le from Model 1 in cent im et ers. Report your answer wit h t he correct num ber of significant figures. ( Circum ference = S × diam et er.)

c. 189 cm + 6.0 in u

5. Consider t he pict ure of t he part ially filled buret t e at t he right . Circle t he reading below t hat shows t he correct level of precision. Explain your answer. 15 m L

15.4 m L

15.40 m L

15

15.400 m L

16

6. Read t he assigned pages in your t ext , and work t he assigned problem s. CA05

- 24 -

Ch e m Act ivit y 6

Density and Temperature ( What is m easurem ent useful for?)

Critical Thinking Question: 1. Which weighs m ore, a t on of bricks or a t on of cot t on balls?

Information: Density CTQ 1 is a “ t rick” quest ion t hat I rem em ber as an early childhood “ j oke.” Of course, t he bricks and t he cot t on balls weigh t he sam e ( 1 t on) . But t he cot t on balls t ake up m ore space ( volum e) , m aking t hem less de n se . Densit y is a m easure of t he m ass of a part icular volum e of som et hing. densit y

m ass volum e

or d

m V

Som et im es, som et hing called spe cific gr a vit y is used t o report densit y. The specific gravit y ( sp gr) is a rat io of t he densit y of som et hing t o t he densit y of wat er. specific gravit y ( sp gr)

densit y of sam ple densit y of wat er

Since t he densit y of wat er at room t em perat ure is 1.00 g/ m L, t he sp gr of a sam ple is t he sam e as t he densit y, except t hat no unit s are given. I f you know t he sp gr, j ust add unit s of g/ m L t o give t he densit y.

Suggested Demonstration: Density tower Critical Thinking Questions: 2. A piece of t in m et al weighing 85.251 g is placed int o 41.1 cm 3 of et hyl alcohol ( d = 0.798 g/ cm 3 ) in a graduat ed cylinder, and it sinks t o t he bot t om . The alcohol level increases t o 52.8 cm 3 . a. What is t he m ass of t he t in? b. What is t he volum e of t he t in? c. What is t he densit y of t he t in? Show work.

d. What is t he specific gravit y of t he t in?

3. I f you have 750 m L of Everclear ( 95% et hyl alcohol, d = 0.80 g/ m L) , what m ass of liquid do you have? ( First , use algebra t o rearrange t he densit y equat ion t o solve for m ass.)

- 25 -

CA06

Model: Three temperature scales, Fahrenheit, Celsius and Kelvin

212 ºF

100 ºC or 373 K

Normal body temperature 98.6 ºF

37 ºC or 310 K

32 ºF

0 ºC or 273 K

Boiling point of water

Freezing point of water

Norm al body t em perat ure is 98.6 degrees Fahrenheit ( 98.6 °F) There are 66.6 Fahrenheit degrees ( F°) bet ween t he freezing point of wat er and body t em perat ure.

Critical Thinking Questions: 1. Writ e an equat ion t o convert Celsius and Kelvins. K = °C + _______ 2. How m any Celsius degrees are t here bet ween t he freezing and boiling point s of wat er?____ 3. How m any Kelvins are t here bet ween t he freezing and boiling point s of wat er? ____ 4. Considering your answers t o CTQs 3 and 4, how is t he size of one Celsius degree relat ed t o t he size of one Kelvin?

5. What is t he boiling point of wat er in Fahrenheit ? ______°F 6. How m any Fahrenheit degrees are t here bet ween t he freezing and boiling point s of wat er?____ 7. Com pare your answers t o CTQs 2 and 6. a. Which has a larger size, a Celsius degree or a Fahrenheit degree?

b. What is t he r a t io of t he larger t o t he sm aller degree? I nclude t he unit s F° or C° where appropriat e.

CA06

- 26 -

8. The freezing point of wat er is t he reference ( zero) point for t he Celsius scale. What num ber of Fa h r e nh e it degrees m ust be subt ract ed from t he t em perat ure at t he freezing point of wat er so t hat it will also be t he ze r o poin t ? 9. So, t o convert from °F t o °C, t wo adj ust m ents are needed. One has t o subt ract 32 t o get t o t he sam e reference point ( 0° C) , and t hen use a conversion fact or ( see CTQ 7) . Using your answers t o CTQ 7 and 8, writ e an equat ion for t his. Do not look it up! °C = 10. To convert °C t o °F, one has t o use t he conversi on fact or t o convert t o °F first , and t hen add 32 °F unit s ( “ adding apples t o apples” ) . Writ e an equat ion for t his. °F = 11. Suppose t he room t em perat ure is 72°F. Repo rt t his t em perat ure in degrees Celsius and Kelvins.

Exercises: 1. The sp gr of m ercury is 13.6. What is t he volum e of m ercury in a barom et er cont aining 2040 g of m ercury?

2. Your long lost great aunt from France left you a fancy French oven in her will. Unfort unat ely, it is calibrat ed in Celsius, and you need t o bake a cake at 350°F. At what t em perat ure do you set t he oven?

3. Som e t ext books give t he rat io 9/ 5 as t he conversion fact or bet ween °F and °C. Explain how t his is t he sam e as t he one in t he equat ion you developed in CTQ 7.

4. Read t he assigned pages in your t ext , and work t he assigned problem s. - 27 -

CA06

Ch e m Act ivit y 7

Electron Configuration and The Periodic Table* ( How are at om s and elem ent s classified?)

Model 1: Diagrams of atoms the first three elements using the shell model shell (n) = 1 shell (n) = 2 nuclear charge = +1

nuclear charge = +2

nuclear charge = +3

valence electron

hydrogen

helium

lithium

The outermost electron shell of an atom is its valence shell. Everything else is the core.

Critical Thinking Questions 1. How m any elect rons are in t he valence shell of H? _____ of He? _____ of Li? _____ 2. How m any inner shell ( core) elect rons does H have? _____ He? _____ Li? _____ 3. Considering everyt hing in t he core of t he lit hium at om , what is t he net ( t ot al) charge in t he core? 4. Explain how you arrived at your answer t o CTQ 3.

Model 2: Diagrams of a lithium atom using the shell model (a) and the core charge concept (b) (a) (b) valence electron

valence electron core electron nuclear charge = +3

core charge = +1

The valence electron "sees" a core charge of +1 * Adapt ed from Chem Act ivit ies 4 and 5, Moog, R.S.; Farrell, J.J. Chem ist ry: A Guided I nquiry, 3 rd ed., Wiley, 2006, pp. 20- 35.

CA07

- 28 -

Critical Thinking Questions: Model 2 shows t wo ways of represent ing t he t hird elem ent , lit hium . CTQs 5- 8 will involve com plet ing a sim ilar diagram ( below) for t he fourt h elem ent , beryllium ( Be) .

(a)

(b)

nuclear charge = +4

core charge = ____

5. Why is t he nuclear charge of Be “ + 4” ?

6. How m any inner shell ( core) elect rons does Be have? _____ 7. How m any valence elect rons would Be have? _____ Add t hem ont o t he shell m odel ( a) in t he diagram above. 8. Com plet e t he diagram of a beryllium ( Be) at om using t he core charge concept ( b) .

Model 3: Diagrams of a sodium (Na) atom using the shell model (a) and the core charge concept (b) (a) (b)

nuclear charge = +11

core charge = +1

9. Label t he shells n= 1, n= 2, and n= 3 in Model 3( a) . 10. How m any elect rons are in a full shell for n= 1? _____ 11. How m any elect rons are in a full shell for n= 2? _____ 12. How m any valence elect rons does Na have? _____ 13. I n one com plet e sent ence, writ e a definit ion of t he t erm valence elect ron.

- 29 -

CA07

14. What do H, Li and Na ( and all group 1 elem ent s) have in com m on?

Model 4: Orbitals and electron capacity in shells 1-4 Elect ron shells are not really t wo- dim ensional circles like t hose shown in t he earlier m odels. Elect rons in shells are act ually found in “ cloud- like” regions of space called subshells, or or bit a ls. Each shell can cont ain one or m ore orbit als. Shell ( n)

Orbit als Available

Tot al Elect ron Capacit y

1 2 3 4

one s one s, t hree p one s, t hree p, five d one s, t hree p, five d, seven f

2 8 18 32

Model 5: Shapes of the first two types of orbitals, s and p z

z

y

x

z

y

x

z

y

x

x

y

three p orbitals

s orbital

Critical Thinking Questions: 15. Explain how Model 4 shows t hat t he t hree p orbit als in shell n = 2 can hold 6 elect rons.

16. How m any elect rons can t he five d orbit als in shell n= 3 hold? ____ 17. How m any elect rons can t he f orbit als in shell n= 4 hold? _____ 18. What is t he m axim um num ber of elect rons t hat can fit int o any part icular orbit al? ____

Model 6: Electron configuration of sodium We can writ e t he e le ct r on con figur a t ion of Na like t his: 1s2 2s2 2p 6 3s1 So t he elect rons are really in clouds shaped like t he pict ures in Model 4. This is good t o know, but m ost of t he t im e, w e w ill st ick t o t h e sim ple r sh e ll m ode l, since it is e a sie r t o de a l w it h .

Critical Thinking Question: 19. Model 3 shows t hat t he n= 2 shell of t he sodium at om cont ains 8 elect rons. What inform at ion is given in Model 6 t hat is not present in Model 3? Give your answer in a com plet e sent ence or t wo.

CA07

- 30 -

Model 7: Electron configuration of lead (optional) Aft er argon ( at om ic num ber 20) , elect rons do not fill neat ly in order from shell 1 o2 o3 et c. Consider, for exam ple, t he order t hat orbit als are filled t o reach t he elect ron configurat ion of t he elem ent lead ( Pb) , wit h 82 elect rons: 1s2 2s2 2p 6 3s2 3p 6 4s2 3d 10 4p 6 5s2 4d 10 5p 6 6s2 4f 14 5d 10 6p 2 20. What is t he largest shell ( n) in lead t hat cont ains elect rons? _____ 21. How m any valence elect rons does lead have ( see above) ? _____ ( Careful! ) 22. Look at t he periodic t able at t ached t o t his act ivit y ( m ost ly blank, but wit h Pb shown) . What is t he relat ionship bet ween t he elect ron shell configurat ion of t he elem ent lead ( Pb) and t he row num ber on t he left side of t he t able?

23. Count t he widt h ( num ber of colum ns) of each of t he four “ blocks.” What is t he relat ionship bet ween t he widt h and t he m axim um num ber of elect rons per subshell? ( Consider your answers t o CTQs 13–16.)

Exercises: 1. Why are t here only t wo elem ent s in t he first row of t he periodic t able?

2. Why does t he second row not have a “ d- block” sect ion?

3. Find t he elem ent “ A” on t he blank periodic t able. a) Based on it s posit ion in t he s- block, what would be t he la st ent ry in it s elect ron configurat ion? _____ ( Do t his wit hout count ing elect rons.) b) How m any valence elect rons does elem ent “ A” have? ____ 4. Find t he elem ent “ B” on t he blank periodic t able. a) Based on it s posit ion in t he p- block, what would be t he la st ent ry in it s elect ron configurat ion? _____ Do t his wit hout count ing elect rons. b) How m any valence elect rons does elem ent “ B” have?____ ( Caut ion: The procedure followed in Exercises 3 and 4 becom es m ore com plicat ed for elem ent s in t he d and f- blocks. Read your t ext if you are int erest ed in t he relat ionship bet ween t he periodic t able and t he elect ron configurat ion.) 5. A st udent m akes t he following st at em ent : “ The elect ron configurat ion of all halogens ends in np 5 .” I s t he st udent correct ? Explain.

6. Read t he assigned pages in your t ext , and work t he assigned problem s. - 31 -

CA07

7

6

5

4

3

2

1

n

A

s-block

‡

*

‡

*

f-block, shell (n-2)

d-block, shell (n-1)

Periodic Table of the Elements

Pb

p-block

B

Ch e m Act ivit y 8

Nuclear Chemistry ( What is radiat ion?)

Model: Nuclide symbols for three isotopes of carbon 12 6C

13 6C

14 6C

carbon- 12

carbon- 13

carbon- 14

Critical Thinking Questions: 1. How m any prot ons are in carbon- 12? ____ How m any neut rons are in carbon- 12? ____ 2. How m any prot ons are in carbon- 13? ____ How m any neut rons are in carbon- 13? ____ 3. How m any prot ons are in carbon- 14? ____ How m any neut rons are in carbon- 14? ____ 4. Make a list of what is t he sam e and what is different am ong isot opes.

5. What does t he subscript ed 6 represent in all t hree nuclide sym bols in t he Model?

Information: Nuclear reactions and ionizing radiation A n u cle a r r e a ct ion is a change in t he com posit ion of t he nucleus of an at om . This is not norm ally considered a chem ical react ion, and does not depend on what m olecule t he at om m ight be in. There are t hree t ypes of nuclear react ions: fusion, fission, and radioact ivit y. Fusion ( com bining of nuclei int o larger nuclei, such as in st ars and t he sun) and fission ( “ split t ing t he at om ,” such as in a nuclear react or) do not concern us m uch in chem ist ry. Som e isot opes are radioact ive, m eaning t hat t heir nuclei break down ( “ decay” ) and give off part icles, “ rays,” or bot h. There is no sim ple way t o predict which isot opes are radioact ive. Ta ble 1 : Som e t ype s of ion izing r a dia t ion pr odu ce d in n u cle a r r e a ct ion s Mass Num ber

Charge

Relat ive penet rat ing abilit y

Shielding required

Biological hazard

4

2+

very low

clot hing

none unless inhaled

e

0

1–

low

heavy clot h, plast ic

m ainly t o eyes, skin

Ȗ

0

0

very high

lead or concret e

whole body

1

0

very high

wat er, lead

whole body

0

1+

low

heavy clot h, plast ic

m ainly t o eyes, skin

Type of Radiat ion

Sym bol

Alpha part icle

D 2 He

Bet a part icle

E,

0 1

Gam m a ray

J,

0 0

Neut ron Posit ron

4

1 0n

E+ ,

0 1

e

- 33 -

CA08

Critical Thinking Questions: 6. What does t he subscript indicat e in t he sym bols in Table 1?

7. Explain how your answer t o CTQ 6 is consist ent wit h your answer t o CTQ 5.

8. Consider t he following nuclear react ion:

238 92 U

o 234 90 Th +

4 2 He

a. What t ype of radioact ivit y is produced? b. How does t he num ber of prot ons in t he react ant com pare wit h t he t ot al num ber of prot ons in t he product s?

c. How does t he num ber of neut rons in t he react ant com pare wit h t he t ot al num ber of neut rons in t he product s?

d. How does t he m ass num ber of t he react ant com pare wit h t he t ot al of t he m ass num bers of t he product s?

e. Show how each side of t he react ion equat ion would change if a gam m a ray were also released in t he process.

9. Balance t he m ass num bers and “ at om ic num bers” t o com plet e t he equat ion. a.

131 53 I

0

o 1 e

+

b. What t ype of radioact ivit y is given off in t his react ion?

CA08

- 34 -

Table 2: Half-lives of some radioisotopes Radioisot ope

Radiat ion t ype

Half- life

Use

barium - 131

J

11.6 days

det ect ion of bone t um ors

carbon- 14

E

5730 yr

carbon dat ing

J, X- rays

27.8 days

m easuring blood volum e

E, J

5.3 yr

food irradiat ion, cancer t herapy

E

8.1 days

hypert hyroid t reat m ent

D,ѽE,J

4.47 × 10 9 yr

dat ing igneous rocks

chrom ium - 51 cobalt - 60 iodine- 131 uranium - 238

The t im e required for half of a sam ple of a radioact ive isot opes t o decay is called t he halflife ( t ½ ) .

Critical Thinking Questions: 10. Consider a 100- gram sam ple of radioact ive cobalt - 60. a. How m uch t im e will it t ake before half t he sam ple has decayed?

b. Approxim at ely how m any gram s of radioact ive cobalt - 60 will rem ain aft er 11 years?

11. Consider a sam ple of iodine- 131. a. How m any half- lives would it t ake for t he sam ple t o decay unt il less t han 1% of t he original isot ope rem ained?

b. How m any days would t his be?

12. Considering only t he half lives of uranium - 238 and iodine- 131, which would be m ore appropriat e for int ernal usage ( ingest ion) for m edical t est s? Explain.

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CA08

Exercises: 1. Aft er an organism dies, it st ops t aking in radioact ive carbon- 14 from t he environm ent . I f 14 6

t he carbon- 14: carbon- 12 rat io (

C / 126 C ) in a piece of pet rified wood is one sixt eent h of

t he rat io in living m at t er, how old is t he rock? ( Hint : How m any half lives have elapsed?)

2. Would chrom ium - 51 be useful for dat ing rocks cont aining chrom ium ? Why or why not ?

3. Suppose t hat 0.50 gram s of barium - 131 are adm inist ered orally t o a pat ient . Approxim at ely how m any m illigram s of t he barium would st ill be radioact ive t wo m ont hs lat er?

4. Com plet e t he equat ions.  o 10 e 

a.

30 15 P

b.

beta decay 113 47 Ag   o

c.

( What t ype of radiat ion is t his?

emission 222 Dand J o 86 Rn 

 00J

5. Read t he assigned pages in your t ext book and work t he assigned problem s.

CA08

- 36 -

)

Ch e m Act ivit y 9

Ions and Ionic Compounds ( What are ionic com pounds?)

Model 1: Common charges (by group) on elements when in ionic compounds. Grp # 2 1 Chg: +1 +2 1

3

4

5

6

7

8

9

10

11

12

13

varies

14

+3

15

16

17

-3

-2

-1 2

H 3

He 4

Li Be 11

18

12

Na Mg 25

8

9

10

B

C

N

O

F

Ne 18

13

14

15

16

17

Al

Si

P

S

Cl Ar

31

32

33

34

35

23

24

V

Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

41

42

27

30

40

Rb Sr

Y

Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te

56

57

72

55

73

74

75

44

28

39

37

43

26

29

7

22

38

21

6

K Ca Sc Ti

19

20

5

45

76

77

Cs Ba La Hf Ta W Re Os Ir 87

88

89

104

105

106

107

108

109

46 78

47 79

48 80

49 81

50 82

51 83

52 84

36

53

54

I

Xe

85

86

Pt Au Hg Tl Pb Bi Po At Rn 110

111

112

114

116

Fr Ra Ac Rf Db Sg Bh Hs Mt Ds A cat ion has a posit ive charge.

An anion has a negat ive charge.

Critical Thinking Questions: 1. I n an ionic com pound, what is t he charge ( sign and m agnit ude) on an ion from a. group 1? _____ b. group 2?_____ c. group 13?_____ 2. Do m et als t ypically form a nions or ca t ion s ( circle one) ? 3. I n an ionic com pound, what is t he charge on an ion from a. group 15? _____ b. group 16?_____ c. group 17?_____ 4. Do nonm et als t ypically form a n ion s or ca t ion s ( circle one) ?

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CA09

Model 2: Formulas and names of some binary ionic compounds Form ula

Group Num ber of Met al

Nam e

1 1 2 8 8 11 13 14

sodium brom ide pot assium oxide m agnesium fluoride iron( I I ) sulfide iron( I I I ) sulfide copper( I ) sulfide indium phosphide lead( I I ) iodide

NaBr K2 O MgF2 FeS Fe 2 S3 Cu 2 S I nP PbI 2

Critical Thinking Questions: 5. I n t he nam e of an ionic com pound, which ion is always writ t en first — t h e a n ion or t h e ca t ion ( circle one) . 6. Circle t he na m e s of com pounds in Model 2 for which t he m et al ion is n ot in groups 1, 2, or 13. How does t he nam e of t he cat ion differ from t he nam e of t he m et al?

7. For m et al ions from groups 1, 2, or 13 in ionic com pounds, how does t he nam e of t he cat ion differ from t he nam e of t he m et al?

8. For nonm et als, how does t he nam e of t he anion differ from t he nam e of t he elem ent ?

9. Consider t he form ula for m agnesium fluoride, MgF2 . a. What is t he charge on t he m agnesium ion? ____ b. What is t he charge on t he fluoride ion? ____ c. What is t he overall ( t ot al) charge on t he com pound? ____ 10. What is t he overall charge on t he com pound sodium brom ide ( NaBr) ? ____ 11. I n general, what is t he overall charge on an ionic com pound? ____ 12. What is t he charge on iron in FeS?_____

I n Fe2 S3 ? _____

13. Based on your answer t o CTQ 12, what do t he Rom an num erals in Model 2 represent ?

14. Com plet e t he rules for nam ing ionic com pounds: x

Nam ing m et al ions: nam e t he m et al [ exam ple: Ca2+ =

x

I f t he m et al is n ot in group 1, 2, or 13, add a Rom an num eral in parent heses t hat

___

]

[ e. g., Fe 3+ = ____________]

represent s x

Nonm et als: change ending of elem ent nam e t o _______ [ e. g., N3- = ___________]

x

Nam ing ionic com pounds: nam e t he cat ion, t hen t he anion [ exam ple: FeN = __________________________

CA09

] - 38 -

Table 1: Formulas and names of some common polyatomic ions to memorize Form ula

Nam e

Form ula

Nam e

NH4 +

am m onium

MnO4 –

perm anganat e

acet at e

CO3 2–

carbonat e

C2 H3 O2 CN

–

–

OH

cyanide

–

OCl

hydroxide

–

hypochlorit e

–

nit rat e

NO3

Cr 2 O7

dichrom at e

2–

sulfat e

3–

phosphat e

SO4 PO4

2–

Table 2: Rules for naming other polyatomic ions Rule

Exam ples +

1. I f adding a H t o a polyat om ic ion result s in a new ion, add t he word “ hydrogen” in front of t he nam e; if adding 2 H+ , add t he word “ dihydrogen.”

HPO4 2– = hydrogen phosphat e H2 PO4 – = dihydrogen phosphat e

2. For ions wit h one fewer oxygen at om s t han t he com m on ion, change t he ending from “ - at e” t o “ - it e”

SO4 2– = sulfa t e , so SO3 2– = sulfit e

Exercises: 1. Com plet e t he following t able. Form ula

Nam e

CoCl 4 ________

pot assium nit rat e

Ba( OH) 2 ________

sodium hydrogen carbonat e

________

beryllium brom ide

Li 2 CO3 ________

copper( I I ) oxide

________

sodium hypochlorit e

Ca( C2 H3 O2 ) 2 ________

m agnesium sulfat e

NaNO2 ________

vanadium ( I I I ) sulfat e

2. Based on t heir elect ron shell configurat ions, give a rat ionalizat ion for why all alkaline eart h m et als in ionic com pounds have a + 2 charge.

3. Based on t heir elect ron shell configurat ions, give a rat ionalizat ion for why all halogens in ionic com pounds have a –1 charge.

4. Read t he assigned pages in t he t ext , and work t he assigned problem s. - 39 -

CA09

Ch e m Act ivit y 1 0

Covalent and Ionic Bonds ( Why do at om s bond t oget her?)

Model 1: Two types of chemical bonding* I ons held t oget her by opposit e charges are said t o be ionically bonded. I onic com pounds cont ain ions—t ypically a m et al ion along wit h nonm et als. At om s sharing valence elect rons are said t o be covalent ly bonded. Covalent m olecules t ypically cont ain only nonm et als.

Model 2: Lewis electron-dot structures for hydrogen and the second row elements

H

Li

Be

B

C

N

O

F

Ne

Critical Thinking Questions: 1. Which t wo elem ent s in Model 2 are m et als? 2. Are t hese t wo elem ent s likely t o be in a covalent m olecule? Explain.

3. Consider Model 2. How is t he num ber of dot s relat ed t o t he num ber of valence elect rons?

4. By ext ension, writ e t he elect ron- dot ( Lewis) st ruct ures for sulfur, chlorine, and sodium .

5. The ions form ed in com pounds from group 1 at om s ( t he alkali m et als) are alm ost always M+ ions ( t hat is, t hey have a + 1 charge) . Considering Model 2, explain t his result .

6. The ions form ed in com pounds from group 2 at om s ( t he alkaline eart h m et als) are alm ost always M2+ ions. Explain t his result .

*

A t hird t ype of bonding, m et allic bonding, will not be considered in t his book.

CA10

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7. The ions form ed in com pounds from group 17 at om s ( t he halogens) are alm ost always X– ions. Explain t his result .

8. Writ e elect ron dot ( Lewis) st ruct ures for each of t he ions F– , Cl – , and Br – .

Model 3: Typical number of covalent bonds for elements common in biology Elem ent

Num ber of Bonds

H

1

O

2

N

3

C

4

Critical Thinking Questions: 9. How m any e le ct r on s in t he dot st ruct ure for O in Model 2 are pa ir e d? ____ 10. How m any unpa ir e d elect rons are in t he dot st ruct ure for O? ____ 11. How m any elect rons in t he dot st ruct ure for N are paired? ____ 12. How m any unpaired elect rons are in t he dot st ruct ure for N? ____ 13. How m any elect rons in t he dot st ruct ure for C are paired? ____ 14. How m any unpaired elect rons are in t he dot st ruct ure for C? ____ 15. How is t he num ber of covalent bonds t hat an at om m akes relat ed t o it s elect ron- dot st ruct ure? 16. Which nonm et al in Model 2 is unlikely t o be in a covalent m olecule? Explain.

Model 4: Covalent bonding (sharing valence electrons) between H and F

F

+

H

F H

lone pair

or

bonding pair (covalent bond)

F

H

usually use this notation

spacefilling model

Model 5: Covalent bonding between H and O

O

+ 2 H

O H H - 41 -

or

O

H

H CA10

Critical Thinking Questions: 17. Given t he shell m odel of t he at om , why do you t hink t hat Lewis proposed a m axim um of t wo elect rons for hydrogen and eight for carbon, nit rogen, oxygen, and fluorine at om s?

18. Answer t he following for t he phosphorus at om : a. How m any valence elect rons does P have? _____ b. What is t he Lewis represent at ion for P?

c. How m any addit ional elect rons does P need when it form s a m olecule? ____ d. What is t he likely form ula for a m olecule com posed of hydrogen at om s and one phosphorus at om ? _____ Draw t he Lewis st ruct ure.

19. Answer t he following for t he sulfur at om : a. How m any valence elect rons does S have? _____ b. What is t he Lewis represent at ion for S?

c. How m any addit ional elect rons does S need when it form s a m olecule? _____ d. What is t he likely form ula for a m olecule com posed of hydrogen at om s and one sulfur at om ? _____ Would t his com pound be ion ic or m ole cula r ( circle one) ? Explain.

e. Draw t he Lewis st ruct ure.

20. a. What is t he likely form ula for a m olecule com posed of sodium at om s and one sulfur at om ?

b. Would t his com pound be ion ic or m ole cula r ( circle one) ? Explain.

c. Why would it be inappropriat e t o draw a Lewis st ruct ure ( such as in Model 5) for t his com pound?

CA10

- 42 -

21. Covalent com pounds form individual unit s called m ole cu le s. Com pare t he m odel of t he HF m olecule in Model 4 wit h t he m odel of a solid sodium chloride ( NaCl) cryst al at t he right [ dot t ed lines shown t o enhance t he t hree- dim ensionalit y of t he cryst al] . Then, explain why you t hink t he word “ m olecule” is not used for ionic com pounds.

Exercises: 1. Draw reasonable Lewis ( elect ron- dot ) st ruct ures for t he following m olecules: a. NH3 ( am m onia)

b. NCl 3 ( nit rogen t richloride)

c. CH4 ( m et hane)

d. CHCl 3 ( chloroform )

2. I n NaCl, what are t he charges on t he ions? Explain your reasoning in t erm s of t he Lewis ( elect ron- dot ) represent at ions of Na and Cl.

3. I n CTQ 21, t he t wo different spheres are different sizes. Which spheres represent t he sodium ions and which represent t he chloride ions? Hint : Consider t he elect ron shell configurat ions of t he t wo ions.

4. Read t he assigned pages in t he t ext , and work t he assigned problem s. - 43 -

CA10

Ch e m Act ivit y 1 1

Electrolytes, Acids, and Bases ( Which com pounds produce ions when dissolved in wat er?)

Suggested Demonstration: Electrolytes Model 1: Electrolytes Only separat e, charged part icles ( such as ions) can carry elect rical current s. Elect rolyt es can carry an elect rical current when dissolved in wat er.

Critical Thinking Question: 1. What happens t o elect rolyt es when t hey dissolve in wat er?

Model 2: Types of electrolytes I n wat er solut ion, st rong elect rolyt es dissociat e com plet ely int o ions, weak elect rolyt es dissociat e only slight ly, and nonelect rolyt es dissociat e undet ect ably or not at all.

Critical Thinking Question: 2. Describe a m et hod by which you could t ell if a part icular solut ion cont ains a st rong elect rolyt e, weak elect rolyt e, or nonelect rolyt e.

Model 3: Some common acids and bases Type of elect rolyt e St r on g ( any acid or base not list ed here is weak) W eak

Acids

Bases

HCl

LiOH

HBr

NaOH

HI

KOH

H2 SO4

Ca( OH) 2

HNO3

Sr( OH) 2

HClO4

Ba( OH) 2

HC2 H3 O2

Mg( OH) 2

HCN Acids dissociat e in wat er t o give hydrogen ( H+ ) ions and an anion. Bases dissociat e in wat er t o give hydroxide ( OH– ) ions and a cat ion. St rong acids and bases dissociat e com plet ely. Not e t hat t he acids are m olecules, while t he bases are ionic com pounds.

CA11

- 44 -

Critical Thinking Questions: 3. Consider Model 3. What do all t he m olecular form ulas of t he acids have in com m on?

4. How can you recognize an acid from t he m olecular form ula?

5. How can you recognize a base from it s form ula?

6. What happens t o st rong acids when dissolved in wat er?

7. What happens t o weak acids when dissolved in wat er?

8. Are all acids st rong elect rolyt es? Explain.

9. Describe what happens t o t he ions in solid sodium hydroxide ( NaOH) during t he process of dissolving in wat er.

Model 4: Solubilities of some ionic compounds I onic com pound

Solubilit y in wat er

Type of elect rolyt e

Maj or species present when dissolved in wat er

MgCl 2

soluble

st rong

Mg 2+ ( aq) , Cl – ( aq)

MgO

insoluble

nonelect rolyt e

MgO( s) +

K ( aq) , S2– ( aq)

K2 S

soluble

st rong

CuS

insoluble

nonelect rolyt e

CuS( s)

Ca( NO3 ) 2

soluble

st rong

Ca ( aq) , NO–3 ( aq)

Ca 3 ( PO4 ) 2

insoluble

nonelect rolyt e

Ca 3 ( PO4 ) 2 ( s)

2+

Molecular com pounds ot her t han acids and bases, and insoluble ionic com pounds do not dissociat e in wat er and so are nonelect rolyt es. The phase label ( aq) , m eaning “ aqueous,” can be used t o show t hat a species is dissolved in wat er. W e w ill not con side r h ow t o pr e dict if a n ionic com poun d is solu ble in w a t e r u n t il la t e r in t h e cou r se . - 45 -

CA11

Critical Thinking Questions: 10. A solut ion of MgCl 2 in wat er could be writ t en as MgCl 2 ( aq) . Besides wat er, what species would act ually be present in t he solut ion? 11. MgO is an ionic com pound t hat does not dissolve in wat er, it would j ust collect as a solid—i. e., MgO( s) —at t he bot t om of t he cont ainer. Would t here be any ions dissolved in t he wat er? Explain.

12. Describe a m et hod by which you could t ell if an ionic com pound is soluble in wat er.

13. Consider Model 4. How is t he solubilit y of an ionic com pound relat ed t o it s classificat ion as an elect rolyt e?

14. The following com pounds from Model 4 are placed in wat er. Add t he phase labels ( s) or ( aq) t o represent whet her t he com pounds are soluble or not : K2 S

CuS

Ca( NO3 ) 2

Ca 3 ( PO4 ) 2

15. Suppose t hat each com pound in t he t able below is placed in wat er. The phase labels given describe whet her t he com pound dissolves or not . Fill in t he t able wit h t he propert ies for each com pound. Com pound

Acid, ot her m ole cu le , ba se , or ot her ionic com pound

HBr( aq) KOH( aq) CH2 O( aq) Ca 3 ( PO4 ) 2 ( s) Al( OH) 3 ( s) HOCl( aq) H2 S( aq) Fe 2 O3 ( s) Na 2 S( aq)

CA11

- 46 -

St r on g, w e a k , or nonelect rolyt e

Exercises: 1. Writ e form ulas for t he m aj or species present in t he solut ions from CTQ 15. Com pound

Maj or species present when dissolved in wat er ( or “ need m ore inform at ion” )

HBr( aq) KOH( aq) CH2 O( aq) Ca 3 ( PO4 ) 2 ( s) Al( OH) 3 ( s) HOCl( aq) H2 S( aq) Fe 2 O3 ( s) Na 2 S( aq) 2. Read t he assigned pages in your t ext , and work t he assigned problem s.

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CA11

Ch e m Act ivit y 1 2

Naming Binary Molecules, Acids, and Bases ( How are chem ical form ulas and nam es relat ed?)

Information: Naming Binary Molecules Recall t hat m olecules ( as opposed t o ionic com pounds) are form ed from nonm et als only. There is a sim ple m et hod t o nam e binary m olecules ( m olecules cont aining only t wo elem ent s) . 1. Nam e t he first elem ent , using a prefix t o indicat e t he num ber of at om s in t he form ula. 2. Nam e t he second elem ent , using a prefix t o indicat e t he num ber of at om s in t he form ula, and changing t he ending t o “ - ide.” 3. Rem ove t he vowel at t he end of t he prefix if it seem s awkward. 4. Do not use t he prefix “ m ono- ” on t he first elem ent .

Table 1: Prefixes to indicate the number of atoms of an element in a binary molecule Num ber

Prefix

Num ber

Prefix

1

m ono-

6

hexa-

2

di-

7

hept a-

3

t ri-

8

oct a-

4

t et ra-

9

nona-

5

pent a-

10

deca-

Exam ples: NI 3 = nit rogen t riiodide; N2 O = dinit rogen m onoxide ( not “ m onooxide” ) ; N2 O5 = dinit rogen pent oxide ( not “ penta oxide” )

Critical Thinking Questions: 1. The following is from “ Ban Dihydrogen Monoxide,” Coalit ion t o Ban DHMO, 1988. Quot ed in ht t p: / / www.dhm o.org/ t rut h/ Dihydrogen- Monoxide.ht m l [ accessed 03 Feb 2006] : Dihydrogen m onoxide is colorless, odorless, t ast eless, and kills uncount ed t housands of people every year. Most of t hese deat hs are caused by accident al inhalat ion of DHMO, but t he dangers of dihydrogen m onoxide do not end t here. Prolonged exposure t o it s solid form causes severe t issue dam age. Sym pt om s of DHMO ingest ion can include excessive sweat ing and urinat ion, and possibly a bloat ed feeling, nausea, vom it ing and body elect rolyt e im balance. For t hose who have becom e dependent , DHMO wit hdrawal m eans cert ain deat h. Explain how t hese claim s are t rue or false.

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2. Explain what is wrong wit h t he following m olecule nam es. Then give t he correct nam e. a. NO

nit rogen oxide

b. CO

m onocarbon m onoxide

c. CS2

carbon disulfat e

Information: Naming the Binary Acids When t he binary acids hydrogen chloride ( HCl) , hydrogen brom ide ( HBr) , and hydrogen iodide ( HI ) are dissolved in wat er ( aq) , t hey are called “ hydrochloric acid,” “ hydrobrom ic acid,” and “ hydroiodic acid” respect ively.

Critical Thinking Questions: 3. Based on t he nam es of HCl, HBr, and HI , what would HF( aq) be called?

4. I s HF a st r on g or w e a k acid ( circle one) ? Explain.

Information: Naming Oxoacids An ox oa cid is an acid t hat cont ains hydrogen, oxygen, and one ot her elem ent . Oxoacids m ay be form ed by adding H+ t o polyat om ic ions. One H+ is added for each negat ive charge on t he ion ( exam ple: t he charge on sulfat e is –2, so add t wo H+ ions t o m ake H2 SO4 ) . A. if t he ion has an ending of “ - at e,” t he acid has an ending of “ - ic” ( exam ple: SO4 2- = “ sulfa t e ,” so H2 SO4 = “ sulfuric acid” ) . x

m em ory device: “ I ck,” I “ a t e ” t he acid!

B. if t he ion has an ending of “ - it e,” t he acid has an ending of “ - ous” ( exam ple: SO3 2- = “ sulfit e ,” so H2 SO3 = “ sulfurous acid” ) .

Critical Thinking Questions: 5. Nam e t he following oxoacids. ( You m ay wish t o consult t he polyat om ic ion nam es in Chem Act ivit y 9, Table 1.) a. HOCl b. HNO3 c. HNO2 d. HC2 H3 O2 - 49 -

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Information: Summary flowchart for naming binary ionic and molecular compounds St a r t ionic

yes

m olecular

I onic or Molecular?

Met al in Group 1, 2, or 13?

I s it an acid?

no

no

yes

N o Rom an num eral, N o prefixes

Add Rom an num eral, N o prefixes

Use acid nam ing rules

Exercises: 1. Fill in t he t able wit h t he nam e of each com pound. Com pound

Nam e

MnS CS2 Na 2 S MgSO4 H2 SO3 H3 PO4 CuOH P2 O5 2. Fill in t he t able wit h t he form ula of each com pound. Com pound

Nam e acet ic acid dinit rogen m onoxide copper( I ) oxide carbonic acid m agnesium brom ide oxygen difluoride sodium nit rit e t in( I I ) hydroxide

3. Read t he assigned pages in your t ext , and work t he assigned problem s. CA12

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Nam e wit h prefixes

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CA12

Ch e m Act ivit y 1 3

Molecular Shapes* ( What shapes do m olecules have?)

Model 1: Bond angle and electron fomains. A bond a ngle is t he angle m ade by t hree connect ed nuclei in a m olecule. By convent ion, t he bond angle is considered t o be bet ween 0q and 180q. Ta ble 1 : Bon d a ngle s a n d e le ct r on dom a ins in se le ct e d m ole cu le s. †

Molecular Formula

Lewis Structure

CO2

O

H

HCCH

C

H

C

No. of Nonbonding Domains (central atom)

‘OCO = 180q

2

0

C

‘HCC = 180q

2

0

‘CCC = 180q

2

0

Cl

‘ClNN = 117.4q

2

1

–

‘ONO = 120q

3

0

3

0

H H

C

H

C

C N

Cl

No. of Bonding Domains (central atom)

O

H

H2CCCH2

Bond Angle (CAChe)

N

ClNNCl + N

–

NO–3

O

O

O

H H

H2CCH2 H

C

C

‘HCH = 121.1q

H

(Table 1 continues on the next page)

*

Adapt ed from Chem Act ivit y 18, Moog, R.S.; Farrell, J.J. Chem ist ry: A Guided I nquiry, 3 rd ed., pp. 102–110, Copyright 2006, John Wiley & Sons, I nc. Models, Tables, Figures, and Crit ical Thinking Quest ions reprint ed wit h perm ission of John Wiley & Sons, I nc. † Bond angles calculat ed wit h MOPAC ( Oxford Molecular, CAChe) . MOPAC calculat ions yield bond orders, bond lengt hs, and bond angles t hat are generally in good agreem ent wit h experim ent al evidence.

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Ta ble 1 ( con t inu e d) : Bon d a ngle s a nd e le ct r on dom a in s in se le ct e d m ole cu le s.

Molecular Formula

Lewis Structure

Bond Angle (CAChe)

No. of Bonding Domains (central atom)

No. of Nonbonding Domains (central atom)

‘HCH = 109.45

4

0

‘HCH = 109.45q ‘HCF = 109.45q

4

0

‘HCH = 109.45q ‘HCCl = 109.45q

4

0

4

0

‘HNH = 107q

3

1

‘HNH =106.95q ‘HNF = 106.46q

3

1

‘HOH = 104.5q

2

2

H

CH4

H

C

H

H

F

CH3F H

C

H

H Cl

CH3Cl H

C

H

H

CCl4

‘ClCCl = 109.45q

Cl Cl

C

Cl

Cl H

N

NH3

H H

N

NH2F

F

H O

H2O

H

H

H

Critical Thinking Questions: 1. How is t he num ber of bonding dom ains on a given at om wit hin a m olecule ( such as t hose in Table 1) det erm ined?

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CA13

2. How is t he num ber of nonbonding dom ains on a given at om wit hin a m olecule ( such as t hose in Table 1) det erm ined?

3. The bond angles in Table 1 can be grouped, roughly, around t hree values. What are t hese t hree values?

4. What correlat ion can be m ade bet ween t he values in t he last t wo colum ns in Table 1 and t he groupings ident ified in CTQ 3?

Model 2: Models for methane, ammonia, and water. Use a m olecular m odeling set t o m ake t he following m olecules: CH4 ; NH3 ; H2 O. I n m any m odel kit s: carbon is black; oxygen is red; nit rogen is blue; hydrogen is whit e; use t he short , gray links for single bonds. Nonbonding elect rons are not represent ed in t hese m odels.

Critical Thinking Questions: 5. Sket ch a pict ure of t he following m olecules based on your m odels: CH4 ; NH3 ; H2 O.

6. Describe ( wit h a word or short phrase) t he shape of each of t hese m olecules: CH4 ; NH3 ; H2 O.

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Model 3: Types of electron domains. A dom ain of elect rons ( t wo elect rons in a nonbon din g dom a in , som et im es called a lon e pa ir ; t wo elect rons in a sin gle bon d dom a in; four elect rons in a dou ble bond dom a in ; six elect rons in a t r iple bond dom a in) t ends t o repel ot her dom ains of elect rons. Dom ains of elect rons around a cent ral at om will orient t hem selves t o m inim ize t he elect ron- elect ron repulsion bet ween t he dom ains. Figur e 1 . M in im iza t ion of e le ct r on - e le ct r on r e pu lsion le a ds t o a u n iqu e ge om e t r y for t w o, t hr e e , a n d fou r dom a ins of e le ct r on s.

180°

two domains of electrons

all angles 120° three domains of electrons

all angles 109.45°

four domains of electrons

another representation of four domains of electrons

Critical Thinking Questions: 7. Based on Figure 1, what bond angle is expect ed for a m olecule cont aining: a. t wo dom ains of elect rons? b. t hree dom ains of elect rons? c. four dom ains of elect rons?

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CA13

8. Writ e t he Lewis st ruct ure for CH4 ( you m ay copy it from Table 1) . a. How m any dom ains of elect rons does t he carbon at om have in CH4 ? b. Which elect ron dom ain geom et ry in Figure 1 applies t o CH4 ? c. Are t he calculat ed bond angles in agreem ent ( see Table 1) ?

9. Draw t he Lewis st ruct ure for NH3 . a. How m any elect ron dom ains does t he nit rogen at om have in NH3 ? b. Which elect ron dom ain geom et ry in Figure 1 applies t o NH3 ? c. Are t he calculat ed bond angles in agreem ent ( see Table 1) ?

10. Draw t he Lewis st ruct ure for H2 O. a. How m any elect ron dom ains does t he oxygen at om have in H2 O? b. Which elect ron dom ain geom et ry in Figure 1 applies t o H2 O? c. Are t he calculat ed bond angles in agreem ent ( see Table 1) ?

11. Draw t he Lewis st ruct ure for NO–3 . a. How m any elect ron dom ains does t he nit rogen at om have in NO–3 ? b. Which elect ron dom ain geom et ry in Figure 1 applies t o NO–3 ? c. Are t he calculat ed bond angles in agreem ent ( see Table 1) ?

12. Draw t he Lewis st ruct ure for CO2 . a. How m any elect ron dom ains does t he carbon at om have in CO2 ? b. Which elect ron dom ain geom et ry in Figure 1 applies t o CO2 ? c. Are t he calculat ed bond angles in agreem ent ( see Table 1) ?

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Information: The nam es for m olecular shapes are based on t he posit ion of t he at om s in t he m olecule— not on t he posit ion of t he elect ron dom ains. Figur e 2 . Th e Le w is st r u ct u r e , e le ct r on dom a in s, a nd m ole cu la r sh a pe of H 2 O.

H

O

H

Lewis structure

H

all angles 109.45°

O H

four electron domains

The water moluecule is said to be “bent” because the three atoms are not in a straight line. The actual bond angle, determined by experiment, is 104.5°.

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CA13

Figur e 3 . Five com m on m ole cula r sha pe s.

180° all angles 120°

linear

trigonal planar

two or more atoms in a straight line

all angles 109.45°

bent angles close to 120° or close to 109°

tetrahedral

trigonal pyramidal angles close to 109°

Critical Thinking Questions: 13. Considering t he geom et ries defined in Figure 1, explain why t he bond angle in bent m olecules is expect ed t o be close t o 109q or close t o 120q.

14. Using gram m at ically correct English sent ences, explain how t he shape of a m olecule can be predict ed from it s Lewis st ruct ure.

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Information: A Step-by-Step Method for Writing Lewis Structures 1. Writ e m olecular skelet on ( t he first elem ent in t he form ula, ot her t han H, goes in t he cent er) . 2. Calculat e t he t ot al num ber of valence elect rons for all at om s in t he com pound ( add ext ra for negat ive charges, subt ract for posit ive charges) . 3. Use elect ron pairs t o form a bond bet ween each pair of at om s. 4. Arrange rem aining elect ron pairs on out er at om s t o sat isfy oct et rule ( duet rule for H/ He) . 5. Any elect ron pairs “ left over” ( if any) are placed on cent ral at om . 6. I f cent ral at om does not have oct et rule sat isfied, share elect ron pairs t o m ake m ult iple bonds.

Example: Draw the Lewis Structure of SO2 St ep 1.

O

St ep 2.

valence elect rons: 1 sulfur ( 6) + 2 oxygens ( 12) = 1 8 valence e – t ot al

St ep 3.

O

St ep 4. St ep 5. St ep 6.

S

O

S

O

O

S

O

O

S

O

O

S

O

( 14 e – rem ain from original 18) ( 2 e – rem ain) ( S does not have oct et , so share from eit her O)

Exercises: 1. Draw t he Lewis st ruct ure for NH3 . How m any elect ron dom ains does t he nit rogen at om have in NH3 ? ____ Make a sket ch of t he elect ron dom ains in NH3 . Exam ine t he drawing for t he m olecular shape of H2 O given in Figure 2; m ake a sim ilar drawing for NH3 . Nam e t he shape of t he NH3 m olecule, and give t he approxim at e bond angles.

2. Draw t he Lewis st ruct ure for CH4 . How m any elect ron dom ains does t he carbon at om have in CH4 ? ____ Make a sket ch of t he elect ron dom ains in CH4 . Exam ine t he drawing for t he m olecular shape of H2 O given in Figure 2; m ake a sim ilar drawing for CH4 . Nam e t he shape of t he CH4 m olecule, and give t he approxim at e bond angles.

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CA13

3. Draw t he Lewis st ruct ure for SO2 . How m any elect ron dom ains does t he sulfur at om have in SO2 ? ____ Make a sket ch of t he elect ron dom ains in SO2 . Exam ine t he drawing for t he m olecular shape of H2 O given in Figure 2; not e t hat t he elect ron dom ain geom et ry is different for SO2 , and m ake a sim ilar drawing for SO2 . Nam e t he shape of t he SO2 m olecule, and give t he approxim at e bond angles.

4. Draw t he Lewis st ruct ure, sket ch t he m olecules, predict t he m olecular shape, and give t he bond angles for: PH3 , CO2 , SO3 2- , H3 O+ , NH2 F, H2 CO. The cent ral at om is t he first at om list ed ( ot her t han hydrogen) .

5. Predict t he bond angles around each at om designat ed wit h an arrow in t he am ino acid glycine, shown below.

H

H

H

N

C

C

H

O

O

H

6. Read t he assigned pages in your t ext book and work t he assigned problem s. CA13

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Ch e m Act ivit y 1 4

Polar and Nonpolar Covalent Bonds ( Are at om s in m olecules com plet ely neut ral?)

Model 1: Relative electronegativity values for selected elements Elem ent

Elect ronegat ivit y

Li

1.0

H

2.2

C

2.4

O

3.5

F

4.0

The elect ronegat ivit y of an at om is it s abilit y t o at t ract elect rons in a covalent bond closer t o it self. Fluorine is t he m ost elect ronegat ive elem ent .

Critical Thinking Questions: 1. When carbon and oxygen are covalent ly bonded, are t he elect rons in t he bond at t ract ed closer t o ca r bon or ox yge n ( circle one) ? 2. Which at om is at t he negat ively polarized end of a bond bet ween carbon and oxygen?____ Which at om is at t he posit ively polarized end? ____ Explain.

3. Which end of a bond bet ween carbon and lit hium is negat ively polarized? ____ Explain.

Model 2: Types of bonds based on difference in electronegativity Bond bet ween

Chem ical Form ula

Elect ronegat ivit y difference

Type of bond

H and H

H2

0

nonpolar, covalent

F and F

F2

0

nonpolar, covalent

C and H

CH4

0.2

slight ly polar, covalent

C and O

CO2

1.1

polar, covalent

H and F

HF

1.8

polar, covalent

Li and F

LiF

3.0

ionic

Critical Thinking Questions: 4. Model 2 st at es t hat a H–H bond is n onpola r . Devise a definit ion for t he t erm nonpolar.

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CA14

5. Considering Model 2, which would you expect t o be m ore polar—a C–O bond, or an H–F bond? Explain.

For t he bond you chose, writ e t he word “ very” in front of “ polar, covalent ” in t he “ t ype of bond” colum n in Model 2. 6. Which would be m ore polar—a C- O bond, or a C- Li bond? Explain.

7. The Greek let t er delt a, G, is oft en used t o m ean “ slight ly” or “ part ially.” So, G- would m ean “ part ially negat ive,” and G+ would m ean “ part ially posit ive.” Place t he sym bols Gor G+ near each elem ent below t hat would have a part ial charge. F O

C

O

F H

F F

F

C

F

F

8. Based on t he difference in t heir elect ronegat ivit ies ( Model 2) , a bond bet ween lit hium and fluorine would be ext rem ely polarized. Which end of t he bond would be negat ively polarized— Li or F ( circle one) ? Explain how t his is consist ent wit h t he guideline learned earlier in t he course t hat t he bond bet ween a m et al and a nonm et al is ionic.

9. We norm ally writ e t he form ula for lit hium fluoride as a lit hium ion ( Li+ ) and a fluoride ion ( F– ) . Why is t his m ore correct t hat writ ing t hem as “ Li G+ ” and “ FG– ” ?

10. Explain t he following st at em ent : The ionic charact er of a bond increases as t he elect ronegat ivit y difference bet ween t he t wo bonded at om s increases.

11. Why m ight we not say t hat t he bond bet ween carbon and lit hium is ionic?

12. A rule of t hum b says t hat t he closer an elem ent is t o fluorine on t he periodic t able, t he m ore elect ronegat ive it is. Based on t his rule, place t he sym bols G- and G+ near each elem ent below t hat would have a part ial charge. F I Cl

F

S O

B

O F

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O F

F

Model 3: Carbon dioxide is overall nonpolar

O

C

O

Consider t he m odel of carbon dioxide above. Bot h C–O bonds are polar, wit h t he negat ive ends on each oxygen. ( The arrows indicat e t hat each elect ronegat ive oxygen is pulling elect rons equally t oward it self.) Since bot h bonds are equally polarized in opposit e direct ions, no part icular side or end of t he CO2 m olecule is negat ively or posit ively polarized. I n ot her words, t he individual bond polarit ies “ cancel out .” This t ype of canceling can occur whenever a ll t he elect rons around a cent ral at om are in ide n t ica l covalent bonds ( t o t he sam e elem ent s) and t herefore are equally polarized. As an analogy, consider t he following: Pict ure t he bonds as ropes, wit h t he t wo oxygens pulling on t he ropes in a “ t ug- of- war” gam e. Since t hey are “ pulling” t he elect rons equally in opposit e direct ions, t here is no net m ovem ent of t he elect rons, and t he m olecule is ove r a ll n onpola r .

Critical Thinking Questions: 13. Place t he sym bols G- or G+ near each elem ent in Model 3 t hat would have a part ial charge. 14. Three m olecules in CTQ 7 have polar bonds, but only one m olecule is polar. Explain.

15. One m olecule in CTQ 12 is nonpolar. Circle it . Explain why it is overall nonpolar.

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CA14

Exercises: 1. Com plet e t he t able below. The first one has been done as an exam ple.

Form ula

Tot al Valence Elect rons

Lewis St ruct ure

Shape Around Most Cent ral At om

Approxim at e Bond Angles

I on, Polar Molecule, or Nonpolar Molecule

t et rahedral

109.5°

nonpolar m olecule

F

CF4

32

F

C

F

F

SO–2 4

CH2 O

CHCl 3

CH3 OH ( bond t he last H t o t he O)

HCN

NO–2

2. Go back t o your Lewis st ruct ures in Exercise 1 and indicat e t he bond polarit ies by placing G– or G+ near all t he at om s. Except ion: The polarit y of a C- H bond is so slight t hat it is norm ally ignored. 3. Read t he assigned pages in your t ext book and work t he assigned problem s. CA14

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Ch e m Act ivit y 1 5

The Mole Concept* ( What is a m ole?)

Model 1: Some selected conversion factors and information 1 dozen it em s = 12 it em s 1 score of it em s = 20 it em s 1 m yriad it em s = 10,000 it em s 1 m ole of it em s= 6.022 × 10 23 it em s ( Avogadro’s Num ber) On e e le pha n t ha s one t r u n k a nd fou r le gs. On e m e t ha n e m ole cu le , CH 4 , cont a in s on e ca r bon a t om a n d fou r h ydr oge n a t om s.

Critical Thinking Questions: Use scient ific not at ion if any answer is a very large or very sm all num ber. I nclude unit s wit h all answers. 1. How m any t runks are found in one dozen elephant s?

2. How m any legs are found in one dozen elephant s?

3. How m any t runks are found in one score of elephant s?

4. How m any legs are found in one m yriad elephant s?

5. How m any t runks are found in one m ole of elephant s?

6. How m any carbon at om s are found in one dozen m et hane m olecules?

7. How m any hydrogen at om s are found in one m yriad m et hane m olecules?

* Adapt ed from Chem Act ivit y 28, Moog, R.S. and Farrell, J.J. Chem ist ry: A Guided I nquiry, 3 rd ed., Wiley, 2006, pp. 158- 160.

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CA15

8. How m any elephant s are t here in one m ole of elephant s?

9. How m any t runks are found in one- half m ole of elephant s?

10. How m any legs are found in one m ole of elephant s?

11. How m any carbon at om s are found in one m ole of m et hane m olecules?

12. How m any hydrogen at om s are found in one- half m ole of m et hane m olecules?

13. How m any m et hane m olecules are t here in one m ole of m et hane?

Model 2: The relationship between average atomic mass and moles The m ass of 1 m ole of any pure subst ance is equal t o t he average at om ic m ass expressed in gram s of t hat subst ance. For exam ple: The average at om ic m ass of a helium at om is 4.003 am u ( at om ic m ass unit s) . Therefore: 4.003 g He = 1 m ol He

Critical Thinking Questions: 14. What is t he average m ass ( in am u) of one carbon at om ?

15. What is t he m ass ( in gram s) of one m ole of carbon at om s?

16. What is t he average m ass ( in am u) of one m et hane m olecule?

17. What is t he m ass ( in gram s) of one m ole of m et hane m olecules?

18. Use a gram m at ically correct English sent ence t o describe how t he m ass in am u of one m olecule of a com pound is relat ed t o t he m ass in gram s of one m ole of t hat com pound.

CA15

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Exercises: Unless ot herwise st at ed, calculat e all m ass values in gram s. 1. Consider CTQ 1. A unit plan for t his problem would be dozens o elephant s o t runks. Since you know t here are 12 elephant s in 1 dozen elephant s and 1 t runk per elephant, you could set up t he problem using conversion fact ors as follows: 1 dozen elephant s u

12 elephant s 1 t runk u 1 dozen elephant s 1 elephant

12 t runks

Com plet e sim ilar unit conversions, showing all work, for CTQs 2- 13.

2. I f you m easure out 69.236 g of lead, how m any at om s of lead do you have? Make a unit plan first . Show work.

3. Consider 1.00 m ole of dihydrogen gas, H2 . How m any dihydrogen m olecules are present ? How m any hydrogen at om s are present ? What is t he m ass of t his sam ple?

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CA15

4. Et hanol has a m olecular form ula of CH3 CH2 OH. a. What is t he average m ass of one m olecule of et hanol?

b. What is t he m ass of 1.000 m oles of et hanol?

c. What is t he m ass of 0.5623 m oles of et hanol, CH3 CH2 OH?

d. How m any m oles of et hanol are present in a 100.0 g sam ple of et hanol?

e. How m any m oles of each elem ent ( C, H, O) are present in a 100.0 g sam ple of et hanol?

f.

How m any gram s of each elem ent ( C, H, O) are present in a 100.0 g sam ple of et hanol?

5. How m any m oles of carbon dioxide, CO2 , are present in a sam ple of carbon dioxide wit h a m ass of 254 gram s? How m any m oles of O at om s are present ?

6. I ndicat e whet her each of t he following st at em ent s is t rue or false, and explain your reasoning. a. One m ole of NH3 weighs m ore t han one m ole of H2 O. b. There are m ore carbon at om s in 48 gram s of CO2 t han in 12 gram s of diam ond ( a form of pure carbon) . c. There are equal num bers of nit rogen at om s in one m ole of NH3 and one m ole of N2 . d. The num ber of Cu at om s in 100 gram s of Cu( s) is t he sam e as t he num ber of Cu at om s in 100 gram s of copper( I I ) oxide, CuO. e. The num ber of Ni at om s in 100 m oles of Ni( s) is t he sam e as t he num ber of Ni at om s in 100 m oles of nickel( I I ) chloride, NiCl 2 . f. CA15

There are m ore hydrogen at om s in 2 m oles of NH3 t han in 2 m oles of CH4 . - 68 -

Ch e m Act ivit y 1 6

Balancing Chemical Equations ( What st ays t he sam e and what m ay change in a chem ical react ion?)

Model: Atoms are conserved in chemical reactions At om s are neit her creat ed nor dest royed when chem ical react ions occur. Two balanced chem ical react ions ( or equat ions) are given below. I n a chem ical react ion equat ion, react ant at om s are shown on t he left side of t he arrow and product at om s are shown on t he right . 2 H2 ( g) + O2 ( g) o 2 H2 O ( g)

( 1)

Fe 2 O3 ( s) + 2 Al ( s) o 2 Fe ( l) + Al 2 O3 ( s)

( 2)

Critical Thinking Questions: 1. Considering react ion ( 1) above: a. What are t he react ant s?

b. What are t he product s?

2. What does t he arrow represent in a chem ical react ion?

3. How m any H at om s are represent ed: a. On t he react ant side of react ion ( 1) ? b. On t he product side? 4. Com pare t he num ber of at om s of each elem ent ( H, O) on t he react ant and product sides of equat ion ( 1) . Are t hese num bers t h e sa m e or diffe r e n t ( circle one) ? 5. Com pare t he num ber of at om s of each elem ent ( Fe, Al, O) on t he react ant and product sides of equat ion ( 2) . Are t hese num bers t he sa m e or diffe r e n t ( circle one) ? 6. For each H2 m olecule consum ed in react ion ( 1) , how m any H2 O m olecules are produced?

7. For each O2 m olecule consum ed in react ion ( 1) , how m any H2 O m olecules are produced?

8. For each m ole of oxygen m olecules consum ed in react ion ( 1) , how m any m oles of wat er m olecules are produced?

9. Considering your answers t o CTQs 7 and 8, what t wo t hings can be indicat ed by t he coefficient ( t he num ber) in front of t he chem ical form ulas?

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CA16

10. For each m ole of H2 O m olecules t hat are produced in react ion ( 1) , how m any m oles of O2 m olecules are required?

11. Describe in a com plet e sent ence or t wo how you arrived at your answer t o CTQ 10.

12. One of t he int erpret at ions for t he coefficient s t hat you list ed in CTQ 9 is consist ent wit h your answer t o CTQ 10, and one is not . Explain.

13. Considering react ion ( 1) , what it t he t ot al num ber of m olecules of wat er produced from t wo m olecules of hydrogen and one m olecule of oxygen?____ 14. I s t he t ot al num ber of m ole cu le s ident ical on t he react ant and product sides of t hese balanced equat ions? ____ 15. Explain how your answer t o CTQ 14 can be consist ent wit h t he idea t hat at om s are neit her creat ed nor dest royed when chem ical react ions t ake place.

Exercises: 1. Balance t he equat ions by adding coe fficie nt s in front of each react ant and product . a.

Cr ( s) +

b.

Fe 3 O4 ( l) +

c.

CH4 ( g) +

S8 ( s) o

Cr 2 S3 ( s)

CO ( g) o O2 ( g) o

FeO ( l) + CO2 ( g) +

CO2 ( g) H2 O ( g)

2. For each m ole of S8 m olecules consum ed in t he react ion in Exercise 1, how m any m oles of Cr 2 S3 m olecules are produced?

3. I n t he balanced react ion:

2 CO ( g) + O2 ( g) o 2 CO2 ( g)

a. How m any m oles of CO2 can be produced from 4 m oles CO and 2 m oles O2 ? b. How m any gram s of CO2 would t his be?

CA16

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4. Add m olecular ( spacefilling) pict ures t o t he boxes below in t he correct rat io so t hat it represent s react ion ( 1) in t he Model. For exam ple, m olecule of H2 .

reactants

could be used t o represent one

products

5. Use m olecular ( spacefilling) pict ures in t he correct rat io t o represent t he react ion in Exercise 1a. Be sure t hat t he represent at ions for Cr at om s and S at om s are dist inguishable. ( You m ay wish t o include a key.)

6. Read t he assigned pages in your t ext , and work t he assigned problem s.

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CA16

Ch e m Act ivit y 1 7

Predicting Binary Reactions ( When will precipit at ion or neut ralizat ion react ions occur?)

Table 1: Solubility rules for ionic compounds 1. A com pound cont aining a cat ion from Group 1 ( alkali m et al) or t he am m onium ion ( NH4 + ) is likely soluble. 2. A com pound cont aining an anion from group 17 ( halides) is likely soluble. x Ex ce pt ion s: Ag + , Hg 2 2+ , and Pb 2+ halides are insoluble. 3. A com pound cont aining nit rat e ( NO3 - ) , acet at e ( CH3 CO2 - ) , or sulfat e ( SO4 2- ) is likely soluble. x Ex ce pt ion s: The sulfat es of Ba2+ , Hg 2 2+ , and Pb 2+ are insoluble. 4. An y ot he r ionic com pounds are likely insoluble. Unless your inst ruct or t ells you ot herwise, you are not required t o m em orize t he inform at ion in Table 1. You m ay refer t o t he t able when needed.

Critical Thinking Questions: 1. Writ e t he chem ical form ula for each com pound. a. barium nit rat e b. sodium sulfat e c. lead( I I ) nit rat e d. sodium iodide e. sodium hydroxide f.

sulfuric acid

g. am m onium chloride h. lead( I I ) iodide 2. Suppose t hat each com pound in CTQ 1 is placed in wat er. Using Model 1 for guidance, place t he sym bol ( aq) or ( s) aft er each form ula t o indicat e it s solubilit y in wat er.

Information: Binary reactions (Double replacement reactions) Chem ical react ions occur bet ween t wo react ant s whenever t wo e le ct r olyt e s can com bine t o form a n on e le ct r olyt e ( i. e., an insoluble solid or a covalent m olecule such as wat er) . Consider t he react ion of barium hydroxide and sulfuric acid, shown below: Ba( OH) 2 ( aq) + H2 SO4 ( aq) o BaSO4 ( s) + 2 H2 O ( l) The react ant s are st rong elect rolyt es ( st rong base and st rong acid) , but t he product s are bot h nonelect rolyt es. This react ion is bot h a neut ralizat ion and a precipit at ion. Ÿ Neut ralizat ion: acid + base o salt + wat er Ÿ Precipit at ion: form at ion of an insoluble solid I f one or m ore nonelect rolyt es can be form ed, a react ion will occur. I f no nonelect rolyt es can be form ed, no react ion occurs. CA17

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Critical Thinking Questions: Writ e a balanced m olecular equat ion st art ing wit h each set of react ant s below. I nclude t he phase labels. Writ e NR for no react ion. Exam ple: Aqueous ba r iu m n it r a t e and aqueous sodium su lfa t e . x

Swit ch t he “ part ners” t o m ake ba r ium sulfa t e and sodium n it r a t e .

x

Ba la nce t he ionic cha r ge s t o get t he right form ula for each product ( as in CTQ 1) . Ba( NO3 ) 2 + Na 2 SO4 o BaSO4 + NaNO3

x

Use t he solubilit y rules in Table 1 t o a ssign t he pha se for each product . ( I f all product s are elect rolyt es, no react ion occurs, and you m ay st op here; writ e NR) . Ba( NO3 ) 2 ( aq) + Na 2 SO4 ( aq) o BaSO4 ( s) + NaNO3 ( aq)

x

Finally, ba la n ce t h e e qu a t ion by assigning coefficient s. Ba( NO3 ) 2 ( aq) + Na 2 SO4 ( aq) o BaSO4 ( s) + 2 NaNO3 ( aq)

3. Aqueous lead( I I ) nit rat e and aqueous sodium iodide.

4. Aqueous sodium hydroxide and aqueous sulfuric acid.

5. Aqueous am m onium chloride and aqueous pot assium nit rat e.

Exercises: 1. Using t he solubilit y rules, writ e a balanced m olecular equat ion st art ing wit h each set of react ant s below. I nclude t he phase labels. a. Aqueous silver nit rat e and aqueous sodium phosphat e ( silver always has a + 1 charge in ionic com pounds) .

b. Aqueous calcium nit rat e and aqueous pot assium carbonat e.

c. Aqueous copper( I I ) nit rat e and aqueous pot assium hydroxide.

d. Aqueous zinc chloride and aqueous sodium sulfide ( zinc always has a + 2 charge in ionic com pounds) .

a. Aqueous lead( I I ) nit rat e and aqueous sodium phosphat e.

b. Aqueous m agnesium sulfat e and aqueous sodium hydroxide.

2. Read t he assigned pages in your t ext , and work t he assigned problem s. - 73 -

CA17

Ch e m Act ivit y 1 8

Oxidation-Reduction Reactions ( What happens when elect rons are t ransferred bet ween at om s?)

Information: Oxidat ion- reduct ion react ions ( oft en called “ redox” react ions) are t hose in which elect rons are t ransferred bet ween at om s. Oxidat ion and reduct ion always occur t oget her, since an at om t hat loses elect rons m ust give t hem t o an at om t hat gains t he elect rons. A way t o keep t rack of elect rons in chem ical react ions is by assigning ox ida t ion n u m be r s t o each elem ent in a react ion. The oxidat ion num ber is t he charge an at om in a subst ance would have if t he elect ron pairs in each covalent bond belonged t o t he m ore elect ronegat ive at om . An a t om t h a t …

is sa id t o be …

a n d it s ox ida t ion num be r …

gains elect rons

reduced

decreases

loses elect rons

oxidized

increases

You are responsible for recognizing ( but not predict ing) redox react ions.

Table 1. General rules for assigning oxidation numbers 1. Pure e le m e nt s ( not in com pounds) have an oxidat ion num ber of 0. This includes diat om ic elem ent s like H2 or O2 . 2. The oxidat ion num ber of an ion equals it s charge. For exam ple: x x

group 1 ( alkali m et als) group 2 ( alkaline eart h)

Æ +1 Æ +2

3. I n m ole cu le s, F is –1; Cl, Br , I are –1 if bonded t o less elect ronegat ive at om 4. Ox yge n is nearly always –2 5. H ydr oge n nearly always + 1 6. Figure any ot hers by calculat ion. The sum of oxidat ion num bers of t he at om s of a m olecule equals zero; t he sum for an ion should equal it s overall charge. x

For exam ple, in CO2 , t he oxidat ion num ber of each O is –2, so t he oxidat ion num ber of C m ust be + 4.

Critical Thinking Questions: 1. Assign an oxidat ion num ber t o each at om individually in t he following subst ances. a. CCl 4

b. CH 4

c. HNO3

d. Ca( NO3 ) 2

e. K2 Cr 2 O7

f. SO4 2–

2. Which elem ent is oxidized and which is reduced in t he following react ion? 5 Fe 2+ ( aq) + MnO–4 ( aq) + 8 H+ ( aq) o Mn 2+ ( aq) + 5 Fe 3+ ( aq) + 4 H2 O( l)

3. Consider t he balanced chem ical equat ion: 5 H2 C2 O4 ( aq) + 2 MnO–4 ( aq) + 6 H+ ( aq) o 10 CO2 ( g) + 2 Mn 2+ ( aq) + 8 H2 O( l) When one m ole of oxalic acid, H2 C2 O4 , react s, ____ m oles of perm anganat e, MnO–4 , will be consum ed. a) 1 CA18

b) 2

c) 2/ 5

d) 5/ 2 - 74 -

4. I s t he react ion in CTQ 3 a redox react ion? Explain your answer in t erm s of oxidat ion num bers.

Information: Redox reactions in biological systems (organic molecules) I t is oft en difficult t o assign oxidat ion num bers in com plex m olecules, but t here is a short cut . I n t hese cases, a m olecule t hat undergoes a ddit ion of ox yge n a t om s or loss of h ydr oge n a t om s is ox idize d. I f oxygen at om s are r e m ove d ( i. e., a product ) or hydrogen at om s are a dde d ( i. e., a react ant ) , t he m olecule is r e du ce d. However, if H2 O ( wat er) is added or rem oved, t his is n ot a redox react ion. Oft en, we do not balance t hese react ions, so t he added or lost hydrogen or oxygen m ay not be shown. Exam ple: CH3 OH Æ CH2 O ( + H2 )

CH3 OH is oxidized ( loss of 2 H)

Exam ple: C4 H10 ( + H2 ) Æ C4 H12

C4 H10 is reduced ( gain of 2 H)

Exam ple: C4 H10 Æ C4 H12 O

neit her ( addit ion of H2 O, not balanced)

Exam ple: CH2 O Æ CH2 O2

oxidat ion ( addit ion of O, not balanced)

Critical Thinking Questions: 5. Glucose ( C6 H12 O6 ) is used for energy in cells in cellular respirat ion by t he net react ion shown: C6 H12 O6 + 6 O2 o 6 CO2 + 6 H2 O I s glucose oxidized, reduced, or neit her? Explain bot h in t erm s of t he oxidat ion num ber of carbon and t he short cut for organic react ions.

6. When linoleic acid, an unsat urat ed fat t y acid, react s wit h hydrogen, it becom es a sat urat ed fat t y acid. C18 H32 O2 o C18 H36 O2 ( not balanced) This process, called hydrogenat ion, is used t o m ake short ening ( “ Crisco” ) out of veget able oil. I n t his hydrogenat ion, is linoleic acid oxidized or reduced? Explain.

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CA18

Exercises: 1. Assign oxidat ion num bers t o each at om in t he react ions below. Then t ell which at om is oxidized and which is reduced in each react ion. a. CuO ( s) + H 2 ( g) o Cu ( s) + H2 O ( g) b. 2 CO ( g) + O2 ( g) o 2 CO2 ( g) 2. I s t he react ion in Exercise 1a an ox ida t ion or a r e du ct ion ( cir cle on e ) of copper( I ) oxide? Explain how you can t ell w it h ou t assigning oxidat ion num bers.

3. I s t he react ion in Exercise 1b an ox ida t ion or r e du ct ion of carbon m onoxide? Explain how you can t ell w it h ou t assigning oxidat ion num bers.

4. When et hanol is ingest ed and m et abolized in t he liver, it is first convert ed t o acet aldehyde, and t hen t o acet ic acid, as shown in t he Lewis st ruct ures below. H H H O H O H

C H

C

O

H

ethanol

H

H

C

C

H

H

acetaldehyde (ethanal)

H

C

C

O

H

H

acetic acid (ethanoic acid)

a. I s t he conversion of et hanol t o acet aldehyde an ox ida t ion or r e du ct ion ( cir cle one ) ? Explain how you can t ell.

b. I s t he conversion of acet aldehyde t o acet ic acid an ox ida t ion or r e du ct ion ( cir cle one ) ? Explain how you can t ell.

5. According t o t he balanced chem ical equat ion: 5 H2 C2 O4 ( aq) + 2 KMnO4 ( aq) + 6 H+ ( aq) o 10 CO2 ( g) + 2 Mn 2+ ( aq) + 8 H2 O( l) + 2 K+ ( aq) How m any m ole s of pot assium perm anganat e, KMnO4 , will exact ly react wit h 0.3500 gr a m s of oxalic acid, H2 C2 O4 , in acidic solut ion. Show work. ( Not e t hat t his is t he sam e react ion as in CTQ 3, but wit h t he spect at or ion pot assium included.) How m any gr a m s of KMnO4 would t his be?

6. Read t he assigned pages in your t ext , and work t he assigned problem s. CA18

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Ch e m Act ivit y 1 9

Mass Relationships (Stoichiometry)* ( How m uch can you m ake?)

Model 1: The S’more: A delicious t reat known as a S’m ore is const ruct ed wit h t he following ingredient s and am ount s: 1 graham cracker 1 chocolat e bar 2 m arshm allows At a part icular st ore, t hese it em s can be obt ained only in full boxes, each of which cont ains one gross of it em s. A gross is a specific num ber of it em s, analogous ( but not equal) t o one dozen. The boxes of it em s have t he following net weight s ( t he weight of t he m at erial inside t he box) : box of graham crackers box of chocolat e bars box of m arshm allows

9.0 pounds 36.0 pounds 3.0 pounds

Critical Thinking Questions: 1. I f you have a collect ion of 100 graham crackers, how m any chocolat e bars and how m any m arshm allows do you need t o m ake S’m ores using all t he graham crackers?

2. I f you have a collect ion of 1000 graham crackers, 800 chocolat e bars, and 1000 m arshm allows: a. How m any S’m ores can you m ake?

b. What ( if anyt hing) will be left over, and how m any of t hat it em will t here be?

Information: Chem ist s refer t o t he react ant which lim it s t he am ount of product which can be m ade from a given collect ion of original react ant s as t he lim it ing r e a ge n t or lim it in g r e a ct a n t .

* Adapt ed from Chem Act ivit y 30, Moog, R.S. and Farrell, J.J. Chem ist ry: A Guided I nquiry. Wiley, 3 rd ed., 2006, pp. 168- 171.

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CA19

Critical Thinking Questions: 3. I dent ify t he lim it ing reagent in CTQ 2. 4. Based on t he inform at ion given: a. Which of t he t hree ingredient s ( a graham cracker, a chocolat e bar, or a m arshm allow) weighs t he m ost ? b. Which weighs t he least ? Explain your reasoning.

5. I f you have 36.0 pounds of graham crackers, 36.0 pounds of chocolat e bars, and 36.0 pounds of m arshm allows: a. Which it em do you have t he m ost of? b. Which it em do you have t he least of? Explain your reasoning.

6. I f you at t em pt t o m ake S’m ores from t he m at erial described in CTQ 5: a. What will t he lim it ing reagent be?

b. How m any gross of S’m ores will you have m ade?

c. How m any gross of each of t he t wo left over it em s will you have?

d. How m any pounds of each of t he left over it em s will you have?

e. How m any pounds of S’m ores will you have?

CA19

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7. Using G as t he sym bol for graham cracker, Ch for chocolat e bar, and M for m arshm allow, writ e a " balanced chem ical equat ion" for t he product ion of S’m ores.

8. Explain in com plet e sent ences why it is not correct t o st at e t hat if we st art wit h 36 pounds each of G, Ch, and M t hen we should end up wit h 3 x 36 = 108 pounds of S’m ores.

Model 2: Water Wat er can be m ade by burning hydrogen in oxygen, as shown below: 2 H 2 + O2 o 2 H 2 O A m ole is a specific num ber of it em s, analogous ( but not equal) t o one dozen. The react ant s in t he equat ion have t he following m olar m asses ( t he weight of t he m at erial in a m ole of m at erial) : m ole of H2 m olecules m ole of O2 m olecules

2.0 gram s 32.0 gram s

Critical Thinking Questions: 9. Based on t he inform at ion given: a. Which of t he t wo ingredient s ( hydrogen m olecules or oxygen m olecules) weighs t he m ost ? b. Which weighs t he least ? Explain your reasoning.

10. I f you have a collect ion of 100 hydrogen m olecules, how m any oxygen m olecules do you need t o m ake wat er wit h all of t he hydrogen m olecule?

11. I f you have 32.0 gram s of hydrogen m olecules and 32.0 gram s of oxygen m olecules, a. which it em do you have t he m ost of? b. which it em do you have t he least of? Explain your reasoning.

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CA19

12. I f you at t em pt t o m ake wat er from t he m at erial described in CTQ 11: a. What is t he lim it ing reagent ?

b. How m any m oles of wat er m olecules will you have m ade?

c. How m uch does one m ole of wat er m olecules weigh? Explain how you can det erm ine t his.

d. Based on your answer t o ( b) , how m any gram s of wat er will you have?

e. A st udent perform ed t he react ion in CTQ 11 and obt ained only 34.0 gram s of wat er. What percent age of t he t ot al possible am ount of wat er did he obt ain?

Information: What if the answers are not in whole numbers? (Example) Pr oble m : Met hane gas ( CH4 ) burns in oxygen ( O2 ) t o produce carbon dioxide and wat er. How m any gram s of CO2 can be produced from 2.0 g of m et hane and 2.0 g of oxygen? x

First , writ e t he t wo product s and ba la nce t he equat ion: ___ CH4 + ___ O2 o

x

Then, using t he m ola r m a ss ( g/ m ol) for each react ant , see how m any m oles of each react ant you have: 1 m ol CH4 2.0 g CH4 ˜ 0.1247 m ol CH4 16.04 g CH4 2.0 g O2 ˜

1 m ol O2 32.00 g O2

0.0625 m ol O2

We need t o det erm ine which reagent will be com plet ely used up. Look at t he balanced equat ion again. I t shows us t hat we need t wo t im es as m any O2 m oles as CH4 m oles. Do we have t his ( is 0.0625 m ol t wice as m uch as 0.1247 m ol) ? Ye s or n o ( circle one) . So t he CH 4 or O 2 ( circle one) will be com plet ely used up ( t he lim it in g r e a ge nt ) . The calculat ion of t he m oles of t he reagent t hat is n ot lim it ing is now useless. Draw a large “ X” t hrough t hat calculat ion above t o indicat e t hat you will not use it . Now, t he quest ion asks how m uch CO2 is produced. Look at t he balanced equat ion yet again. I t shows t hat t he m oles of CO2 produced will be half t he num ber of m oles of O2 used up. Since O2 was t he lim it ing reagent , you can m ult iply t he m oles of O2 by ½ , or use a conversion fact or, as follows: 0.0625 m ol O2 ˜

1 m ol CO2 2 m ol O2

0.03125 m ol CO2

See how t he coefficient s from t he equat ion m ake a conversion fact or ( “ ½ ” ) for you? ( I nclude t he unit s! ) Now j ust use t he m ola r m a ss of CO2 t o change t he m oles int o gram s, and you’re done! CA19

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0.03125 m ol CO2 ˜

44.01 g CO2 1 m ol CO2

1.4 g CO2

Pa r t 2 : I f 1.0 g of CO2 are produced in t he above react ion, what is t he percent yield? The 1.4 g is how m uch you can t h e or e t ica lly get . But we only got 1.0 g, so m aybe som e gas escaped, or perhaps our react ion was incom plet e. Our pe r ce n t yie ld is: 1.0 g produced u100 71% yield 1.4 g t heoret ica l

Exercises: 1. The t herm it e react ion, once used for welding railroad rails, is oft en used for an excit ing chem ist ry dem onst rat ion because it produces red- hot m olt en iron. The react ion is: Fe 2 O3 ( s) + 2 Al ( s) o 2 Fe ( l) + Al 2 O3 ( s) I f you st art wit h 50.0 g of iron( I I I ) oxide and 25.0 g of alum inum , what is t he lim it ing reagent ? What is t he m axim um m ass of alum inum oxide t hat could be produced? How m uch alum inum oxide would be produced if t he yield is 93% ?

2. How m uch aspirin can be m ade from 100.0 g of salicylic acid and 100.0 g of acet ic anhydride? I f 122 g of aspirin are obt ained by t he react ion, what is t he percent yield? C7 H6 O3 + C4 H6 O3 o C9 H8 O4 salicylic acid acet ic anhydride aspirin

+

HC2 H3 O2 acet ic acid

3. Com plet e Chem Worksheet 1: St oichiom et ry Pract ice Worksheet 1. 4. Read t he assigned pages in your t ext , and work t he assigned problem s. - 81 -

CA19

Ch e m Act ivit y 2 0

Thermochemistry ( What does heat have t o do wit h chem ist ry?)

Suggested demonstration: Exothermic and endothermic reactions Information: The chem icals in a react ion syst e m can eit her absorb heat energy from t he su r r ou ndings or release heat energy t o t he surroundings. A react ion in which heat energy m oves from t he syst em t o t he surroundings ( and feels warm ) is said t o be exot herm ic. A react ion in which heat energy m oves from t he surroundings t o t he syst em ( and feels cool) is said t o be endot herm ic. Heat energy is m easured in calories ( English syst em ) or Joules ( m et ric syst em ) . 1 cal = 4.184 j oules ( exact ) Diet ary calories report ed on food product s are act ually kilocalories, writ t en as “ kcal” or “ Cal” ( wit h a capit al “ C” ) .

Model 1: Heating curve of a pure substance The graph ( heat ing curve) below shows t he t em perat ure of 19 g of a pure subst ance t hat is heat ed by a const ant source supplying 500.0 calories per m inut e.

gas

liquid

solid

temp, °C 120 110 100 90 80 70 60 50 40 30 20 10 0 -10 -20

Z X Y

V W

U time (not to scale)

Regions: UV = 0.76 m in, VW = 3.04 m in, WX = 3.8 m in, XY = 20.5 m in, YZ = 0.63 m in

Critical Thinking Questions: 1. At t he coldest t em perat ure ( point U) , what is t he physical st at e of t he pure subst ance? Circle one: solid

CA20

liqu id

ga s.

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2. I dent ify t he region on t he graph ( UV, VW, et c.) t hat corresponds t o t he pure subst ance: a. being warm ed while rem aining a solid _____ b. being warm ed while rem aining a liquid _____ c. being warm ed while rem aining a gas _____ d. changing from a solid t o a liquid _____ e. changing from a liquid t o a gas _____ 3. What is t he boiling point of t he liquid? _____ 4. What is t he m elt ing point of t he liquid? _____ 5. Why is t he t em perat ure not changing from point V t o W, even t hough heat is being added?

6. Why is t he t em perat ure not changing from point X t o point Y?

7. How m any calories were needed t o m elt t he solid?

8. From your answer t o CTQ 7, calculat e t he calories used per gram ( cal/ g) of t his subst ance in order t o m elt it . I nclude unit s. This quant it y is known as t he he a t of fu sion of t he subst ance.

9. How m any calories were needed t o change t he liquid t o a gas?

10. From your answer t o CTQ 9, calculat e t he calories used per gram ( cal/ g) of t his subst ance t o vaporize it . I nclude unit s. This quant it y is known as t he h e a t of va por iza t ion of t he subst ance.

11. How m any calories were needed t o warm t he liquid from point W t o point X?

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CA20

12. How m any degrees Celsius did t he t em perat ure ch a n ge from point W t o point X?

13. From your answers t o CTQs 11 and 12, calculat e t he calories used per gram of t his cal

subst ance per degree Celsius ( g qC ) t o change it s t em perat ure. I nclude unit s. This quant it y is known as t he spe cific he a t ca pa cit y ( or “ specific heat ” ) of t he subst ance.

14. Where on t he curve do t he m olecules have t he highest kinet ic energy? ___________ 15. What do you t hink is t he ident it y of t he subst ance being heat ed? ___________ 16. Com plet e t he definit ions of t he t erm s in boldfa ce in CTQs 8, 10, and 13. a.

The heat of fusion of a subst ance is

b.

The heat of vaporizat ion of a subst ance is

c.

The specific heat capacit y of a subst ance is

Model 2: Specific heat capacity (or “specific heat”) The heat energy in calories ( q) needed t o raise t he t em perat ure of a subst ance m ay be calculat ed using t he following equat ion: q ( heat in calories) = s ( specific heat , cal ) × ( m ass in gram s) × ( 'T in °C) g qC

or sim ply: q = s·m ·'T 17. How m uch heat energy is required t o com plet ely change 25 gram s of liqu id w a t e r at 0°C t o wat er at body t em perat ure ( 37°C) ?

CA20

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18. Consider how m uch heat energy is required t o com plet ely change 25 gram s of ice at 0°C t o w a t e r at 0°C. a. Circle t he const ant t hat is needed t o perform t his calculat ion: h e a t of fusion

h e a t of va por iza t ion

spe cific h e a t

b. Perform t he calculat ion.

19. How m uch t ot al heat energy is required t o com plet ely change 25 gram s of ice at 0°C t o wat er AND also raise it s t em perat ure t o body t em perat ure ( 37°C) ?

20. A hot iron skillet ( 178°C) weighing 1.51 kg is s it t ing on a st ove. How m uch heat energy ( in j oules) m ust be rem oved t o cool t he skillet t o room t em perat ure, 21°C? The specific heat of iron is 0.450 J/ ( g·°C) .

21. Where does t he heat energy in CTQ 20 go? 22. I s t he process in CTQ 20 exot herm ic or endot herm ic?

Exercises: 1. Convert your answer t o CTQ 19 int o kilocalories. Would eat ing ice be a good way t o lose weight ? Explain why or why not .

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CA20

2. I t t akes 5.85 kJ of heat energy t o raise t he t em perat ure of 125.6 g of m ercury from 20.0°C t o 28.3°C. Calculat e t he specific heat capac it y of m ercury.

3. When 23.6 g of calcium chloride were dissolved in 300 m L of wat er in a calorim et er, t he t em perat ure of t he wat er rose from 25.0°C t o 38.7°C . What is t he heat energy change in kcal for t his process? [ The specific heat of H2 O = 1.00 cal/ g °C.]

4. I n Exercise 3 you calculat ed t he heat energy produced by dissolving 23.6 g of calcium chloride in wat er. How m any m ole s of calcium chloride is t his? How m uch heat energy is produced pe r m ole of calcium chloride t hat dissolves in wat er? Report your answer in k iloca lor ie s/ m ole ( kcal/ m ol) .

5. Read t he assigned pages in your t ext , and work t he assigned problem s.

CA20

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Ch e m Act ivit y 2 1

Equilibrium* ( Do react ions really ever st op?)

Model 1: The conversion of cis-2-butene to trans-2-butene. Consider a sim ple chem ical react ion where t he forward react ion occurs in a single st ep and t he reverse react ion occurs in a single st ep: A qwe B The following chem ical react ion, where cis- 2- but ene is convert ed int o t rans- 2- but ene, is an exam ple. H CH3 CH3 H 3C C

C

C

H

H3C

H

C H

cis-2-butene trans-2-butene I n t his exam ple, one end of a cis- 2- but ene m olecule rot at es 180° t o form a t rans- 2- but ene m olecule. Rot at ion around a double bond rarely happens at room t em perat ure because t he collisions are not sufficient ly energet ic t o weaken t he double bond. At higher t em perat ures, around 400°C for cis- 2- but ene, collisions are sufficient ly energet ic and an appreciable react ion rat e is det ect ed.

Critical Thinking Questions: 1. Make a m odel of cis- 2- but ene wit h a m odeling kit . ( I n m ost m odel kit s, black = C, whit e = H. Use short bonds for single bonds and t wo of t he longer, flexible bonds for double bonds.) What m ust be done t o convert cis- 2- but ene t o t rans- 2- but ene?

2. Make a m odel of t rans- 2- but ene wit h a m odeling kit . What m ust be done t o convert t rans- 2- but ene t o cis- 2- but ene?

3. A large num ber of cis- 2- but ene m olecules is placed in a cont ainer. a. Predict what will happen if t hese m olecules are allowed t o st and at t em perat ure for a long t im e.

room

b. Predict what will happen if t hese m olecules are allowed t o st and at 400°C for a long t im e.

* Adapt ed from Chem Act ivit y 37, Moog, R.S. and Farrell, J.J. Chem ist ry: A Guided I nquiry, 3 rd ed., Wiley, 2006, pp. 205- 209.

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CA21

Model 2. The number of molecules as a function of time. Consider t he sim ple react ion: A qwe B The syst em is said t o be at equilibrium when t he concent rat ions of react ant s and product s st op changing. I m agine t he following hypot het ical syst em . Exact ly 10,000 A m olecules are placed in a cont ainer which is m aint ained at 800°C. We have t he abilit y t o m onit or t he num ber of A m olecules and t he num ber of B m olecules in t he cont ainer at all t im es. We collect t he dat a at various t im es and com pile Table 1.

Table 1. Number of A and B molecules as a function of time Tim e ( seconds)

Num ber of A Molecules

Num ber of B Molecules

Num ber of A Molecules t hat React in Next Second

Num ber of B Molecules t hat React in Next Second

Num ber of A Molecules Form ed in Next Second

Num ber of B Molecules Form ed in Next Second

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 30 40 50

10000 7500 5875 4819 4132 3686 3396 3207 3085 3005 2953 2920 2898 2884 2874 2868 2864 2862 2860 2859 2858 2858 2858 2857 2857 2857 2857 2857 2857

0 2500 4125 5181 5868 6314 6604 6793 6915 6995 7047 7080 7102 7116 7126 7132 7136 7138 7140 7141 7142 7142 7142 7143 7143 7143 7143 7143 7143

2500 1875 1469 1205 1033 921 849 802 771 751 738 730 724 721 719 717 716 715 715 715 715 714 714 714 714 714 714 714 714

0 250 413 518 587 631 660 679 692 699 705 708 710 712 713 713 714 714 714 714 714 714 714 714 714 714 714 714 714

0 250 413 518 587 631 660 679 692 699 705 708 710 712 713 713 714 714 714 714 714 714 714 714 714 714 714 714 714

2500 1875 1469 1205 1033 921 849 802 771 751 738 730 724 721 719 717 716 715 715 715 715 714 714 714 714 714 714 714 714

CA21

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Critical Thinking Questions: 4. During t he t im e int erval 0–1 s: a. How m any A m olecules react ? _____ b. How m any B m olecules are form ed? _____ c. Why are t hese t wo num bers equal?

5. During t he t im e int erval 10–11 s: a. How m any B m olecules react ? ____ b. How m any A m olecules are form ed? ____ c. Why are t hese t wo num bers equal?

6. a. During t he t im e int erval 0–1 s, what fract ion of t he A m olecules react ?

b. During t he t im e int erval 10–11 s, what fract ion of t he A m olecules react ?

c. During t he t im e int erval 24–25 s, what fract ion of t he A m olecules react ?

d. During t he t im e int erval 40–41 s, what fract ion of t he A m olecules react ?

7. Based on t he answers t o CTQ 6, verify t hat 921 m olecules of A react during t he t im e int erval 5–6 s.

8. During t he t im e int erval 100–101 s, how m any m olecules of A react ? Explain your reasoning.

9. a. During t he t im e int erval 1–2 s, what fract ion of t he B m olecules react ?

b. During t he t im e int erval 10–11 s, what fract ion of t he B m olecules react ?

c. During t he t im e int erval 24–25 s, what fract ion of t he B m olecules react ?

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CA21

d. During t he t im e int erval 40–41 s, what fract ion of t he B m olecules react ?

10. Based on t he answers t o CTQ 9, verify t hat 631 m olecules of B react during t he t im e int erval 5–6 s.

11. During t he t im e int erval 100–101 s, how m any m olecules of B react ? Explain your reasoning.

12. For t he react ion described in Table 1: a. How long did it t ake for t he react ion t o com e t o equilibrium ?

b. Are A m olecules st ill react ing t o form B m olecules at t = 500 seconds? c. Are B m olecules st ill react ing t o form A m olecules at t = 500 seconds?

Information: For t he process in Model 2, rat e of conversion of B t o A = num ber of B m olecules t hat react per second =

' num ber of B m olecules 't

where “ '” m eans “ change in.” The relat ionship bet ween t he rat e of conversion of B t o A and t he num ber of B m olecules is given by equat ion ( 1) : rat e of conversion of B t o A = ( fract ion) × ( num ber of B m olecules)

( 1)

where ( fract ion) is a specific value called a rat e const ant , k B.

Critical Thinking Questions: 13. Rewrit e equat ion ( 1) , replacing “ ( fract ion) ” wit h “ k B.”

14. What is t he value of k B in equat ion ( 1) ? Be sure t o include t he unit s in your answer.

15. Rewrit e equat ion ( 1) , replacing “ k B” wit h t he value of k B t hat you det erm ined in CTQ 14.

CA21

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16. Writ e a m at hem at ical equat ion analogous t o equat ion ( 1) for t he rat e of conversion of A m olecules int o B m olecules. This equat ion should include a const ant k A.

17. What is t he value of k A in your equat ion in CTQ 16? I nclude unit s.

Exercises: 1. I n t his act ivit y, you calculat ed a rat e const ant for t he forward react ion ( k A) of ________ ( include unit s) . 2. I n t his act ivit y, you calculat ed a rat e const ant for t he reverse react ion ( k B) of ________ ( include unit s) . 3. The rat io of t he forward rat e const ant t o t he reverse rat e const ant (i. e., k A/ k B) is called t he equilibrium const ant . Calculat e t he equilibrium const ant , K eq , for t he react ion in Model 2.

4. When equilibrium is reached, what fract ion of t he m olecules in t he above react ion exist as product s ( B) ?

5. Does t his equilibrium favor ( have m ore of) t he pr odu ct s or t he r e a ct a n t s ( circle one) ? 6. For a react ion wit h

K eq > 1, does t he equilibrium favor t he product s or t he react ant s?

7. For a react ion wit h K eq < 1, would t he equilibrium favor t he product s or t he react ant s?

8. For a react ion wit h K eq = 1, would t he equilibrium favor t he product s or t he react ant s?

9. Read t he assigned pages in your t ext book, and work t he assigned problem s.

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CA21

Ch e m Act ivit y 2 2

Rates of Reactions ( What det erm ines how fast a chem ical react ion proceeds?)

Information: Consider Model 1 of Chem Act ivit y 21 concerning t he equilibrium of cis- 2- but ene and t rans- 2 but ene.

Critical Thinking Questions: 1. I n what way( s) are t he collisions bet ween m olecules at 400°C different from t he collisions at room t em perat ure?

2. What condit ion( s) m ust be sat isfied for a m olecular collision t o result in a chem ical react ion?

Model 2: Energy diagrams for three model reactions a) an exothermic reaction energy of transition state

b) the same exothermic reaction in the presense of a catalyst

c) an endothermic reaction

energy of reactants energy of products progress of reaction

energy

energy

energy

activation energy energy of reactants energy of products progress of reaction

energy of products energy of reactants progress of reaction

A cat alyst increases t he rat e of a chem ical react ion wit hout being consum ed in t he react ion. The t ransit ion st at e is t he part icular arrangem ent of atom s in t he react ant s at t he m axim um energy level.

Critical Thinking Questions: 3. What is act ivat ion energy?

4. Label t he t ransit ion st at e energy in Model 2, part s ( b) and ( c) . 5. Draw vert ical arrows ont o Models 2 ( b) and ( c) t hat represent t he m agnit ude of t he act ivat ion energy for conversion of t he react ant s t o t he product s. CA22 - 92 -

6. What happens t o react ant m olecules t hat have le ss energy t han t he energy of t he t ransit ion st at e?

7. How does t he addit ion of a cat alyst affect t he m agnit ude of t he act ivat ion energy of a chem ical react ion?

8. What is t he role of act ivat ion energy in affect ing t he rat e of a chem ical react ion?

9. On average, do m olecules have m ore energy at h ighe r or low e r t em perat ures ( circle one) ? 10. Why do react ions proceed fast er at higher t em perat ures?

11. Com plet e t he sent ence: A cat alyst increases t he rat e of a chem ical react ion by

12. Considering Model 2, does a cat alyst affect t he energies of t he react ant s? _____ 13. Considering Model 2, does a cat alyst affect t he energies of t he product s? _____

Information: The lower t he energy of a species, t he m ore favorable it is t o be form ed. I n exot herm ic react ions, t he product s have less energy t han t he react ant s, and so t hese react ions favor conversion of m ost of t he react ant s t o product s. The exact equilibrium am ount s of react ant s and product s are det erm ined by t heir relat ive energies.

Critical Thinking Questions: 14. Considering your answer t o CTQs 12 and 13, does a cat alyst affect t he equilibrium am ount s of react ant s or product s? Why or why not ?

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CA22

15. I n a chem ical equilibrium , what is it t hat is equal?

16. The st at e of equilibrium is oft en called a dynam ic equilibrium . Explain t he m eaning of t he t erm dynam ic in t his cont ext .

Exercises: 1. Most exot herm ic react ions are considered spont aneous, m eaning t hat t he product s are m ore st able ( have lower energy) t han t he react ant s. However, spont aneous does not necessarily equat e t o fast . Explain why a react ion t hat is considered spont aneous m ay nevert heless not show any observable react ion.

2. Sket ch a plot of energy ( y axis) versus react ion progress ( x axis) for a t ypical exot herm ic react ion. Then, draw anot her line ont o t he sam e plot which shows how t he curve changes when t he react ion occurs in t he presence of a cat alyst . Label each line. Explain how a cat alyst increases t he rat e of a chem ical react ion in t erm s of t he m eaning of your plot .

CA22

- 94 -

3. Consider t he react ion in Exercise 2: a. Would t his react ion favor form at ion of m ost ly pr oduct s or m ost ly r e a ct a n t s? b. Would t he equilibrium const ant , K eq , for t his react ion be le ss t ha n, e qua l t o, or gr e a t e r t ha n 1? Circle one, and explain your choice.

c. When t his react ion had reached equilibrium , would t he forward rat e be le ss t ha n , e qua l t o, or gr e a t e r t h a n t he reverse rat e? Circle one, and explain your choice.

4. The burning of a piece of paper in air ( t o produce CO2 and H2 O vapor) is a spont aneous exot herm ic react ion. a. What would be necessary t o st art t his react ion?

b. Aft er t he react ion st art s, what provides t he act ivat ion energy for t he react ion t o cont inue?

5. Read t he assigned pages in your t ext book and work t he assigned problem s.

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CA22

Ch e m Act ivit y 2 3

Gases ( Do all gases behave t he sam e?)

Model 1: Representation of the molecular view of water in a closed container at 20°C (293 K).

Key:

|

= 1 × 10 21 m olecules of wat er.

Vapor (gas)

Liquid

Critical Thinking Question: 1. Considering Model 1, what are t he m ain differences bet ween a gas and a liquid at t he sam e t em perat ure?

Model 2: Representation of some gases and their properties under different conditions.

A

C

B

D E

T = 273 K (0ºC) P = 1.0 atm V = 0.224 L

T = 273 K P = 1.0 atm V = 0.224 L

T = 273 K P = 2.0 atm V = 0.224 L

Key: T = temperature; P = Pressure; V = volume;

T = 546 K P = 2.0 atm V = 0.224 L

= 1.0×10 21 molecules N 2;

T = 273 K P = 2.0 atm V = 0.112 L = 1.0×10 21 atoms He

The standard air pressure at sea level (about 15 lb/in2) is 1 atmosphere (atm) 1 atm = 760 mm Hg = 760 torr CA23

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Critical Thinking Questions: 2. Com pare cont ainers A and B in Model 2. How do t he propert ies of t em perat ure ( T) and pressure ( P) change when t he ident it y of t he gas in t he cont ainer is changed at const ant volum e?

3. Com pare cont ainers A and C in Model 2. How does t he pressure of a gas change when t he num ber of m ole cule s in t he cont ainer is doubled at const ant volum e?

4. How does t he pressure of a gas change when t he num ber of m ole s of gas in t he cont ainer is doubled at const ant volum e?

5. Considering Model 2, how does t he pressure of a gas change when t he t em perat ure is doubled at const ant volum e?

6. Considering Model 2, how does t he pressure of a gas change when t he volum e of t he cont ainer is doubled at const ant t em perat ure?

7. The sym bol & m eans “ is proport ional t o,” and t he sym bol “ n” m eans “ num ber of m ole s of gas.” Based on your answer t o CTQ 4, circle t he correct expression. P& n

P &

1 n

8. Based on your answer t o CTQ 5, circle t he correct expression. P &T

P &

1 T

9. Based on your answer t o CTQ 6, circle t he correct expression. P &V

P &

1 V

10. Add each of t he t hree t erm s from t he right - hand side of t he expressions t hat you circled in CTQs 7, 8, and 9 t o t he expression below ( all m ult iplied t oget her) . P &

×

×

CTQ 7 · CTQ 8 · CTQ 9

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CA23

11. Since & m eans “ proport ional t o” som et hing, it also m eans “ equals a const ant m ult iplied by” som et hing. Rewrit e t he equat ion from CTQ 10 and replace t he “&” wit h “ = ( const ant ) t im es” .

12. The proport ionalit y const ant im plied in CTQ 11 is called t he ide a l ga s const a nt , and is given t he sym bol R. Rewrit e t he expression, replacing “ = ( const ant ) t im es” wit h “ = R” .

13. Use algebra t o rearrange t he equat ion from CTQ 12 so t hat all t he variables are in t he num erat or. This equat ion is called t he ide a l ga s la w .

14. Using t he num ber of m olecules in cont ainer A in Model 2, calculat e t he num ber of m oles ( n) .

15. Use t he values of P, T, n, and V from container A in Model 2 t o calculat e R. Give t he correct unit s.

16. Repeat CTQ 15 using t he values of P, T, n, and V from cont ainer C, D, or E in Model 2 ( your choice) .

17. Com pare your answers t o CTQs 15 and 16. Are t hey t he sam e? Why or why not ?

CA23

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Model C : Mixtures of gases I n a m ixt ure of gases, t he t ot al pressure, PT, is t he sum of t he pressures of t he individual gases. The individual pressures are called pa r t ia l pr e ssu r e s.

PT

¦ P , m eaning i

PT

P1  P2  P3  

( 1)

i

The pressure of each gas in t he m ixt ure obeys t he ideal gas law:

Pi

ni

RT V

( 2)

Critical Thinking Questions: 18. I s equat ion ( 2) in agreem ent wit h t he ideal gas law? Explain.

19. A st udent m akes t he following st at em ent : “ The rat io of t he part ial pressures of t wo gases in a m ixt ure is t he sam e as t he rat io of t he num ber of m oles of t he t wo gases.” I s t he st udent correct ? Explain your answer.

Exercises: 1. Calculat e t he volum e of 359 g of et hane ( C2 H6 ) at 0.658 at m and 75°C. ( Wat ch t he unit s! )

2. A 10.00- L SCUBA t ank holds 18.0 m oles of O2 and 12.0 m oles of He at 289 K. What is t he part ial pressure of O2 ? What is t he part ial pressure of He? What is t he t ot al pressure?

3. Com plet e Chem Worksheet 2, Gases: Pract ice Worksheet . 4. Read t he assigned pages in your t ext , and work t he assigned problem s. - 99 -

CA23

Ch e m Act ivit y 2 4

Solutions and Molarity ( When is it dissolved, and when is it suspended?)

Suggested demonstration: Solutions, colloids, and suspensions Information: When t wo subst ances are m ixed t oget her and t he m ixt ure is hom ogeneous, we call t he m ixt ure a solu t ion . The com ponent of a solut ion in t he great er am ount is t he solve n t , and t he com ponent in t he lesser am ount is t he solu t e . The m ost com m on solvent on eart h is wat er. Solut es t hat dissolve in wat er m ay be solids ( e. g., salt or sugar) , liquids ( e. g., alcohol) or gases ( e. g., am m onia or oxygen) . Each solut e has a part icular solu bilit y in wat er—oft en report ed as t he m axim um num ber of gram s of t he solut e t hat will dissolve in 100 gram s of wat er. A solut ion cont aining t he m axim um am ount of dissolved solut e is called a sa t ur a t e d solut ion. I f t he solut e part icles are large enough t o be seen ( t hey scat t er light , m aking t he m ixt ure cloudy) but not t o set t le out , t he m ixt ure is called a colloid. I f t he part icles are larger st ill and can set t le out over t im e, t he m ixt ure is called a su spe n sion .

Critical Thinking Question: 1. Label each of t he following as a solut ion, colloid, or suspension. a. t om at o j uice

e. t ea

b. fog

f.

c. apple j uice

g. hom ogenized m ilk

d. I t alian salad dressing

h. cola

m uddy wat er

2. I n each of t he following m ixt ures, nam e t he solvent and at least one solut e. M ix t u r e

Solve n t

Solut e ( s)

fog apple j uice cola NaCl( aq)

Information: The m easure of t he am ount of solut e dissolved in a specified am ount of solut ion is called t he con ce n t r a t ion of t he solut ion. The m ost com m on m easure of concent rat ion is m ola r it y. Molarit y ( M) – t he m oles of solu t e per lit er of solu t ion: M

moles of solute L of solution

mol L

For exam ple, bat t ery acid is approxim at ely 6 M H2 SO4 . This m eans t hat t here are 6 m oles of sulfuric acid in every lit er of solut ion. We can writ e t wo conversion fact ors: 6 m oles H2 SO 4 and 1 L

1 L 6 m oles H2 SO 4

Not e how we assum e—but didn’t act ually writ e—t he word “ solut ion” wit h t he “ 1 L.” CA24

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Suggested Demonstration: How to make a solution of a particular molarity Model 1: How to make 100 mL of a 2.0-molar (2.0 M) aqueous solution of a solid

100 mL

100 mL

100 mL

empty 100-mL volumetric flask

place 0.200 moles of solid in flask

dilute to the mark with water and mix well

Note: volumetric flasks come in standard sizes, such as 25 mL, 50 mL, 100 mL, 250 mL, 500 mL, 1 L, etc.

Critical Thinking Questions: 3. I n Model 1, only t wo- t ent hs of a m ole of a solid was used t o m ake a 2.0- m olar solut ion. Explain how t his is possible.

4. An experim ent calls for 50 m L of a 0.50 M aqueous solut ion of sodium hydrogen carbonat e ( sodium bicarbonat e, or baking soda) . Describe ( wit h am ount s) t he st eps you would use in order t o m ake up such a solut ion, such t hat you have none left over.

Information: When dilu t in g a solut ion, t he m oles of t he solut e do not change. Therefore, m oles ( before) = m oles ( aft er)

or

m ol 1 = m ol 2

Since M= m ol/ L, ( M × L) is equal t o m oles. So we can writ e: M × L ( before) = M × L ( aft er) or M1 V1 = M2 V2 ( where V = volum e)

( 1)

This is an equat ion t hat can be used for dilu t ion s. I t m ay also be rearranged t o give: M1 V1 ( 2) M2 V2 - 101 -

CA24

Model 2: How to make 100 mL of 0.060 M CuSO4 from 0.60 M CuSO4

take 10 mL of 0.60 M CuSO4

100 mL

100 mL

place the 10 mL into 100-mL volumetric flask

dilute to the mark with water and mix well

Critical Thinking Questions: 5. Consider Model 2. Verify t hat dilut ing 10 m L of 0.60 M copper( I I ) sulfat e t o 100 m L will produce a 0.060 M solut ion. ( I dent ify M1 , V1 , and V2 and calculat e M2 ) .

6. How m any m L of 0.60 M copper( I I ) sulfat e would need t o be added t o t he flask and dilut ed t o a t ot al volum e of 100 m L t o m ake an 0.1 M solut ion? ( Hint : Which variable in equat ion ( 1) is not known?)

Exercises: 1. I f 50 m L of concent rat ed ( 18 M) sulfuric acid is dilut ed t o a t ot al volum e of 1.0 L, what is t he new concent rat ion?

2. I odine dissolves in various organic solvent s, such as dichlorom et hane ( CH2 Cl 2 ) , in which it form s an orange solut ion. What is t he m olarit y of I 2 when 5.00 g iodine is dissolved in enough dichlorom et hane t o m ake 50.0 m L of solut ion?

3. A pat ient ’s blood calcium level is 9.2 m g/ dL. What is t his concent rat ion in m olarit y? Make a unit plan first .

4. Read t he assigned pages in your t ext book, and work t he assigned problem s. CA24

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Ch e m Act ivit y 2 5

Hypotonic and Hypertonic Solutions ( I s it m ore concent rat ed or m ore dilut e?)

Information: Units of concentration The concent rat ion of a solut ion is a m easure of t he am ount of solut e dissolved in a specified am ount of solut ion. A solut ion t hat is m ore concent rat ed has m ore solut e per unit of volum e t han one t hat is m ore dilut e. We saw in Chem Act ivit y 24 t hat t he m ost com m on m easure of concent rat ion is m olarit y. Molarit y ( M)

– t he m oles of solut e per lit er of solut ion:

M

moles of solute L of solution

mol L

The ot her com m on m easures of concent rat ion are not done using m oles, but using eit her m ass ( weight ) or volum e. These are com m only report ed in percent . “ Percent by weight ” and “ percent by volum e” are com m on t erm s, and m ay be represent ed as % ( w: w) or % ( v: v) . Weight s ( or m asses) are in gram s, and volum es in m illilit ers. The first let t er in t he parent hesis represent s t he unit s of t he solu t e , and t he second is for t he solu t ion. Since percent m eans “ per hundred,” a 3.2% ( v: v) aqueous solut ion of alcohol would m ean 3.2 m L of alcohol are in every 100 m L of solut ion. We can writ e t his as a conversion fact or: 3.2 m L alcohol 100 m L and 100 m L 3.2 m L alcohol

Not e how we assum e—but didn’t act ually writ e—t he word “ solut ion” wit h t he “ 100 m L.”

Critical Thinking Questions: 1. Writ e t he t wo conversion fact ors for each of t he following concent rat ions. a. 10% ( v: v) acet one in wat er b. 0.90% ( m : v) NaCl( aq) c. 5.0% ( w: w) NaHCO3 ( aq) d. 5.0% ( w: v) NaHCO3 ( aq) 2. The solut ions in CTQ 1 ( c) and ( d) are oft en considered t o be t he sam e concent rat ion even t hough t he unit s are different . Explain how t his can be. ( Hint : What is t he densit y of wat er?)

3. Som et im es when concent rat ions are report ed for aqueous solut ions of solids, t hey do not say whet her t hey are by m ass or by volum e. Explain why it is probably okay t o assum e t hey are % ( m : v) .

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CA25

4. Percent m eans part s per hundred. We can also report concent rat ions in part s per t housand ( ppt ) , part s per m illion ( ppm ) , or part s per billion ( ppb) . The EPA safe lim it for lead in drinking wat er is 15 ppb. Writ e t he t wo conversion fact ors for t his concent rat ion.

Information: Hypotonic and hypertonic solutions Osm ot ic pressure ( S) is t he pressure t hat wat er exert s against a sem iperm eable m em brane, such as a cell m em brane. The high e r t he solu t e concent rat ion, t he low e r t he osm ot ic pressure. Red blood cells have an osm ot ic pressure equal t o t hat of a 0.90% ( m : v) solut ion of sodium chloride. I n t his solut ion, since t here are 2 m oles of ions per m ole of NaCl, t he concent rat ion of ion s is t wice as m uch, or 1.8% ( m : v) . A solut ion t hat has a higher ion concent rat ion t han t his is called h ype r t on ic; one wit h a lower ion concent rat ion is called h ypot on ic. I t is im port ant when giving fluids int ravenously t hat t hey be isot onic, t o avoid hem olysis ( burst ing) or crenat ion ( shriveling) of red blood cells.

Critical Thinking Questions: 5. Given t he definit ions of t he t erm s hypot onic and hypert onic, what do you suppose an isot onic solut ion would be?

6. I s a 1.0 M solut ion of NaCl hypot onic, hypert onic, or isot onic t o red blood cells? ( Hint : Convert t o % m : v—m ake a unit plan! )

7. An experim ent calls for 10.0 m L of a 4.00% aqueous solut ion of sodium t et raborat e ( “ borax” ) . Describe how you would m ake up such a solut ion so t hat you will not have any left over.

CA25

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Information: Elect rolyt es in blood are oft en m easured in equivalent s per lit er ( Eq/ L) . An equivalent of an ion is t he am ount of t hat ion t hat gives 1 m ole of posit ive or negat ive charge. This gives rise t o conversion fact ors such as: 1 m ol Na + = 1 Eq Na +

1 m ol Ca 2+ = 2 Eq Ca 2+

1 m ol SO24– = 2 Eq SO24–

Critical Thinking Questions: 8. A pat ient has a blood calcium level of 4.6 m Eq/ L. a. Writ e t he t wo conversion fact ors for convert ing m oles of calcium int o equivalent s of calcium .

b. Writ e a u n it pla n t o convert m Eq/ L int o m ol/ L.

c. What is t he pat ient ’s blood calcium level in m olarit y?

Exercises: 1. What is t he blood calcium level of t he pat ient in CTQ 8 in m g/ dL? ( Unit plan! )

2. I f t he lead level in som e drinking wat er is 15 ppb, how m any m L of t he wat er would a person have t o drink in order t o ingest one gram of lead? ( Unit plan! )

3. I s a 0.9 M CaCl 2 solut ion isot onic t o red blood cells? Explain why or why not , wit hout doing a calculat ion.

4. Read t he assigned pages in your t ext book, and work t he assigned problem s. - 105 -

CA25

Ch e m Act ivit y 2 6

Acids and Bases ( What happens when hydrogen ions are t ransferred bet ween species?)

Information: Two definitions of acids and bases Arrhenius definit ions An acid is a species t hat dissociat es int o H+ ( hydrogen) ions and anions when dissolved in wat er. A base is a species t hat dissociat es int o OH– ( hydroxide) ions and cat ions when dissolved in wat er. Brønst ed- Lowry definit ions An acid donat es a prot on ( H+ ion) t o anot her species. A base accept s a prot on ( H+ ion) from anot her species. Th e Br øn st e d- Low r y de fin it ion e x pla in s w h y t he h ydr oge n ion s ( H + ) in w a t e r a r e a ct ua lly h ydr oniu m ion s—a w a t e r m ole cu le ( H 2 O) ha s a cce pt e d t he pr ot on t o be com e h ydr oniu m ( H 3 O + ) . Ta ble 1 : Som e com m on a cids a n d ba se s Type of elect rolyt e St r on g ( all not list ed here are weak)

W eak

Acids

Bases

HCl

LiOH

HBr

NaOH

HI

KOH

H2 SO4

Ca( OH) 2

HNO3

Sr( OH) 2

HClO4

Ba( OH) 2

HC2 H3 O2

Mg( OH) 2

HCN Recall t hat when st rong acids and bases dissolve in wat er, t hey dissociat e com plet ely int o ions, while weak acids and bases dissociat e only slight ly.

Critical Thinking Questions: 1. Writ e t he t hree species t hat act ually exist in significant am ount s in a one- t ent h m olar aqueous solut ion of HCl.

2. Explain why hydrogen ion ( H+ ) is not one of t he t hree species in CTQ 1.

3. Writ e t he t hree species t hat act ually exist in significant am ount s in a one- t ent h m olar aqueous solut ion of LiOH.

CA26

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Information: Hydronium-hydroxide balance I n pure wat er, a sm all am ount of self ionizat ion occurs, wit h one wat er m olecule act ing as an acid ( donat ing a prot on) and anot her as a base ( accept ing a prot on) : H2 O( l) + H2 O( l) qwe H3 O+ ( aq) + OH– ( aq) I n pure wat er, t he concent rat ions of hydronium ion and hydroxide ion are each 1.0× 10 –7 M. Furt herm ore, t he product of t he concent rat ions of hydronium ion and hydroxide ion in aqueous solut ion at 25°C is always 1.0× 10 –14 M. [ H3 O+ ] [ OH– ] = 1.0× 10 –14 This m eans t hat if t he hydronium ion concent rat ion [ H3 O+ ] increases, t he hydroxide ion concent rat ion [ OH– ] decreases. This relat ionship allows us t o calculat e t he am ount s of hydronium ion and hydroxide ion in any solut ion of st rong acid or base.

Critical Thinking Questions: 4. What is t he hydronium ion concent rat ion in a 1.0× 10 –5 M aqueous solut ion of HCl?

5. What is t he hydroxide ion concent rat ion in an aqueous solut ion if t he hydronium ion concent rat ion is 1.0× 10 –5 M?

Information: pH pH ( t he “ power of hydrogen” ) is defined as t he negat ive of t he logarit hm of t he m olar concent rat ion of hydronium ions ( wit hout unit s) : pH = –log[ H3 O+ ] Therefore, in pure wat er, t he pH is –log( 1.0× 10 –7 ) , which equals 7. pH values below 7 are called acidic; t hose above 7 are t erm ed basic or alkaline. pH values can act ually go below 0 and above 14, t hough t his is not com m only seen. Ta ble 2 : Th e r e la t ionsh ip be t w e e n a cidit y, pH , a nd t h e h ydr on iu m a n d h ydr ox ide ion con ce nt r a t ion s of a solu t ion. Relat ive Concent rat ions

pH

Solut ion

[ H3 O+ ] > [ OH– ]

< 7

acidic

+

–

> 7

basic

+

–

= 7

neut ral

[ H3 O ] < [ OH ] [ H3 O ] = [ OH ]

Critical Thinking Questions: 6. What is t he pH of t he 1.0× 10 –5 M aqueous solut ion of HCl from CTQ 3? ( Be sure you can ent er t his int o your calculat or correct ly, e. g., 1 . 0 EE 5 ± log ±)

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CA26

7. According t o Table 2, which ion does t he solut ion in CTQ 5 cont ain m ore of: h ydr on iu m or h ydr ox ide ( circle one) ? 8. Does your answer t o CTQ 7 agree wit h your answers t o CTQs 4 and 5?

9. a. I f t he pH of a cola drink is 3.2, what is t he hydronium ion concent rat ion? Be sure you can ent er t his int o your calculat or correct ly, e. g., 3 . 2 ± 10x ( t he 10x key is oft en an inverse or 2nd log) .

b. What is t he hydroxide ion concent rat ion in t he cola?

10. a. What is t he hydroxide ion concent rat ion in a 1.0× 10 –5 M aqueous solut ion of N a OH ?

b. What is t he pH of t his solut ion? ( Careful! )

Model: Conjugate acid-base pairs According t o t he Brønst ed- Lowry t heory, a react ion of an acid and a base involves a prot on ( i. e., hydrogen ion) t ransfer from t he acid t o t he base. Two ions or m olecules t hat differ only by t hat one hydrogen ion m ake up a con j u ga t e a cid- ba se pa ir . Three exam ple pairs are shown below: H3 O+ and H2 O

H2 O and OH–

NH4 + and NH3

The con j u ga t e a cids have one m ore prot on ( H+ ) t han t he con j u ga t e ba se s. For exam ple, consider t he react ion below: conj ugat e acid- base pair

CH3 COOH + H2 O acet ic acid + wat er

qwe

CH3 COO– + H 3 O+ acet at e + hydronium ion

conj ugat e acid- base pair

Som et im es we sim plify t he nam ing by saying an acid and a base react t o give a conj ugat e acid and conj ugat e base. CH3 COOH + H2 O acid

qwe

base

CH3 COO– conj ugat e base

+

H 3 O+ conj ugat e acid

x

The conj ugat e base is what result s aft er t he acid gives up a hydrogen ion; so we say t hat acet at e is t he conj ugat e base of acet ic acid.

x

The conj ugat e acid is what result s aft er t he base picks up a hydrogen ion; so we say t hat hydronium ion is t he conj ugat e acid of wat er.

CA26

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Critical Thinking Questions: 11. For t he equat ions below, ident ify t he acid and base on t he react ant side and t he conj ugat e acid and conj ugat e base on t he product side. Draw a line t o connect conj ugat e acid- base pairs t oget her.

a. HCN( aq) + OH – ( aq) qwe H2 O( l) + CN– ( aq)

b. F– ( aq) + H3 O+ ( aq) qwe H2 O( aq) + HF( aq)

c. NH3 ( aq) + HCl( aq) qwe NH4 + ( aq) + Cl – ( aq)

Exercises: 1. Which definit ion of an acid—Arrhenius or Brønst ed—is m ore com plet e for aqueous solut ions? Explain.

2. Calculat e t he pH of each of t he following aqueous solut ions. a. 1.0× 10 –4 M nit ric acid b. 5.0× 10 –3 M hydrobrom ic acid c. a 1.0× 10 –6 M solut ion of t he diprot ic acid H2 SO4 ( diprot ic m eans t hat each m olecule of H2 SO4 donat es t wo hydrogen ions t o wat er m olecules) d. 0.0012 M Ca( OH) 2 3. Calculat e t he hydroxide ion concent rat ions of t he solut ions in Exercise 2.

4. Writ e form ulas for t he conj ugat e bases of t he acids in Exercises 2a and 2b.

5. Read t he assigned pages in your t ext book, and work t he assigned problem s. - 109 -

CA26

Ch e m Act ivit y 2 7

Buffers ( How do acids and bases react t oget her?)

Model 1: A buffer system Consider a solut ion cont aining bot h t he w e a k acid acet ic acid and it s con j u ga t e ba se , sodium acet at e. This set s up t he equilibrium shown below. ( The sodium ion is j ust a spect at or, and does not show up in t he equilibrium expression.) CH3 COOH

+

acet ic acid ( acid)

H2 O wat er ( base)

CH3 COO-

qwe

+

acet at e ( conj ugat e base)

H 3 O+ hydronium ion ( conj ugat e acid)

Critical Thinking Questions: 1. Suppose som e st rong base ( hydroxide ion) is added t o t he buffer syst em in Model 1. Circle t he t wo species in t he m odel t hat could react wit h t he hydroxide.

2. Writ e a chem ical equat ion for t he react ion of hydroxide ion being neut ralized by react ing wit h acet ic acid, producing acet at e and wat er.

3. Explain why addit ion of st rong base ( hydroxide ion) t o t he solut ion in Model 1 will not cause a great change in t he pH of t he solut ion, as long as t he acet ic acid is not used up.

4. Draw a box around t he one species in Model 1 t hat could react wit h and neut ralize any added hydronium ion. 5. Explain why addit ion of st rong acid will not cause t he pH of t he solut ion in Model 1 t o change m uch, as long as plent y of acet at e is present .

Model 2: A solution of a strong acid and its conjugate base Consider a solut ion cont aining bot h t he st r on g acid hydrochloric acid and it s conj uga t e ba se , sodium chloride. This set s up t he syst em shown below. HCl

+

hydrochloric acid ( acid)

CA27

H2 O wat er ( base)

ssd

Cl –

+

chloride ( conj ugat e base)

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H 3 O+ hydronium ion ( conj ugat e acid)

Critical Thinking Questions: 6. Recalling t hat st rong acids dissociat e com plet ely in wat er, draw a large ‘X’ t hrough t he species in Model 2 t hat is not present in any significant am ount . 7. Why is a forward arrow used in Model 2 ( ssd) inst ead of an equilibrium arrow ( qwe) ?

8. Explain t he follow ing st at em ent : There is no species in Model 2 t hat can neut ralize added hydronium ion.

Information: A summary A solut ion cont aining bot h a w e a k acid and it s con j u ga t e ba se is resist ant t o changes in pH when sm all am ount s of acid or base are added. This solut ion is called a bu ffe r . A solut ion of a st r on g acid and it s conj ugat e base is n ot a buffer, since any added st rong acid will not be neut ralized and will j ust increase t he H3 O+ concent rat ion. Sim ilarly, a solut ion of a st r on g ba se and it s conj ugat e acid is n ot a buffer.

Exercises: 1. St at e if each solut ions would be useful as a buffer or not . Then explain t he reason for your choice. a. A solut ion cont aining 0.08 M NaCN and 0.10 M HCN

b. A solut ion cont aining 0.05 M NaOH in H2 O

c. A solut ion cont aining 0.25 M HCl and 0.20 M NaCl

d. A solut ion cont aining 0.05 M NH4 Cl and 0.10 M NH3

e. A solut ion cont aining 0.20 M KF and 0.15 M HF

2. Considering t he solut ion in Exercise 1b. a. Writ e t he t hree species t hat would be present in significant am ount s.

b. I s t here any species present t hat can neut ralize added hydroxide? Explain.

3. Com plet e Chem Worksheet 3, St oichiom et ry Pract ice Worksheet 2. 4. Read t he assigned pages in your t ext book, and work t he assigned problem s.

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CA27

Ch e m Act ivit y 2 8

Alkanes, Cycloalkanes and Alkyl Halides ( What m akes a m olecule “ organic?” )

Information: Organic m olecules are based on carbon backbone st ruct ures. Of t hese, hydrocarbons cont ain only carbon and hydrogen. The hydrocarbons cont aining no m ult iple bonds are called a lk a n e s. The first t en “ st raight - chain” alkanes are shown in Table 1. Ta ble 1 . N a m e s a n d st r u ct u r e s of t h e fir st t e n a lk a ne s ( a n d a lk yl gr oups) Condensed St ruct ure

Nam e of alkane

“ St ick” St ruct ure

Condensed St ruct ure

Nam e of alkyl group subst it uent

CH4

m et hane

none

- CH3

m et hyl

CH3 CH3

et hane

- CH2 CH3

et hyl

CH3 CH2 CH3

propane

- CH2 CH2 CH3

propyl

CH3 CH2 CH2 CH3

but ane

- CH2 CH2 CH2 CH3

but yl

CH3 ( CH2 ) 3 CH3

pent ane

- CH2 ( CH2 ) 3 CH3

CH3 ( CH2 ) 4 CH3

hexane

- CH2 ( CH2 ) 4 CH3

CH3 ( CH2 ) 5 CH3

hept ane

- CH2 ( CH2 ) 5 CH3

CH3 ( CH2 ) 6 CH3

oct ane

- CH2 ( CH2 ) 6 CH3

CH3 ( CH2 ) 7 CH3

nonane

- CH2 ( CH2 ) 7 CH3

CH3 ( CH2 ) 8 CH3

decane

- CH2 ( CH2 ) 8 CH3

The st ruct ures shown in Table 1 are called conde n se d st r uct ur e s. They can be writ t en as a m olecular form ula by adding all t he carbons and hydrogens ( e. g., et hane = C2 H6 ) or expanded int o a com plet e Lewis st ruct ure (e. g., et hane, shown below) :

H

H

H

C

C

H

H

H

Critical Thinking Questions: 1. Draw a com plet e Lewis st ruct ure for propane.

2. Com plet e Table 1 by filling in t he m issing “ st ick” st r u ct ur e s and na m e s of t he alkyl group subst it uent s. CA28

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3. Using gram m at ically correct English sent ences, describe how one can derive t he nam e of an alkyl group subst it uent from t he nam e of t he corresponding alkane.

Information: Once organic m olecules get large, it is convenient t o writ e t hem as “ st ick st r u ct u r e s.” Each “ end” and “ bend” is a carbon, and t he hydrogens are not shown. St ick st ruct ures are com m only used for cycloa lk a ne s—alkanes in which t he carbons are connect ed t o m ake a ring ( “ head” t o “ t ail.” ) Som e com m on cycloalkanes are shown in Table 2. Ta ble 2 . N a m e s a n d st r u ct u r e s of som e com m on cycloa lk a ne s Condensed St ruct ure

Nam e of alkane

“ St ick” St ruct ure

Ball- and- St ick Model

CH2 cyclopropane

H2C

CH2

H2C

CH2

H2C

CH2

cyclobut ane

CH2 CH2

CH2

cyclopent ane

CH2 CH2 CH2 H2C

CH2

H2C

CH2

cyclohexane

CH2

Critical Thinking Question: 4. Draw condensed and “ st ick” st ruct ures for cyclohept ane.

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CA28

Information: Rules for naming branched alkanes Alkanes are br a nche d when t he carbons are not connect ed in a cont inuous m anner. For exam ple, if you cannot t race all t he carbons wit h your pencil ( or finger) from one carbon t o t he next wit hout lift ing your pencil or ret racing a port ion, t here is a branch. The branches are called a lk yl gr ou ps. Any such side ch a in , even if it is not an alkyl group, is called a su bst it ue nt . The nam e of t he m olecule is t hen det erm ined as follows: 1. Find t he longest cont inuous chain of carbons t o get t he ba se na m e . 2. Nam e each su bst it ue nt alphabet ically. I f t here are t wo equivalent subst it uent s, use t he prefixes di- , t ri- , t et ra, et c. in front of t hem . 3. Ment ally num ber each carbon of t he base chain st art ing at t he end closest t o t he branch. Finally, give e a ch su bst it u e n t a n um be r t hat corresponds t o t he num ber of t he carbon of t he base chain t o which it is connect ed. For exam ple, consider t he t wo m olecules in Figure 1: Figur e 1 : Ex a m ple of n a m in g t w o diffe r e nt five - ca r bon a lk a n e s 4

3

2

1

CH3 CH2 CH CH3

CH3 CH2 CH2 CH2 CH3 pentane

2-methylbutane

CH3

The first m olecule in Figure 1 is pe nt a ne . The second has only four cont inuous carbons, and so t he base nam e is but a ne . The carbons of t he but ane are num bered st art ing from t he end closest t o t he - CH3 group ( called a m e t h yl gr oup, see Table 1 above) . This num bering is shown above t he m olecule in t he Figure. Then t he num ber “ 2” is assigned t o t he m et hyl group. So t he nam e is 2 - m e t h ylbu t a ne . Not e t hat t here is no space bet ween “ m et hyl” and “ but ane,” and t here is a hyphen bet ween t he num ber and let t er. See your t ext book for m ore exam ples. Not e som et hing else about t he t wo m olecules in Figure 1: t hey have t he sam e m olecular form ula ( C5 H12 ) . Molecules t hat have t he sam e m olecular form ula but are arranged different ly are called isom e r s. So, pent ane and 2- m et hylbut ane are isom ers. We will learn m ore about isom ers in Chem Act ivit ies 29- 31. Som e com m on subst it uent nam es are shown in Table 3. I f t he subst it uent is a halogen, t hen t he m olecule is called a h a loa lk a n e or a lk yl ha lide . Ta ble 3 . Com m on subst it u e n t n a m e s in a lk a n e s a n d h a loa lk a n e s. “R” in dica t e s w h e r e t he su bst it ue nt is a t t a che d t o t h e “Re st ” of t he m ole cu le ( t h e m a in ch a in) . Subst it uent

Nam e

St ick st ruct ure

Subst it uent

Nam e

–F

fluoro

–Cl

chloro

–Br

brom o

–I

iodo

CHCH3 isopropyl

CH3 CH2 CH CH3 CH3 CHCH2CH3

CH3 CH3 C CH3 CH3

CA28

isobut yl sec- but yl

R

R

R

t ert - but yl R

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Critical Thinking Question: 5. Com plet e t he following t able: Nam e

Lewis st ruct ure

condensed form ula

st ick st ruct ure

2- chloropropane

CH2 CH2 CH3

CH3

CH3 CH2 CH CH CH CH3 CH2CH3

isopropylcyclohexane

Exercises: 1. Draw ( any st yle) and nam e all 5 of t he different ly branched isom ers of C6 H14 .

2. Com plet e Chem Worksheet 4: Funct ional Groups. 3. Read t he assigned pages in your t ext book and work t he assigned problem s. - 115 -

CA28

Ch e m Act ivit y 2 9

Conformers ( How and why do m olecules “ t wist ?” )

Model 1: Representations of ethane in its most favorable conformation Ball- and- st ick m odel

Spacefilling m odel

Lewis dash- andwedge st ruct ure H H H C

H H

H

C

H H

Newm an proj ect ion

H

H

H H

Critical Thinking Questions: 1. Each represent at ion in Figure 1 shows et hane ( CH3 CH3 ) in it s lowest pot ent ial energy ( m ost favorable) conform at ion. I f you have a m odel set available, m ake a m odel of et hane and rot at e t he single bonds unt il it is in t his conform at ion. a. Const ruct an explanat ion for why t his conform at ion is t he m ost favorable.

b. Consider t he Newm an Proj ect ion in Figure 1. What at om is at t he cent er of t he pict ure?

c. The at om you nam ed in CTQ 1b is represent ed as a large circle or disc. What single at om is hidden from view behind t he disc?

2. Consider t he Newm an proj ect ion shown below of et hane in ( nearly) it s least favorable conform at ion. I f you have a m odel set available, rot at e t he single bonds unt il it is in t his conform at ion. Draw a dash- and- wedge st ruct ure for t his conform at ion. HH

H

H

H

H

3. I n your own words, explain what t he t erm conform at ion m eans, as applied t o et hane.

CA29

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4. St ruct ures t hat represent t he sam e m olecule in different conform at ions are t erm ed conform ers. Const ruct an explanat ion for why t he conform er in Model 1 is called st a gge r e d and t he conform er in CTQ 2 is called e clipse d.

5. Look at your m odel “ end- on” and com pare wit h t he Newm an proj ect ions below. The angle you observe bet ween t he hydrogens at t he t op of each st ruct ure in boldfa ce is called t he t orsional angle. Circle t he correct t orsional angle for each conform er. H H

H

HH

Torsional angle ( st aggered)

Torsional angle ( eclipsed)

0° H H st aggered

H

0° H

60°

H

H

60°

H

120°

120° eclipsed

180°

180°

6. What repulsive forces will cause et hane in t he eclipsed conform at ion t o quickly adopt t he st aggered conform at ion?

Model 2: Some representations of alkanes with 2, 3 and 4 carbons Skelet al or “ St ick” st ruct ure

Ball- and- st ick st ruct ure

Lewis st ruct ure

H H

H

H

H

Dash- and- wedge st ruct ure

H

C

C

H

H

H

H

H

H

H

H

C

C

C

H

H

H

H

H

H

HH C

H

H

H

C

C

HH

H

H

H

H

C

C

C

C

H

H

H

H

H

H

H

H

C

C

C

C

H

H

H

H

HH

H H

H

H

H H C

C C

C

HH

H

HH

HH

H

H

C

H C

C

HH

C H

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H

C

C

H

H H

CA29

Critical Thinking Questions: 7. Underneat h each st ick st ruct ure in Model 2, writ e t he nam e of t he alkane. 8. The ball- and- st ick st ruct ure below m at ches one of t he t wo but ane conform at ions shown in Model 2, looking “ end- on” down t he bond bet ween carbons 2 and 3. Circle t he st ick st ruct ure in Model 2 t hat m at ches t he st ruct ure below.

9. Com plet e t he Newm an proj ect ion at t he right in CTQ 7 by adding H or CH3 groups so t hat it represent s t he conform at ion shown in t he ball- and- st ick st ruct ure at t he left . 10. Consider t he ot her st ick st ruct ure for but ane in Model 2 ( t he one you did not circle in CTQ 7) , sight ing down t he C2–C3 bond. a. I s t his st ruct ure st a gge r e d or e clipse d ( circle one) ? You m ay use your m odel t o help you. b. Draw a Newm an proj ect ion for t his conform at ion of but ane.

c. At room t em perat ure, t he single bonds in but ane are cont inually rot at ing t hrough t he st aggered and eclipsed conform at ions. However, t he m olecule spends m ore t im e closer t o one of t he t wo ext rem es. Which conform at ion is it likely t o spend m ore t im e in— st a gge r e d or e clipse d ( circle one) ?

Model 3: Some conformations of pentane

Critical Thinking Questions: 11. Circle t he t wo st ruct ures of pent ane in Model 3 t hat are in t heir m ost favorable conform at ion. Explain why t hese are t he m ost favorable.

12. Draw a line t o connect t he t wo st ruct ures t hat you circled in Model 3. These t wo st ruct ures are in t he sam e conform at ion, but t he m olecule as a whole is rot at ed.

CA29

- 118 -

13. Find t wo ot her st ruct ures in Model 3 t hat are ident ical ( i. e., in t he sam e conform at ion) , and draw a second line t o connect t hem . 14. What is t he t ot al num ber of dist inct conform ers of pent ane shown in Model 3?

Exercises: 1. Draw a wedge and dash- bond represent at ion of pent ane in it s m ost favorable conform at ion.

2. Consider t he m olecule 2- m et hylbut ane. a. Using t he t em plat es at t he right , com plet e t wo st aggered Newm an proj ect ions for 2m et hylbut ane: one sight ing down t he C1–C2 bond and t he second sight ing down t he C2–C3 bond.

4

H 3C

H2 C

2

3

CH

CH3 1

CH3

b. Com plet e t he eclipsed Newm an proj ect ion for 2- m et hylbut ane sight ing down t he C2– C3 bond.

3. Sugars and ot her com plex m olecules are oft en depict ed using a represent at ion called a Fisher proj ect ion. I n a Fisher proj ect ion all horizont al bonds are assum ed t o com e out of t he page t oward you ( wedge bonds) and all vert ical bonds are assum ed t o go back int o t he page, away from you ( dash bonds) . Draw a wedge and dash represent at ion of t he Fisher proj ect ion of glyceraldehyde shown below. O

H C

HO

H CH2OH

4. Read t he assigned sect ions in your t ext and work t he assigned problem s. - 119 -

CA29

Ch e m Act ivit y 3 0

Constitutional and Geometric Isomers ( Are t hey ident ical, or are t hey isom ers?)

Model 1: Representations of some organic molecules Skelet al or “ St ick” st ruct ure

Lewis st ruct ure

H

H

H

C

C

H

H

H

H

H C

H

Ball- and- st ick st ruct ure

H

C

C

H

H

H

H H

H

H

C C

C

H

H

H

Critical Thinking Questions: 1. Consider t he Lewis st ruct ures in Model 1. How m any covalent bonds does each carbon have? 2. I n skelet al represent at ions, t he hydrogens are not shown. Explain how it is st ill possible t o t ell how m any hydrogens t here are on each carbon.

3. Draw a Lewis st ruct ure represent at ion of t he m olecule for which a skelet al represent at ion is shown below.

CA30

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Model 2: Constitutional Isomers

Column 1 structure

Column 2 molecular formula

structure

molecular formula

C5H12

Critical Thinking Questions: 4. Com plet e Model 2 by writ ing in t he m issing m olecular form ulas in bot h colum ns. 5. What do t he m olecules in a given colum n ( 1 or 2 in Model 2) have in com m on wit h t he ot her m olecules in t hat colum n?

6. What do t he m olecules in a given colum n not have in com m on wit h t he ot her m olecules in t hat colum n?

7. All t he st ruct ures in a given colum n are const it ut ional isom ers of one anot her, but t he st ruct ures in Colum n 1 are not const it ut ional isom ers of st ruct ures in Colum n 2. Based on t his inform at ion, writ e a definit ion for t he t erm const it ut ional isom ers.

8. I f t he m olecule shown below were placed int o Model 2, would it belong in Colu m n 1 or Colum n 2 ( circle one) ? Explain your choice.

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CA30

Model 3: Representations of methylcyclobutane

Skelet al ( “ st ick” )

Wedge and dash

H

H

H

H

H

CH3

H

H

CH3

H

Wedge and dash wit h explicit hydrogens

Perspect ive

Ball- and- st ick

Model 4: 1,2-dimethylcyclobutane, shown with ring carbons numbered 1–4 ( b)

( a) 4

1

CH3

4

1

CH3

3

2

CH3

3

2

CH3

Critical Thinking Questions: 9. Are t he m olecules in boxes ( a) and ( b) of Model 4 const it ut ional isom ers of each ot her? Explain.

10. Ot her t han bonds t o carbons wit hin t he ring, what t wo groups are bonded t o t he following carbons? a. carbon 1 in box ( a) ? b. carbon 1 in box ( b) ? c. carbon 2 in box ( a) ? d. carbon 2 in box ( b) ? 11. I f you have access t o a m odel kit , m ake m odels of t he t wo m olecules in Model 4 ( C = black; H = whit e; use t he short bonds for single bonds) . I s it possible t o rot at e single bonds in t he m odels such t hat t he m olecule in box ( a) is t he sam e as t he one box ( b) ?

Information: Since each carbon in t he m olecule in box ( a) in Model 4 is bonded t o t he sam e four groups as t he corresponding carbon in t he m olecule in box ( b) , t he m olecules are said t o have t he sam e con ne ct ivit y. You confirm ed in CTQ 11 t hat t he t wo st ruct ures of 1,2- dim et hylcyclobut ane shown above are not sim ply con for m e r s of each ot her. I m agine t hat t he four carbons of t he cyclobut ane ring define a plane. I n one st ruct ure, t he t wo m et hyl groups are on t he sam e side of t his plane, and in t he ot her t hey are on opposit e sides of t he plane. The single bonds in t he ring cannot rot at e wit hout breaking t he ring. Two groups on t he sam e side of t he plane are considered t o be cis t o one anot her. Groups on opposit e sides are called t rans. Ge om e t r ic isom e r s ( cis- t rans isom ers) are m olecules t hat have t he sam e connect ivit y and differ only in t he geom et ric arrangem ent of groups. CA30

- 122 -

Critical Thinking Questions: 12. Label one box in Model 4 wit h t he nam e “ cis- 1,2- dim et hylcyclobut ane” and t he ot her wit h t he nam e “ t rans- 1,2- dim et hylcyclobut ane.” Then add perspect ive represent at ions int o each box. 13. Draw wedge- and- dash and perspect ive represent at ions of cis- and t rans- 1,3dim et hylcyclobut ane. ( Not e: t hat is “ 1,3- dim et hyl,” not “ 1,2- dim et hyl.” )

Exercises: 1. I ndicat e if t he following pairs of st ruct ures are ident ical, conform ers, geom et ric isom ers, const it ut ional isom ers, or not isom ers. a.

1,2- dim et hylcyclobut ane and 1,3- dim et hylcyclobut ane

b.

and

c.

and

d.

and

e.

and

f.

and

g.

and

h.

and

i.

and

j.

and

k.

and

2. Draw a st ruct ure for a m olecule not shown in t his act ivit y t hat would belong in Colum n 2 of Model 2.

3. Read t he assigned pages in your t ext book and work t he assigned problem s. - 123 -

CA30

Ch e m Act ivit y 3 1

Isomers ( What are som e different t ypes of isom ers?)

Information: Review of some types of isomers, from least to most similar 1. Con st it u t iona l isom e r s: m olecules wit h t he sam e m olecular form ula but different st ruct ures ( different connect ivit y) . 2. Ge om e t r ic isom e r s ( cis- t rans isom ers) : m olecules t hat have t he sam e connect ivit y and differ only in t he geom et ric arrangem ent of groups. 3. Confor m a t iona l isom e r s ( conform ers) : m olecules t hat can be int erconvert ed by rot at ion around single bonds.

Model 1: Alkenes As we saw in Chem Act ivit y 21, t here is no free rot at ion around double bonds at room t em perat ure. This m eans t he t wo m olecules below are not t he sam e, as t hey cannot be int erconvert ed via rot at ion of single bonds. H

CH3 C

H 3C

H 3C

C

CH3 C

H

H

C H

Critical Thinking Questions: 1. Recall t he m eanings of t he t erm s cis and t rans as applied t o cycloalkanes. By analogy, label one of t he m olecules in Model 1 “ cis- 2- but ene” and t he ot her “ t rans- 2but ene.” 2. According t o t he definit ions in t he I nform at ion, what is t he relat ionship bet ween cis2- but ene and t rans- 2- but ene?

3. Draw skelet al ( “ st ick” ) represent at ions of cis and t rans- 2- but ene.

4. Draw skelet al ( “ st ick” ) represent at ions of cis and t rans- 3- hexene.

CA31

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5. I dent ify t he t ype of isom eric relat ionship bet ween each pair of m olecules below, from t he following five choices ( arranged from least t o m ost sim ilar) : not isom ers, const it ut ional isom ers, geom et ric isom ers, conform ers, ident ical. a. CH3

Cl and H 3C

Cl

b.

and

c. Cl

Cl

H

Cl

Cl

H

and H

H

d. H

HO H3C

C C

CH2

C

H

H C

and

C

H3C

C

CH2

HO

H

e. H

CH3

H

H

CH3

H3 C

H3 C

H

H

H

and CH3

CH3

f. CH3

CH2 CH3

C

C

H

and

H

g. and

h. H

CH3 H3 C

CH3

H

H

H

and CH3

H

H

CH3

H

i. H3C C H

H

CH2 CH3

C

and

C H

CH3

CH3CH2

C H

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CA31

Exercises: 1. Redraw t he st ruct ures from CTQ 5b as condensed st ruct ures.

2. Redraw t he st ruct ures from CTQ 5e as condensed st ruct ures.

3. Redraw t he st ruct ures from CTQ 5g as condensed st ruct ures.

4. Redraw t he st ruct ures from CTQ 5h as condensed st ruct ures.

5. Redraw t he st ruct ures from CTQ 5d as skelet al ( “ st ick” ) st ruct ures.

6. Nam e t he m olecules in CTQ 5abceh. a. b. c. e. h. CA31

- 126 -

7. Draw a represent at ion ( any st yle) of 1- but ene. Can t his m olecule have cis and t rans isom ers? I f not , explain. I f so, draw t hem .

8. Read t he assigned pages in your t ext book and work t he assigned problem s.

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CA31

Ch e m Act ivit y 3 2

Properties of Organic Molecules ( Which organic m olecules are soluble in wat er?)

Model: Polarity and properties of organic molecules Bond polarit ies are det erm ined by t he elect ronegat ivit y difference bet ween t he bonded at om s. Recall t hat C- H bonds are nonpolar, but C- N and C- O bonds are polar. Since wat er is a polar solvent , t he m ore polar an organic m olecule, t he m ore soluble it will be in wat er. Ta ble 1 : Pola r it y, solu bilit y a nd boiling poin t s of se le ct e d com poun ds* Alkane

St ruct ure

Molar m ass, g/ m ol

Dipole m om ent , Debyes

Boiling point , °C

Wat er solubilit y, gram s per 100 m L H2 O

propane

CH3 CH2 CH3

44

0

- 42

0.007

but ane

CH3 CH2 CH2 CH3

58

0

0

0.006

pent ane

CH3 CH2 CH2 CH2 CH3

72

0

36

0.04

hexane

CH3 ( CH2 ) 4 CH3

86

0

69

0.001

hept ane

CH3 ( CH2 ) 5 CH3

100

0

98

0.01

Alcohol

St ruct ure

Molar m ass, g/ m ol

Dipole m om ent , Debyes

Boiling point , °C

Wat er solubilit y, gram s per 100 m L H2 O

et hanol

CH3 CH2 OH

46

1.7

78

f

1- propanol

CH3 CH2 CH2 OH

60

1.7

82

f

1- but anol

CH3 CH2 CH2 CH2 OH

74

1.67

118

6.3

1- pent anol

CH3 ( CH2 ) 4 OH

88

1.7

137

2.7

1- hexanol

CH3 ( CH2 ) 5 OH

102

1.8

157

0.6

1- hept anol

CH3 ( CH2 ) 6 OH

116

1.7

176

0.1

Et her

St ruct ure

Molar m ass, g/ m ol

Dipole m om ent , Debyes

Boiling point , °C

Wat er solubilit y, gram s per 100 m L H2 O

dim et hyl et her

CH3 OCH3

46

1.3

- 23

f

diet hyl et her

CH3 CH2 OCH2 CH3

74

1.15

35

6.9

dipropyl et her

CH3 ( CH2 ) 2 O( CH2 ) 2 CH3

102

1.2

89

0.25

Am ine

St ruct ure

propyl am ine

CH3 CH2 CH2 NH2

Molar m ass, g/ m ol 59

Dipole m om ent , Debyes 1.3

Boiling point , °C 48

Wat er solubilit y, gram s per 100 m L H2 O f

but yl am ine

CH3 CH2 CH2 CH2 NH2

73

1.3

77

f

hexyl am ine

CH3 ( CH2 ) 5 NH2

101

–

131

1.2

oct yl am ine

CH3 ( CH2 ) 7 NH2

129

–

180

0.02

*

Sources: CRC Handbook of Chem ist ry and Physics, 47 t h ed., 1967; Chem Finder .com ; I PCS I NCHEM, www.inchem .org; Korea t her m ophysical propert ies Dat a Bank, www .cher ic.org/ kdb/ [ accessed June 2006]

CA32

- 128 -

Not es: The dipole m om ent is a m easure of t he polarit y of a m olecule; f m eans infinit ely soluble ( i. e., t he liquids are m iscible) ; a dash m eans t he dat a were unavailable.

Critical Thinking Questions: 1. Considering t he dipole m om ent s of t he m olecules in Table 1, which funct ional group is m ost non pola r ? 2. I n general, do polar com pounds have h igh e r or low e r boiling point s t han nonpolar com pounds ( circle one) ? 3. For each funct ional group list ed below, indicat e how t he boiling point changes as t he m olar m ass increases. a. alkane b. alcohol c. am ine 4. The boiling point of a liquid increases as t he at t ract ions bet ween m olecules increase. These at t ract ions are called int erm olecular forces. Based on your answers t o CTQ 3, how do t he int erm olecular forces bet ween m olecules change as t he m olar m ass increases?

5. Find one m olecule from each funct ional group ( alkane, alcohol, et her, am ine) wit h roughly t he sam e m olar m ass ( wit hin 5 g/ m ol) , and writ e t heir nam es below. Rank t hese com pounds from t he highest t o lowest boiling point .

6. Repeat CTQ 5 wit h anot her set of four com pounds.

7. Based on relat ive boiling point s, writ e t he num bers from 1 t o 4 under t he funct ional group nam es below, wit h t he num ber 1 being t he group wit h t he m ost int erm olecular at t ract ions, and 5 being t he least . alkane

alcohol

et her

- 129 -

am ine

CA32

8. For each funct ional group below, circle each t ype of bond cont ained in t he m olecule. You m ay refer t o Table 1. funct ional group

Type of bond

alkane

C- H

C- O

O- H

N- H

alcohol

C- H

C- O

O- H

N- H

et her

C- H

C- O

O- H

N- H

am ine

C- H

C- O

O- H

N- H

9. Based on your answer t o CTQ 8, which t wo t ypes of bonds are present in t he m olecules wit h t he st rongest int erm olecular at t ract ions?

Information: The t ypes of bonds you ident ified in CTQ 8 can exhibit what is known as hydr oge n bonding. A “ hydrogen bond” is sim ply a part icularly st rong at t ract ion bet ween t he bonded hydrogen at om and a lone pair on anot her at om . This at t ract ion causes m olecules t o st ick t oget her, but is m uch weaker t han a covalent bond ( up t o one- t ent h as st rong) , and so can be broken and reform ed cont inually at room t em perat ure. Wat er has O- H bonds, and t he O has t wo lone pairs, m eaning t hat wat er m eet s t he requirem ent s for hydrogen bonding. Therefore, wat er is part icularly suit able for dissolving organic m olecules t hat can exhibit hydrogen bonding, since t he wat er and organic m olecule can “ hydrogen- bond” t oget her. Consider t he m olecule et hanol, CH3 CH2 OH. I t has one polar funct ional group ( t he –OH) and a t wo- carbon nonpolar alkyl group ( CH3 CH2 - ) . Since t he nonpolar group is not at t ract ed t o t he wat er, it is reasonable t o say t hat t he –OH is t he reason t hat et hanol is m iscible w it h wat er.

Critical Thinking Question: 10. Suppose t hat we consider anyt hing over about 1 gram per 100 m L wat er t o be “ soluble.” Then, considering t he wat er solubilit ies of t he alcohols, et hers, and am ines in Table 1, t he presence of one polar funct ional group is sufficient t o dissolve a m olecule cont aining about how m any nonpolar carbons? Circle one of t he following choices: 1- 2

3- 4

5- 6

7- 8

Ta ble 2 : W a t e r solubilit y of se le ct e d k e t one s

Ket one

St ruct ure

Wat er solubilit y, gram s per 100 m L H2 O

acet one

O H3C C CH3

f

2hexanone

O H3C C (CH2)3CH3

1.4

2but anone

O H3C C CH2CH3

25.6

2hept anone

O H3C C (CH2)4CH3

0.4

2pent anone

O H3C C CH2CH2CH3

4.3

CA32

- 130 -

Ket one

St ruct ure

Wat er solubilit y, gram s per 100 m L H2 O

Exercises: 1. Consider t he wat er solubilit ies of t he ket ones shown in Table 2. Are t hey consist ent wit h your answer t o CTQ 10? Writ e a sent ence t hat generalizes how m any nonpolar carbons can be dissolved by one polar funct ional group.

2. A hydrogen bond is usually depict ed by a dashed or dot t ed line. Circle t he pict ure t hat correct ly represent s a hydrogen bonding int eract ion.

H

O H C H H

H O H C H H

H

O H C H H

H O H C H H

H O H C H H

H O H C H H

H O H C H H

H

O H C H H

3. Describe what is wrong wit h each of t he ot her t hree pict ures in Exercise 1.

4. Draw a represent at ion of one wat er m olecule part icipat ing in a hydrogen bond wit h anot her wat er m olecule. Then place t he sym bols G+ and G– near each at om t o indicat e t he polarit y of t he bonds.

5. Wit hout looking up any inform at ion in a t able, ident ify t he m olecule in each group t hat would have t he highest boiling point , and explain your answer. a. CH3 CH2 OH O H C C CH3 3 b. c.

OH H3C CHCH3

d. CH3 CH2 NHCH2 CH3

e.

OCH3

CH3 CH2 CH2 OH

HOCH2 CH2 OH

CH3 CH2 OH OH H3C CHCH2CH2CH3

CH3 CH2 NH2 OH CH3CH2 CHCH3

CH3 CH2 CH2 CH2 CH3 CH3 CH3 O

OH

6. Benzoat e ion is very soluble in wat er, but benzoic acid is not . Based on t his inform at ion which species do you t hink has a larger dipole m om ent —benzoic acid or benzoat e? Explain.

O C

OH

benzoic acid

O C

O

benzoat e

7. Read t he assigned pages in your t ext , and work t he assigned problem s. - 131 -

CA32

Ch e m Act ivit y 3 3

Reactions of Organic Molecules ( What are t he m ain t ypes of organic react ions in biological syst em s?)

Information: Organic reactions common in biochemistry There are seven com m on react ion classes of organic m olecules t hat we will consider. The first class—acid- base react ions—is one we have seen before. Then t here are t hree pairs of react ions t hat are opposit es of each ot her—addit ion and elim inat ion; reduct ion and oxidat ion; condensat ion and hydrolysis. Ta ble 1 : Se ve n com m on t ype s of or ga n ic r e a ct ion s 1. a cid- ba se : t ransfer of a prot on ( H+ ) OH

x exam ple:

O

+

+

H2O

H3O

2. a ddit ion : addit ion of a sm all m olecule ( usually H2 O) across a double bond x exam ple ( hydrat ion) :

CH3 CH

H+ catalyst

+

CH2

H2O

CH3 CH OH

CH2 H

3. e lim ina t ion : rem oval of a sm all m olecule ( oft en H2 O) t o form a double bond x exam ple ( dehydrat ion) :

CH3 CH OH

conc. acid

CH2

CH3 CH

and heat

CH2

+

H2O

H

4. r e du ct ion : addit ion of 2 H at om s t o or rem oval of an O at om from a m olecule. Can be sym bolized “ [ r] ” x exam ple:

Pt catalyst

CH3 CH

CH2

+

H2

CH3 CH2 CH3

5. ox ida t ion: addit ion of an O at om t o or rem oval of 2 H at om s from a m olecule. Can be sym bolized “ [ o] ” x exam ple wit h prim ary alcohol: form s an aldehyde, t he n an acid

CH3

OH C H

[o]

H ethanol

O CH3 CH

+

[o] H2O

acetaldehyde (ethanal)

x exam ple wit h secondary alcohol: form s an ket one

OH CH3 C H CH3 2-propanol

CA33

[o]

O CH3 C CH3 2-propanone (acetone)

- 132 -

O CH3 C OH acetic acid (ethanoic acid)

x exam ple wit h t ert iary alcohol: no react ion

OH

[o]

CH3 C CH3

NR (no H's on alcohol carbon to be removed)

CH3 tert-butyl alcohol x exam ple wit h t hiol: form s a disulfide

[o] 2 CH3-SH

CH3-S-S-CH3 (+ H2O)

methanethiol

dimethyl disulfide

6. con de n sa t ion: coupling of t wo m olecules wit h t he loss of a sm all m olecule ( usually H2 O)

O x exam ple ( est erificat ion) :

CH3 C OH

+

acetic acid (an acid)

O

H+ catalyst

H-O-CH3

CH3 C OCH3

methanol (an alcohol)

methyl acetate (an ester)

+

H2O

7. h ydr olysis: split t ing a m olecule in t wo wit h t he addit ion of wat er

O x exam ple ( est er hydrolysis) :

+

CH3 C OCH3 methyl acetate

acid or base catalyst

H2O

O

+

CH3 C OH H-O-CH3 new water added

Just as it is beneficial t o m ake a sum m ary of t he t ypes of react ions, it is also useful t o m ake a list of t he funct ional groups t hat com m only react by each of t he seven pat hways. Such a sum m ary is shown in Table 2. Ta ble 2 : Fu n ct iona l gr oups t ha t com m on ly r e a ct by t h e se ve n r e a ct ion pa t hw a ys acid- base

– carboxylic acids, phenols ( acids) – am ines ( bases)

addit ion

– alkenes, alkynes

elim inat ion

– alcohols

reduct ion

– alkenes, ket ones

addit ion

– carboxylic acids, phenols ( acids)

oxidat ion

– aldehydes, prim ary and secondary alcohols

condensat ion

– acid + alcohol, acid + am ine

hydrolysis

– est ers, am ides ( pept ides)

Critical Thinking Questions: 1. For t he acid- base react ion in Table 1, ident ify t he acid and base on t he react ant side and t he conj ugat e acid and conj ugat e base on t he product side. Draw a line t o connect conj ugat e acid- base pairs t oget her. 2. Tell which of t he seven com m on t ypes of organic react ions is illust rat ed by each of t he following.

a.

O CH3CH2 C CH2CH3

+

H2

Pt

- 133 -

OH CH3CH2 CHCH2CH3

CA33

O C NHCH2CH3

b.

CH3CH2

c.

OH CH3CH2 CHCH2CH3

+

H 2O

H 3 O+

conc. H3O+

CH3CH2

O C OH

+

CH3CH2NH2

CH3CH CHCH2CH3

d. CH3 OH2 + + CH3 NH2 ssd CH3 NH3 + + CH3 OH O OH KMnO4

OH

e.

f.

O CH3CH2 C OH

+

H2O

+

CH3CH2OH

H3O+

H 3O+

O CH3CH2 C OCH2CH3

+

H 2O

OH

g. 3. Classify each of t he m olecules below as a prim ary, secondary, or t ert iary alcohol. OH

a. OH

b. c.

CH3CH2OH

CH3 HO

C

CH2CH2CH3

CH3

d.

4. For each of t he alcohols in Exercise 3, show t he oxidat ion product s, if any. I f t he oxidat ion t akes place in t wo st eps, show t he product of each st ep.

Exercises: 1. Each of t he following carboxylic acids can donat e a hydrogen at om t o wat er. Com plet e a chem ical react ion for each. O C

OH +

H2O

a. b. acet ic acid O C

c.

H

OH

2. Read t he assigned pages in your t ext book and work t he assigned problem s. CA33

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Ch e m Act ivit y 3 4

Carbohydrates ( What m akes a sugar?)

Model 1: Glyceraldehyde Sim ple sugars have a m olecular form ula of Cn ( H2 O) n , m eaning t hat for n carbon at om s in t he sugar, t here are n oxygen at om s and 2n hydrogen at om s. The sim plest sugar ( a m onosaccharide) is glyceraldehyde, wit h t he m olecular form ula C3 H6 O3 and t he st ruct ure shown below. O OH H

C

CH CH2OH

Critical Thinking Questions: 1. Circle t he aldehyde funct ional group in t he st ruct ure of glyceraldehyde above. 2. What is t he value of n for glyceraldehyde in t he form ula Cn ( H2 O) n ? 3. Hypot hesize on t he origin of t he t erm carbohydrat e.

4. Using a m olecular m odel kit , m ake a m olecule of glyceraldehyde. ( I n m any m odel kit s, black = C, red = O, whit e = H. Use short bonds for single bonds and t wo of t he longer, flexible bonds for double bonds.) Hold your m odel wit h t he aldehyde carbon at t he t op and t he CH2 OH at t he bot t om , and arrange it so t hat t he m iddle carbon is closer t o you t han t he ot her t wo carbons. Now, com pare your m odel wit h t he t wo st ruct ures below. Circle t he one t hat m at ches your m odel. ( Rem em ber what t he dashes and wedges m ean?) O

H

O

C H

C

H C

OH

HO

CH2OH

C

H

CH2OH

5. Now m ake a second m odel of glyceraldehyde, so t hat you have one of each of t he t wo st ruct ures shown above. Can you rot at e or t wist t hem so t hat all t he at om s are in t he sam e place in bot h m olecules? _____

Model 2: Types of stereoisomers St ereoisom ers are m olecules wit h t he sam e connect ivit y, but different arrangem ent s in space. This cat egory includes geom et ric ( cis- t rans) isom ers and a t ype of isom ers we have not seen before called enant iom ers. Enant iom ers are a pair of non- ident ical m olecules t hat are m irror im ages of each ot her. Whenever a carbon is bonded t o fou r dissim ila r gr oups, it is said t o be chiral. An obj ect or m olecule t hat is chiral is not ident ical t o it s m irror im age. The t erm chiral lit erally m eans “ handedness.”

Critical Thinking Question: 6. Confirm t hat t he cent er carbon of eit her glyceraldehyde is connect ed t o four different ( or dissim ilar) groups. - 135 -

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7. Are t he t wo form s of glyceraldehyde ge om e t r ic isom e r s one) ? Explain in one com plet e sent ence.

or

e na n t iom e r s

( circle

Figur e 1 : M e m or y de vice for D - glyce r a lde h yde To dist inguish t he t wo isom ers of glyceraldehyde in CTQ 4, chem ist s call t he one on t he left and t he one on t he right D - glyceraldehyde. One way t o rem em ber which is which is t o im agine a line from C1 t o C3 t hrough t he m iddle OH. The D isom er m akes t he let t er “ D.” Nearly all com m on sugars of biological significance have t he D configurat ion. O H C L- glyceraldehyde

H

C

OH

CH 2OH Figur e 2 : W e dge - a nd- da sh st r u ct u r e s a nd Fish e r pr oj e ct ion s of glyce r a lde hyde O

H

H

C wedge-and-dash structures

H

C

OH

HO

H

C

HO CH2

CH2OH mirror plane

chiral carbon O

H

H

C Fisher projections

O C

H

O C

OH CH2OH

D-glyceraldehyde

HO

H

HO CH2 L-glyceraldehyde

Critical Thinking Question: 8. The Fisher proj ect ion of L- glyceraldehyde shown below does not exact ly m at ch t he Fisher proj ect ion in Figure 2. Explain how t he t wo st ruct ures are different , and why bot h are correct ly nam ed. You m ay wish t o look at and m anipulat e your m odel. O

H C

HO

H CH2OH

L-glyceraldehyde

Information: The cent er carbon of glyceraldehyde is connect ed t o four different ( or dissim ilar) groups. This m akes it a chir a l carbon and leads t o t he t wo enant iom ers. Each addit ional chiral carbon doubles t he num ber of pairs of enant iom ers, so t hat in 6- carbon carbohydrat es wit h four chiral carbons, t here are 16 possible isom ers ( 2 4 = 16) . I n t he Fisher proj ect ion, t he last ( bot t om ) chiral carbon det erm ines whet her t he isom er is L or D. CA34

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Definitions: aldose ket ose

– –

a sim ple sugar cont aining an aldehyde a sim ple sugar cont aining a ket one

Most of t he com m on sugars have 5 or 6 carbons, and so are called pe n t ose s or h e x ose s. You can m ix t he t erm s wit h a ldose and k e t ose t o get nam es such as k e t ope n t ose and a ldoh e x ose . Figur e 3 : Com m on sim ple su ga r s H H O O C C H

C

OH

H

C

OH

HO

C

H

HO

C

H

H

C

OH

HO

C

H

H

C

OH

H

C

OH

CH2OH D-glucose

CH2OH

CH2OH

C

O

C

O

HO

C

H

H

C

OH

H

C

OH

HO

C

H

H

C

OH

HO

C

H

CH2OH

CH2OH D-galactose

D-fructose

O

H C H

C

OH

H

C

OH

H

C

OH

CH2OH

CH2OH

D-ribose

L-fructose

Critical Thinking Questions: 9. Label each sugar in Figure 3 wit h t he appropriat e nam e from t his list : aldohexose, aldopent ose, ket ohexose, ket opent ose. 10. Circle each chiral carbon in Figure 3. ( Hint : There are 17 t ot al) . 11. Consider t he t wo isom ers of fruct ose in Figure 3. What is t he difference bet ween t he L and D isom er of a sim ple sugar?

Information: The sim ple 5- and 6- carbon sugars oft en react wit h t hem selves t o m ake cyclic st ruct ures. When t his happens, a new funct ional group is form ed—a hem iacet al. A hem iacet al has an alcohol ( - OH) and an et her ( - OR) at t ached t o t he sam e carbon. An exam ple is shown in Figure 4. Figur e 4 : Cycliza t ion of D - glu cose t o m a k e D- D - glu cose . O H C1 2

H

3

HO

4

H

5

H

H OH OH

CH2OH 6

6

6

OH

CH2OH bend around

H

5

H OH

4

HO

CH2OH H

O

make new H bond

H 1

C

H

4

O HO

2

3

H

OH

5

H OH 3

H

anomeric carbon

O

H 1

H 2

OH

D-D-glucose

D-glucose

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OH OH "down" = alpha (D)

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When t his happens, a new chiral carbon is creat ed. This carbon ( t he a n om e r ic carbon) is designat ed Dor E depending on whet her it s OH is “ down” or “ up” when t he cyclic st ruct ure is drawn in t he orient at ion shown in Figure 4. I n disaccharides and polysaccharides, sim ple sugars are connect ed t oget her wit h at least one of t hem bonding at t he anom eric carbon. Bonds t o t he anom eric carbon are called glycosidic bonds. Figur e 5 : D a n d E glycosidic bonds in t w o disa ccha r ide s. ( a ) m a lt ose , or m a lt su ga r , a n d ( b) la ct ose , or m ilk suga r . (b) CH2OH (a) CH2OH O

H H OH

CH2OH H

O

H H OH

H

H

OH

H OH OH

H

O

H

H OH

H H

H OH

H

H

OH

Maltose contains two glucose monomers joined by an D-1,4 glycosidic bond

O

HO

H

O

HO

H

O

H

CH2OH

H

OH

OH

Lactose contains one galactose and one glucose monomer joined by a E-1,4 glycosidic bond

Critical Thinking Questions: 12. Consider Figures 4 and 5. Explain what is m eant by “ D,” “ 1,” and “ 4” in t he t erm D- 1,4 glycosidic bond.

13. Explain what is m eant by “E,” “ 1,” and “ 4” in t he t erm E- 1,4 glycosidic bond.

14. By ext ension, explain what would be m eant by t he t erm D- 1,6 glycosidic bond.

Exercises: 1. What is t he difference bet ween t he st ruct ures of D- glucose and D- galact ose?

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2. Considering your answer t o Exercise 1 and Figures 4 and 5, explain how you can det erm ine which part of t he disaccharide lact ose in Figure 5 cam e from t he glucose m onom er, and which cam e from galact ose. Then label t he Figure accordingly.

3. Find descript ions in your t ext book of t he st ruct ures of t he polysaccharides am ylose, am ylopect in, and cellulose. Describe t he sim ilarit ies and differences am ong t hese polym ers. Be sure t o consider t he t ypes of glycosidic bonds as exam ined in CTQs 12- 14.

4. Read t he assigned pages in your t ext book and work t he assigned problem s. - 139 -

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Ch e m Act ivit y 3 5

Lipids ( What are t he com ponent s of cell m em branes?)

Information: Fatty Acids Fa t t y a cids have a carboxyl group “ head” and a hydrocarbon “ t ail” t hat is 11- 19 carbons in lengt h. Figur e 1 : St r u ct u r e of pa lm it ole ic a cid a nd it s con j uga t e ba se O

O O-

OH

palmitoleic acid

palmitoleate

Critical Thinking Questions: 1. Which would be m ore wat er soluble: —palm it oleic acid, or sodium palm it oleat e? Explain.

2. Select a carbon in Figure 1 t hat is in a C= C bond, and circle it . To how m any ot her a t om s is t his carbon covalent ly bonded?

Information: We say t hat a carbon is sa t u r a t e d when it is bonded t o 4 different at om s. A C= C in a m olecule is referred t o as a sit e of u n sa t u r a t ion . Fat t y acids are oft en u n sa t u r a t e d, t hat is, t hey cont ain one or m ore cis carbon- carbon double bonds.

Critical Thinking Questions: 3. Com pare t wo or m ore fat t y acids from Table 1 t hat have t he sam e num ber of carbons ( and roughly t he sam e m olar m ass) . a. How does t he m elt ing point change as t he num ber of sit es of unsat urat ion increases?

b. How do t he int erm olecular at t ract ions bet ween m olecules change as t he num ber of sit es of unsat urat ion increases?

c. Given t he shapes of t he m olecular st ruct ures shown in Table 1, devise an explanat ion for t he effect t hat you described in CTQ 3b.

4. Give a nam e of a fat t y acid from Table 1 t hat fit s t he class list ed. a. a sat urat ed fat t y acid b. a m onounsat urat ed fat t y acid c. a polyunsat urat ed fat t y acid CA35

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5. Which class of fat t y acid list ed in CTQ 4 would be a solid at room t em perat ure?

Information: Lipids Ta ble 1 : St r u ct u r e s a n d m e lt ing poin t s of som e com m on fa t t y a cids* Alkane

Num ber of carbons

palm it ic acid

16

st earic acid

18

St ruct ure

Molar m ass, g/ m ol

Melt ing point , °C

256

62

284

69

313

75

254

0

282

13

280

-5

278

- 11

304

- 49

O OH

O OH

arachidic acid

O

20 OH

O

palm it oleic acid

16

OH

O

oleic acid

18

OH O

linoleic acid

18

OH O

linolenic acid

18

arachidonic acid

20

OH

O OH

Pr ost a gla ndin s are form ed from t he 20- carbon fat t y acid arachidonic acid. Various prost aglandins act like horm ones in t he body. St eroids and pharm aceut icals such as nonst eroidal ant i- inflam m at ory drugs ( NSAI Ds) inhibit t he form at ion of prost aglandins responsible for producing inflam m at ion and pain. W a x e s are m odified fat t y acids in which t he hydrogen of t he carboxyl head is replaced wit h a second hydrocarbon t ail. Many plant s and anim als produce or secret e waxes as a prot ect ive, wat er- repellent barrier. *

Source: Chem Finder.com [ accessed June 2006]

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Figur e 2 : Ca r na uba w a x ca n be isola t e d fr om pa lm t r e e s.

O O

Critical Thinking Question: 6. What organic funct ional group ( ot her t han alkane) is cont ained in a wax m olecule?

Information: Chem ically, fa t s a n d oils are t r ia cylglyce r ols ( t riglycerides) . They are com posed of a glycerol “ backbone” est erified wit h t hree fat t y acids. Figur e 2 : Com posit ion of a ge ne r ic fa t m ole cu le ( t r ia cylglyge r ol) O

CH2 OH CH OH CH2 OH

O

HO

+

for energy storage

O

CH2 O

O

+

CH O

HO

for energy usage

O

3 H2O

O

CH2 O

HO glycerol

3 fatty acids

a fat molecule

Critical Thinking Questions: 7. The m ain difference bet ween a fat and an oil is t hat oils are liquids at room t em perat ure. What conclusion could you m ake about t he likely class of t he t hree fat t y acyl groups in an oil versus t hose in a fat ?

Information: Cell Membranes The m ain lipid com ponent of m em branes is glyce r oph osph olipids. The st ruct ure of a glycerophospholipid is like a fat m olecule ( t riacylglycerol) , but wit h one fat t y- acyl “ t ail” replaced wit h a phosphat e group plus an am ino alcohol. This gives t he glycerophospholipid one very polar “ head” group wit h t w o nonpolar “ t ail” groups. Figur e 4 : A ph ospha t idyle t h a n ola m in e , a t ypica l glyce r oph ospholipid H

O NH3 CH2 CH2

O

O

C

H

O

O

H

C

O

C CH2 CH2 CH2 CH2 CH2 CH2 CH2

C

O

H H

H polar head

H

P

C H

O

C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 H C CH2 CH2 CH2 CH2 CH3

C CH2 CH2 CH2 CH2 CH2 CH2 CH2 C C two nonpolar tails

H

CH2

C H

The ot her lipid com ponent of m em branes is t he st eroid chole st e r ol, shown in Figure 5. Ot her st eroids wit h t he sam e ring st ruct ure are horm ones such as est rogen and t est ost erone. CA35

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Figur e 4 : St r u ct u r e of ch ole st e r ol a s com m on ly dr a w n ( a ) , a nd in t he t ypica l “r igid” ch a ir confor m a t ion ( b) H (a) (b) CH3 CH3 H H H

HO H "head"

HO

"tail"

Cell m em branes are com posed of lipid bila ye r s wit h associat ed prot eins. The bilayers are form ed when lipids associat es in t wo layers wit h t he hydrophobic ( nonpolar) t ails on t he inside, and t he hydrophilic ( polar) heads on t he out sides. See Figure 3 for a schem at ic of a cell ( plasm a) m em brane. ( Not ice in t he cart oon in Figure 3 how each head has t wo t ails?) Cholest erol has a very sm all polar head wit h a large, nonpolar t ail. Because of t he four fused rings in t he st ruct ure, cholest erol is very rigid, and t herefore adds rigidit y t o t he m em brane. Figur e 3 : M ode l of a lipid bila ye r a s a ce ll m e m br a ne . M e m br a n e pr ot e ins floa t in a “se a ” of lipids, bu t ca n not u nde r go t r a n sve r se m ove m e n t ( “flip- flop”) .

[ From Essent ial Biochem ist ry; Prat t , C.W. and Cornely, K.; Copyright © ( 2004) by John Wiley & Sons, I nc. Reprint ed wit h perm ission of John Wiley & Sons, I nc.]

Critical Thinking Questions: 8. A cell m em brane can vary in t erm s of how “ fluid” or flexible it is. Describe if you would expect each of t he following changes t o m ake a part icular m em brane m ore fluid or m ore rigid. Explain your choices. a. an increase in t he cholest erol cont ent

b. an increase in t he fract ion of unsat urat ed fat t y- acyl t ails in t he m em brane glycerophospholipids

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9. Would you expect a reindeer living in nort hern Canada t o have a cell m em brane com posit ion sim ilar t o a red- t ailed deer in t he sout hern Unit ed St at es? I f so, explain why. I f not , how would you expect t he m em branes t o differ?

Exercises: 1. Describe t he differences bet ween t riacylglycerols and glycerophospholipids.

2. Explain why a leaf coat ed wit h wax m akes t he leaf wat er- repellent .

3. A com m on ingredient in m any prepared foods is part ially hydrogenat ed soybean oil. “ Part ially hydrogenat ed” m eans t hat som e ( but not all) of t he double bonds in t he fat t y acid t ails have been convert ed t o single bonds. I n t his process, som e of t he rem aining cis double bonds are also convert ed t o t rans double bonds. a. What would be t he purpose of convert ing som e of t he double bonds in t he fat t y acid t ails t o single bonds? ( How would t his affect t he propert ies of t he oil?)

b. How can you reduce t he am ount of t rans fat s in your diet ?

4. Read t he assigned pages in t he t ext and work t he assigned problem s.

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Ch e m Act ivit y 3 6

Amino Acids and Proteins ( What does it m ean for a prot ein t o be denat ured?)

Information: The building blocks of prot eins are D- a m in o a cids, sm all m olecules t hat cont ain a carboxylic acid and an am ino group. The am ino group is connect ed t o t he carbon next t o t he carboxyl group, designat ed t he D carbon. There are 20 different am ino acids found in prot eins, differing only in t he side ch a in. Prot eins are m ade of long chains of am ino acids bonded t oget her and folded int o a part icular shape. Prot eins m ay be eit her fibrous or globular, and t he specific shape of each prot ein is individualized t o help it perform a specific funct ion. Ta ble 1 : Pr ot e in s pe r for m m a ny im por t a n t fu n ct ion s Funct ion

Type

Exam ples

st ruct ure

fibrous

m icrofilam ent s are part of t he cyt oskelet on

cat alysis

globular

sucrase is t he enzym e t hat aids t he hydrolysis of sucrose t o fruct ose and glucose

cont ract ion

globular and fibrous

act in and m yosin in m uscle fibers

t ransport

globular

hem oglobin in blood; various m em brane prot eins perform act ive t ransport

Model 1: General structure of an D-amino acid in acidic, neutral, or basic solution. O R

CH C

O OH

R

CH C

O O-

R

CH C

NH3+

NH3+

NH2

acidic

neutral

basic

O-

Am ino acids cont ain bot h a carboxylic acid ( prot on donor) and a basic am ino group ( prot on accept or) . The cent er “ zwit t erionic” form is com m only found in wat er solut ions wit h a pH near neut ralit y. The side chain “ R” m ay be one of 20 choices.

Critical Thinking Questions: 1. Using Table 2 on t he following page t o help you, draw t he st ruct ure of t he given am ino acid as it would com m only exist under each of t he following condit ions. a. Valine; in t he st om ach at pH 1.5

b. Serine; in t he sm all int est ine at pH 8

c. Glut am at e; in t he blood plasm a at pH 7.4

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Ta ble 2 : St r u ct u r e s a n d a bbr e via t ion s of t h e 2 0 st a nda r d a m ino a cids in zw it t e r ion ic for m , cla ssifie d a ccor din g t o t h e ir side ch a in ( R gr ou p) ch e m ist r y

[ From Essent ial Biochem ist ry; Prat t , C.W. and Cornely, K.; Copyright © ( 2004) by John Wiley & Sons, I nc. Reprint ed wit h perm ission of John Wiley & Sons, I nc.]

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Critical Thinking Questions: 2. Which of t he t hree classes of am ino acids ( hydrophobic, polar, charged) have R groups t hat would be at t ract ed t o wat er?

3. What is t he opposit e of hydrophobic? 4. Consider your answer t o CTQ 3. To which of t he t hree classes of am ino acids could you apply t his t erm ?

Information: The am ino group of one am ino acid and t he acid group of anot her can undergo a condensat ion react ion t o form a new bond, called eit her a pe pt ide bond or an a m ide bond. The new m olecule is called a dipept ide. More condensat ions can lead t o t ripept ides, t et rapept ides, et c., and finally polype pt ide s.

Model 3: The condensation reaction of glycine and alanine to make the dipeptide glycylalanine. O O + +

H3N

+

C CH2

H3N

C

+

O-

CH

H3N CH2

OCH3

glycine

alanine

O

CH3

C

CH

N H

glycylalanine

O-

+ H2O

C O

Critical Thinking Questions: 5. Draw a box around t he part of t he st ruct ure for glycylalanine in Model 3 t hat originat ed in a la n ine . 6. Draw anot her box around t he part of t he st ruct ure for glycylalanine in Model 3 t hat originat ed in glycine . 7. Draw an arrow t o t he bond t hat was form ed in t he condensat ion react ion t o connect glycine t o alanine. This bond is called a pe pt ide bon d. Label it in t he m odel. 8. Would t he dipept ide Ala- Gly ( alanylglycine) be t he sam e m olecule as Gly- Ala ( glycylalanine) ? Explain.

Model 4: A tripeptide O +

H 3N

CH C

O NH

CH C

CH CH3

CH2

CH3

OH

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O NH

CH C H2C

C

OO-

O

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Critical Thinking Questions: 9. Circle t he t hree side chains ( R groups) in Model 4. What t hree am ino acids condensed t oget her t o m ake t he t ripept ide? Label t hem in t he Model. 10. Const ruct an explanat ion for why t he cent ral am ino acid in Model 4 is called an am ino acid residue.

Information: A polypept ide wit h a specific biological funct ion is called a pr ot e in. The am ino acid sequence of a prot ein is called t he pr im a r y st r u ct u r e . Prot ein st ruct ure is usually classified int o four levels.

Figure 1: Levels of protein structure in hemoglobin.

[ From Essent ial Biochem ist ry; Prat t , C.W. and Cornely, K.; Copyright © ( 2004) by John Wiley & Sons, I nc. Reprint ed wit h perm ission of John Wiley & Sons, I nc.]

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Information: Secondary st ruct ures are norm ally held in place by hydrogen bonding. Two exam ples of secondary st ruct ure are shown in Figure 2. Figur e 2 : Th e D h e lix ( a ) a n d t he E she e t ( b) a r e se con da r y st r u ct u r e s of a pr ot e in ( a)

I n t he D helix, t he polypept ide backbone t wist s so t hat hydrogen bonds ( dashed lines) form bet ween a C= O and an N–H group four residues furt her along in t he sequence.

I n E sheet s, t he polypept ide backbone is ext ended, and hydrogen bonds form bet ween t he C= O and N–H groups on adj acent st rands. I n parallel sheet s, t he prim ary sequence of bot h st rands proceeds in t he sam e direct ion. I n ant iparallel sheet s, t he prim ary sequences proceed in opposit e direct ions.

( b)

[ From Essent ial Biochem ist ry; Prat t , C.W. and Cornely, K.; Copyright © ( 2004) by John Wiley & Sons, I nc. Reprint ed wit h perm ission of John Wiley & Sons, I nc.]

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Critical Thinking Question: 11. Describe how t he conform at ion of t he prim ary st ruct ure of a prot ein is affect ed when it adopt s a part icular secondary st ruct ure.

Information: Globular proteins have hydrophobic cores An act ive prot ein m ust be in it s nat ive conform at ion. The prim ary st ruct ure is com posed of am ino acids connect ed wit h covalent pept ide bonds t hat are not easily broken. However, t he secondary, t ert iary, and, quat ernary st ruct ures are held t oget her wit h weaker, noncovalent int eract ions. These can be dist urbed by heat , agit at ion, drast ic pH changes, det ergent s, salt s, organic solvent s, et c., in a process called de n a t u r in g. Denat ured prot eins are oft en said t o be u n folde d, and are no longer able t o perform t heir int ended funct ion. ( You can see t his visually by frying an egg. As t he prot eins denat ure, t hey becom e opaque. Good luck hat ching a hard- boiled egg! ) Som e prot eins can spont aneously refold, but m ost denat urat ions are essent ially irreversible, inact ivat ing t he prot ein. Secondary st ruct ures ( D helix, E sheet ) are held t oget her wit h only hydrogen bonding. Tert iary and quat ernary st ruct ures also m ake use of hydrogen bonding, but t he m ost im por t a n t force holding globular prot eins in place is t he h ydr ophobic e ffe ct —a process by which t he wat er solvent at t ract s t he polar am ino acids t o t he out side of t he prot ein and “ squeezes” t he nonpolar am ino acids int o t he cent er of t he st ruct ure.

Critical Thinking Questions: 12. About half of t he 223 am ino acid residues of t he digest ive enzym e t rypsin are hydrophobic. Describe where in t he t ert iary st ruct ure you would expect t o find t hose residues.

13. Describe where in t he t ert iary st ruct ure you would expect t he polar am ino acid residues in t rypsin t o be locat ed.

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14. Anot her digest ive enzym e, chym ot rypsin, has t wo polypept ide chains, each wit h 245 am ino acid residues. Again, about half of t he residues have nonpolar side chains. The quat ernary st ruct ure is shown below ( not e t he t wo separat e polypept ide chains) . Describe t hree dist inct locat ions in t he overall st ruct ure you would expect t he nonpolar am ino acid residues t o be locat ed.

Exercises: 1. Explain why a pept ide bond m ay also be called an am ide bond.

2. Rank t he solubilit y of t he following am ino acids in wat er at pH 7, from m ost t o least soluble: Val, Ser, Phe, Lys. Explain your answer.

3. Suppose a polypept ide cont aining 150 am ino- acid residues ( chosen at random from t he 20 com m on ones) is synt hesized in t he laborat ory. Why is it not correct t o call t his polypept ide a prot ein?

O 4. A det ergent , sodium Ododecyl sulfat e, is shown at Na+ S t he right . Label t he O hydrophilic and O hydrophobic part s of t he det ergent . Which class of am ino acid side chains would be at t ract ed t o t he hydrophobic part ? ______________ Considering your answers t o CTQs 12 and 14, explain how adding a det ergent t o a prot ein solut ion m ight cause t he prot ein t o denat ure.

5. Read t he assigned pages in your t ext , and work t he assigned problem s. - 151 -

CA36

Ch e m Act ivit y 3 7

Energy and Metabolism ( What are t he energy currencies of t he cell?)

Definitions: m et abolism – all biochem ical react ions involving t he use, product ion, and st orage of energy anabolism – synt het ic ( reduct ive) m et abolic react ions which require energy cat abolism – degradat ive ( oxidat ive) m et abolic react ions which produce energy

Information: Met abolic processes are all t hose relat ed t o t he product ion and use of energy. I n t he broadest definit ion, t his includes all biochem ical react ions of t he cell, as all react ions eit her require or produce energy. We will m ost ly concern ourselves wit h cat abolic react ions—t hose which provide t he energy required for life ( anabolic) processes. Non- phot osynt het ic organism s such as hum ans m ust find all of t heir required energy in t heir diet , and so t he first st age of energy product ion begins wit h digest ion. The m ain classes of food m olecules are carbohydrat es, prot eins, and lipids. All com plex carbohydrat es, prot eins, and lipids m ust be broken down int o t heir com ponent s in order t o be absorbed int o t he bloodst ream . Enzym es in t he digest ive t ract cat alyze t heir breakdown int o sim ple sugars, am ino acids and sm all pept ides, and fat t y acids. These m olecules m ake t heir way int o t he cells via t he lym phat ic and circulat ory syst em s. Once inside cells, t hey can be broken down ( oxidized) and t heir energy st ored. Ta ble 1 : En zym e s t ha t ca t a lyze t h e hydr olysis of food m ole cu le s, a n d t h e ir loca t ion s a n d pr odu ct s Su bst r a t e

En zym e s

Loca t ion

Pr odu ct s

com plex carbohydrat es

am ylase, glycosidases

saliva, sm all int est ine

sim ple sugars and disaccharides

prot eins

pept idases, prot eases

st om ach, sm all int est ine

am ino acids and short pept ides

lipids

lipases

sm all int est ine

fat t y acids and m onoacylglycerols

Critical Thinking Question: 1. What t ype( s) of enzym es would be involved in beginning digest ion of t he following food sources? Where in t he digest ive t ract would t his occur? a. Pot at o st arch b. Corn oil c. Soy prot ein d. Short ening from a pie crust

Information: Once food m olecules have been digest ed and ent er t he cells, t hey ent er cat abolism . Cat abolic pat hways are t hose t hat involve t he breakdown and oxidat ion of food m olecules t o produce energy. This process is oft en called ce llu la r r e spir a t ion . The energy produced in cat abolic react ions is st ored in m any different m olecules, but t he m ost im port ant of t hese is ATP, a denosine t riphosphat e. I t is said t hat an organism m ust produce and use it s weight in ATP each day. CA37

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Under st andard condit ions, t he hydrolysis of ATP t o m ake ADP and wat er releases 7.3 kcal/ m ol of energy. Since t he react ion is reversible, t his m eans t hat 7.3 kcal/ m ol of energy from food m olecules are r e qu ir e d t o m a k e ATP from ADP and wat er. Figur e 2 : “H igh - e ne r gy” phospha t e bon ds ( indica t e d in bold) in ATP ca n be h ydr olyze d t o r e le a se e n e r gy. O HO

P O-

ATP

O

P O-

NH2

O

O O

P

N

N

O N

O- CH2 H

ATP

HO

N

O

O-

O

H

H

OH

+

HO

P

H OH

Pi

O-

H2 O

Pi

+

P

OH

ADP ADP

+

NH2

O

O O

P

N

N

O

O- CH2 H

N

N

O

H

H

OH

H OH

7.3 kcal/mol

Not e t he abbreviat ions in Figure 2 and what t hey m ean: ATP is an adenosine wit h t hree phosphat es connect ed t oget her via phosphoanhydride bonds; ADP is an adenosine wit h t wo phosphat es connect ed t oget her via a phosphoanhydride bond; and Pi m eans inorganic phosphat e, shown as dihydrogenphosphat e ( H2 PO–3 ) . You will som et im es see P ( a P in a circle) as a short cut t o designat e a phosphat e group. I t is im port ant t o not e t hat energy is not released by br e a k ing t he P- O bond, but by t he m a k in g of t he new O- H and P- O bonds during hydrolysis ( am ong ot her t hings we need not concern ourselves wit h) . This is why t he t erm “ high- energy” bond is shown in quot at ion m arks.

Critical Thinking Questions: 2. The high energy phosphat e bond in ADP can be hydrolyzed t o release anot her 7.3 kcal/ m ol of energy, as shown: ADP + H2 O qwe AMP + Pi + 7.3 kcal/ m ol By analogy t o Figure 2, draw t he t wo product s of t his hydrolysis.

3. I f an ATP m olecule is hydrolyzed ( by t wo wat er m olecules) all t he way t o AMP ( and t wo Pi ) in t wo st eps, how m any t ot al kcal/ m ol of energy could be released? Explain.

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Model: ATP is the energy currency of the cell Consider t he phosphorylat ion of glucose react ion below. The react ion is unfavorable ( endot herm ic) because 3.3 kcal/ m ol of energy are required. Glucose + Pi + 3.3 kcal/ m ol qwe glucose- 6- phosphat e + H2 O

( 1)

I f ATP is present , it can be hydrolyzed t o bot h provide t he needed Pi and t he energy. ATP + H2 O qwe ADP + Pi + 7.3 kcal/ m ol

( 2)

When t hese t wo react ions are coupled ( added) t oget her, t he net react ion is favorable ( exot herm ic) : Glucose + ATP qwe glucose- 6- phosphat e + ADP + 4.0 kcal/ m ol

( 3)

Critical Thinking Questions: 4. Copy a ll of t he react ant s from bot h react ions ( 1) and ( 2) in t he Model int o t he space below. Repeat for t he product s. Confirm t hat what you have writ t en is equivalent t o react ion ( 3) in t he Model.

reactants

products

5. Consider t he react ion below: fruct ose- 6- phosphat e + Pi + 3.4 kcal/ m ol qwe fruct ose- 1,6- bisphosphat e + H2 O a. I n t his react ion, is energy r e quir e d, or is energy r e le a se d ( circle one) ? b. I f energy from ATP is required for t his react ion, in order t o provide sufficient energy, would t he product of ATP hydrolysis need t o be AD P or AM P ( circle one) ? Explain your choice.

6. Consider t he react ion below: phosphoenolpyruvat e + H 2 O qwe pyruvat e + Pi + 14.8 kcal/ m ol a. I n t his react ion, is energy r e quir e d, or is energy r e le a se d ( circle one) ? b. Would energy from ATP be required for t his react ion? c. I s sufficient energy produced by t his react ion t o m ake ATP from ADP + Pi ? Explain.

Information: Other energy currencies We have defined oxidat ion as t he loss of elect rons, and we recognize t hat t he addit ion of oxygen at om s or loss of hydrogen at om s are signs of oxidat ion. Since cat abolic react ions are m ainly oxidat ive, t hen ox idizin g a ge n t s ( i. e., e le ct r on a cce pt or s) m ust be present . The ult im at e oxidizing agent in m et abolism is O2 , but int erm ediat e st eps require coenzym es such as FAD and NAD + t o accept elect rons. The reduced form s of t hese cofact ors represent anot her locat ion ( “ currency” ) in which energy m ay be st ored for lat er release.

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Figur e 4 : Re du ce d a n d ox idize d for m s of N AD + . N AD + ( for m e d fr om vit a m in B 3 , n ia cin) is in volve d in ox ida t ions pr odu cing C= O double bonds. O

H

H NH2

ADP O CH2 H

O

H

OH

H OH

N

CH2

O

H

NH2

ADP

+ H+ + 2 e –

N+

O

H

H

O

H

H

OH

H OH

NAD+ (Nicotinamide Adenine Dinucleotide)

NADH

For exam ple, et hanol is oxidized t o acet aldehyde in liver cells according t o t he react ion: alcohol dehydrogenase

CH3 CH2 OH + NAD +

CH3 CHO + NADH + H +

Critical Thinking Question: 7. Consider Figure 4, including t he exam ple react ion at t he bot t om . a. Why is t he react ion of NAD + t o NADH and H+ called a r e duct ion of NAD + ?

b. Due t o t he short hand not at ion for t he aldehyde in t he exam ple react ion of Figure 4, t he locat ion of t he C= O t hat is produced in t he react ion is not obvious. Draw a chem ical st ruct ure t hat shows t he carbonyl group.

Figur e 5 : Re du ce d a n d ox idize d for m s of FAD ; AD P = a de nosin e diph osph a t e . FAD ( for m e d fr om vit a m in B 2 , r ibofla vin ) is involve d in r e a ct ion s t h a t pr odu ce a C= C double bon d. H

O N NH N

N

N NH

+ 2 H+ + 2 e – O

N

CH2

N

CH2

H

C

OH

HO

C

H

H

C

OH

H2C

O

O

ADP

H

H

C

OH

HO

C

H

H

C

OH

H2C

O

O

ADP

FADH2

FAD (Flavin Adenine Dinucleotide)

For exam ple, succinat e is convert ed t o fum arat e by t he following react ion: –

O2 C- CH2 - CH2 - CO2 – + FAD

succinate dehydrogenase

–

O2 C- CH= CH- CO2 – + FADH2

Critical Thinking Questions: 8. Where are t he t wo hydrogen at om s added t o FAD? Circle t hem in t he st ruct ure of FADH2 in Figure 5. - 155 -

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9. What cofact or ( NAD + , NADH, FAD, FADH2 ) would be required in t he following react ion? Explain. O O H3C C C O pyruvate

lactate dehydrogenase

OH O H3C CH C O lactate

10. Which is in t he m ore oxidized form —pyruvat e or lact at e? Explain how you can t ell.

Exercises: 1. Considering t he react ions in t he Model and your answer t o CTQ 4, writ e a sent ence t hat explains how coupling an unfavorable react ion wit h ATP hydrolysis can m ake t he react ion favorable.

2. Redraw t he exam ple react ion of Figure 5 wit hout t he short hand ( t hat is, use chem ical st ruct ures t hat explicit ly show t he locat ions of t he C= O in t he carboxyl groups.

3. Draw a com plet e balanced react ion for t he conversion of pyruvat e t o lact at e ( CTQ 8) , showing all react ant s, product s, enzym es and cofact ors.

4. We saw in t his act ivit y t hat oxidat ion react ions of alcohol groups in carbohydrat es can provide enough energy t o t ransfer elect rons t o NAD + . a. Which form of t his cofact or is t he m ore oxidized form —NAD+ or NADH? b. Which form is t he m ore reduced form ? c. Given t hat in an oxygen- cont aining environm ent , oxidat ion react ions are generally favorable ( spont aneous) , which form (ox idize d or r e du ce d) is at a higher pot ent ial energy level? d. I n general, would oxidized cofact ors or reduced cofact ors provide a “ st ockpile” of energy? Explain.

5. The enzym e nucleoside diphosphat e kinase cat alyzes t he following react ion, wit h an equilibrium const ant , Keq , equal t o 1: GTP + ADP qwe GDP + ATP Explain why ATP and GTP “ st ore” equivalent am ount s of energy.

6. Read t he assigned pages in your t ext book and work t he assigned problem s. CA37

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Ch e m Act ivit y 3 8

Enzymes ( Why are biochem ical react ions so fast ?)

Information: Cat alyst s increase t he rat e of a chem ical react ion wit hout being changed t hem selves. Most biological cat alyst s are prot ein e n zym e s t hat change t he way a react ion t akes place so t hat it occurs fast er. The react ant s in enzym e- cat alyzed react ions are called subst r a t e s. Enzym es lower t he act ivat ion energy of a react ion by binding one or m ore subst rat es int o an a ct ive sit e, using hydrophobic or hydrophilic int eract ions, hydrogen bonding, et c. This binding st ret ches each subst rat e int o a react ive form and aligns it properly for t he chem ical react ion t o t ake place. Figur e 1 ( a t r igh t ) : Te r t ia r y st r u ct u r e of t h e dige st ive e n zym e t r ypsin fr om St r e pt om yce s gr ise us. Th e pr ot e in su bst r a t e bin ds in t o t h e a ct ive sit e , in dica t e d w it h da r k e r color .

Critical Thinking Questions: 1. Circle t he side chains in each am ino acid below. To which class ( hydrophobic, polar, charged) does each belong? ( You should be able t o do t his wit hout looking at a t able) .

O

O O C

CH2 CH

NH3

C

O O

CH3 CH C

+

H N

O O

HO CH2 CH

NH3+

C

NH3+

O

H

N+

O CH2 CH

C

O

NH3+

2. Suppose t hat t he first am ino acid ( on t he left ) in CTQ 1 is part of a subst rat e. Hypot hesize on which of t he ot her t hree am ino acid residues m ight be present in t he act ive sit e of an enzym e in order t o bind t o t he side chain of t he subst rat e. Explain.

Model 1: Energy diagrams for uncatalyzed and catalyzed exothermic reactions a) an exothermic reaction

b) the same exothermic reaction in the presense of a catalyst

energy of transition state

energy

energy

activation energy energy of reactants energy of products

energy of reactants energy of products

progress of reaction

progress of reaction

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CA38

Critical Thinking Questions: 3. Does an enzym e change t he energies of t he react ant s or product s? 4. Considering your answer t o CTQ 3, does an enzym e change t he equilibrium am ount s of react ant s and product s? Explain.

5. Draw a vert ical arrow ont o Model 1 ( b) t hat represent s t he m agnit ude of t he act ivat ion energy. 6. Explain t he funct ion of an enzym e in a com plet e sent ence or t wo.

Information: Six classes of enzymes Enzym es are classified according t o t he t ype of react ion t hey cat alyze. These react ions are oft en divided int o six basic t ypes. Enzym es are oft en nam ed aft er t heir subst rat e and react ion class, and have t he suffix “ - ase.” For exam ple, t he enzym e t riose phosphat e isom erase cat alyzes t he isom erizat ion of glyceraldehyde 3- phosphat e and dihydroxyacet one phosphat e—t wo “ t riose phosphat es.” Since all react ions are reversible, t he enzym e m ay be nam ed according t o t he reverse of t he norm al react ion, i. e., using t he nam e of a product . Find t he descript ion of t he six classes of enzym es in your t ext book. You are responsible for ident ifying react ions in t he six classes. Here are som e t ricks t o help you keep t he six classes st raight : 1. ox idor e duct a se : cat alyzes a redox react ion; adds oxygen or rem oves 2 hydrogen at om s from subst rat e; requires a cofact or such as NAD+ , NADP+ or FAD; 2. t r a n sfe r a se : t ransfers a funct ional group ( e. g., NH2 or phosphat e) bet ween subst rat es; 3. h ydr ola se : cat alyzes a hydrolysis react ion; subst rat e + H2 O o t wo product s; 4. lya se : adds or rem ove groups involving a double bond ( no ATP required) ; 5. isom e r a se : m akes an isom er of t he subst rat e by rearrangem ent ; 6. liga se : form s a bond t o j oin t wo subst rat es using ATP hydrolysis for energy.

Critical Thinking Questions: 7. Which of t he six classes of enzym es cat alyze each of t he following react ions?

COO CH3 CH COO +

NH3 a.

-

C

+

-

O

alanine transaminase

COO C

CH2 CH2 COO

CH3

COO

-

O

HC

+

b. CA38

urease

NH2

2 NH3 H2O - 158 -

+

NH2

CH2 CH2 COO

O H2N C

-

H2CO3

O

c.

CH2

HO

OPO3-2

CH2

C

O CH3 C COO

-

H

O

H

OH

HO

OH H OH

f.

-

NADH

OH

H

O 2C CH2 C COO

CH3 CH

+

NAD

H OH

OPO3-2

O

alcohol dehydrogenase

CH3 CH2 OH

H

CH CH2

ADP

ATP

e.

C -

CO2

d.

OH

O

pyruvate carboxylase

+

O

hexokinase ATP

H

ADP HO

O H OH

H

H

OH

H OPO3-2

8. What is t he nam e of t he subst rat e m olecule shown for hexokinase in CTQ 7f? Be specific.

Information: Inhibition of enzymes Many pharm aceut icals are enzym e inhibit ors. These inhibit ors m ay be eit her com pe t it ive, if t hey bind t o t he act ive sit e t o keep t he subst rat e from ent ering, or n on com pe t it ive , if t hey bind elsewhere t o t he enzym e and inact ivat e it by changing it s shape. An exam ple is a t reat m ent for t he disease em physem a. The enzym e e la st a se is a prot ease which helps t o m aint ain connect ive t issue in t he lungs, and elsewhere. I t does t his by cat alyzing t he breakdown of old prot eins in t hese t issues. Em physem a is a lung disease caused when t he body’s abilit y t o produce a nat ural inhibit or of elast ase is com prom ised. When t his nat ural inhibit or is absent , t he elast ase get s out of cont rol and dest roys large am ount s of healt hy t issue. Prolonged st udy by chem ist s has enabled t he synt hesis of a m olecule t hat resem bles t he shape of t he nat ural inhibit or, and can be given in an inhaler t o t hose suffering from em physem a. This m olecule is a com pet it ive inhibit or of elast ase, and slows it s act ion by filling up t he act ive sit e and prevent ing it from binding t o ( and hydrolyzing) t he prot eins in t he lungs.

Critical Thinking Question: 9. Glucosam ine ( shown below) is an inhibit or of hexokinase. Would you expect it t o be com pet it ive or noncom pet it ive? Explain. OH H

O H OH

H

H

NH2

HO

H OH

Exercises: 1. Oft en, an enzym e requires a cofact or or prost het ic group in it s act ive sit e in order t o be an efficient cat alyst . Many such cofact ors are synt hesized by plant s and obt ained in t he diet as vit am ins. Find a descript ion in your t ext of a part icular B vit am in. Give it s nam e, and describe it s biological funct ion in t erm s of t he nam e( s) and funct ion( s) of t he enzym e( s) t hat require it . 2. Read t he assigned pages in your t ext book and work t he assigned problem s. - 159 -

CA38

Ch e m Act ivit y 3 9

Nucleic Acids ( What is DNA m ade of?)

Information: Nucleic acids were so nam ed because t hey were acidic m olecules found in t he nucleus of at om s. Like ot her biom olecules such as polysaccharides and prot eins, nucleic acids are polym ers m ade from a sm all num ber of building blocks. The t wo m ain t ypes of nucleic acids are r ibon ucleic a cid ( RNA) and 2’- deoxyr ibonucleic a cid ( DNA) . DNA cont ains t he genet ic m at erial of a cell, and is found in t he nucleus. ( The chrom osom es which form during cell replicat ion cont ain all t he cellular DNA, along wit h prot eins.) During t he life of a cell, t he inform at ion in DNA is copied t o RNA, which cont ains t he inform at ion needed t o synt hesize each prot ein t hat t he cell requires. The building blocks of RNA and DNA are nucleot ides ( See Figure 1) . Not e t hat t he carbons in t he sugar residue are num bered wit h “ prim es” —1’, 2’, 3’, 4’, and 5’ ( read one prim e, t wo prim e, et c.) t o dist inguish t hem from t he num bers of t he carbons in t he base. RNA cont ains t he sugar ribose, which has hydroxyl ( - OH) groups at t he 2’ and 3’ posit ions. I n DNA, t he 2’ hydroxyl group is m issing. Figur e 1 : A n u cle ot ide con t a in s a n it r oge ncon t a in ing ( “n it r oge n ou s”) ba se , su ga r r e sidu e , a n d one or m or e ph ospha t e gr ou ps.

Figur e 2 : The or ga nic m ole cule s ( ba se s) pyr im idin e a n d pu r in e

N

N

N H

N pyrimidine

N N purine

Figur e 3 : Th e n it r oge n - con t a in in g ba se s in D N A a nd RN A NH2 N

N H

NH2

O N

N

Adenine (A) DNA and RNA

N

N H

N

NH N Guanine (G) DNA and RNA

O

N H

NH2

H 3C

O

Cytosine (C) DNA and RNA

O NH

N H Thymine (T) DNA only

NH O

N H

O

Uracil (U) RNA only

Critical Thinking Question: 1. Based on Figures 2 and 3, which nit rogen- cont aining bases in DNA and RNA are purines? Which are pyrim idines?

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Information: A nucleoside is a nit rogen- cont aining base connect ed t o a sugar ( ribose or 2’- deoxyribose) residue. A nucleot ide is a nit rogen- cont aining base connect ed t o a sugar residue a nd at least one phosphat e group. Mem ory device: “ A nucleot ide has a phosphat e t ie d on.” Ta ble 1 : N a m e s of nu cle oside s a n d som e n u cle ot ide s RNA Base Adenine ( A) Guanine ( G) Cyt osine ( C) Uracil ( U)

Nucleoside Adenosine ( A) Guanosine ( G) Cyt idine ( C) Uridine ( U)

Nucleot ide Adenosine- 5’- m onophosphat e ( AMP) Guanosine- 5’- diphosphat e ( GDP) Cyt idine- 5’- t riphosphat e ( CTP) Uridine- 5’- m onophosphat e ( UMP)

DNA Base Adenine ( A) Guanine ( G) Cyt osine ( C) Thym ine ( T)

Nucleoside Deoxyadenosine ( A or dA) Deoxyguanosine ( G or dG) Deoxycyt idine ( C or dC) Deoxyt hym idine ( T or dT)

Nucleot ide Deoxyadenosine- 5’- m onophosphat e ( dAMP) Deoxyguanosine- 5’- diphosphat e ( dGDP) Deoxycyt idine- 5’- t riphosphat e ( dCTP) Deoxyt hym idine- 5- m onophosphat e ( dTMP)

Critical Thinking Question: 2. Refer t o Table 1. Writ e t he full nam e of t he m olecule wit h t he following abbreviat ions. a. ATP b. UDP c. dA

Information: Levels of structure of nucleic acids Sim ilar t o t he levels of prot ein st ruct ure t hat we have seen, nucleic acids also have prim ary, secondary, and t ert iary st ruct ures. The pr im a r y st r u ct u r e is t he nucleic acid sequence. The nucleosides are connect ed from t he 5’ phosphat e on one nucleoside t o t he 3’ hydroxyl group on t he next , via a phosph odie st e r linkage ( see Figure 4) .

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Figu r e 4 : N u cle ot ide s con de n se t oge t h e r via phosph odie st e r lin k a ge s. NH2 N

NH2 N

N

-

O O

P

O

N O

N

5'

CH2

OH

– H2O

O

4'

H

H

2'

3'

O-

N

5'

CH2 4'

H

3'

OH

O

N O

CH2

phosphodiester O linkage

2'

O

1'

P O

H

2'

3'

O

H

N

H

-

NH

N

NH2

N

5'

CH2 4'

1'

OH

O

4'

O NH2

N

5'

NH

N O

P

H

N

H

O P

O

O

1'

OH

HO

N

-

H

H

O

3'

1' 2'

OH

H

H

Critical Thinking Questions: 3. I s t he dinucleot ide in Figure 4 RNA or DNA? Explain how you can t ell.

4. The 5’ end of a polynucleot ide has a free 5’- phosphat e, and t he 3’ end has a free 3’hydroxyl group. Nam e t he nucleot ide at t he 5’ end of t he dinucleot ide in Figure 4.

Information: When nucleosides are connect ed t oget her, t here is only one free 5’ phosphat e and one free 3’ hydroxyl group. By convent ion, t he sequences ( prim ary st ruct ures) of DNA and RNA are always given from t he 5’ end t o t he 3’ end. Therefore, AAAAGT is not t he sam e as TGAAAA. The m ost im port ant se con da r y st r u ct u r e of DNA is t he fam ous double helix. I n t his st ruct ure, t wo sugar- phosphat e backbones spiral around each ot her, and are held t oget her by hydr oge n bonds bet ween pairs of nit rogen bases, called ba se pa ir s. Figure 5 illust rat es how t he hydrogen bonding part nership works. Not e t hat bases alm ost never pair wit h t he wrong part ner, because t he hydrogen bonds would not line up correct ly. Figur e 5 : H ydr oge n bonds be t w e e n AT a n d CG ( com ple m e n t a r y) ba se pa ir s h old t h e D N A dou ble h e lix t oge t he r . H H H

N

CH3

O

H N

N N

HN N Adenine (A)

CA39

N

O

H

N NH

N

H

N

NH

HN N

O

N

Thymine (T)

Guanine (G)

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H

H

O Cytosine (C)

Critical Thinking Questions: 5. Review: I n general, t wo condit ions are required for hydrogen bonding t o occur: a. There m ust be a hydrogen at om bonded t o a _______________ or ______________ at om ( t o donat e t he hydrogen bond) . b. There m ust be a ______________________________________________________ ( t o accept t he hydrogen bond) . 6. Use dashed lines t o draw hydrogen bonds bet ween A and T in t he appropriat e locat ions in Figure 5. Do t he sam e for t he CG pair. 7. DNA double helices wit h m ore CG cont ent are m ore heat st able t han t hose wit h m ore AT cont ent . Explain why t his is so. ( See Figure 5.)

8. Words or phrases t hat read t he sam e forwards and backwards are called palindrom es. Exam ples are “ radar” or t he phrase “ A m an, a plan, a canal: Panam a! ” Suppose t hat one st rand of DNA is palindrom ic ( such as ATGGTA) . Would t he com plem ent ary st rand also be palindrom ic? ( Hint : Wwrit e t he com plem ent ary sequence.) Explain.

Information: Tertiary structure of nucleic acids I ndividual st rands of RNA or DNA can also have t ert iary st ructure. Consider t he t ypical RNA m olecule shown in Figure 6.

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Figur e 6 : Th e se con da r y ( a ) a nd t e r t ia r y ( b) st r u ct u r e s of a t ypica l t r a nsfe r RN A m ole cu le . Th e t hr e e - ba se a nt icodon h ydr oge n - bon ds t o a m e sse n ge r RN A codon , a n d t h e fr e e 3 ’ e n d for m s a n e st e r bond t o a pa r t icu la r a m in o a cid ( in t his ca se , ph e nyla la n in e ) . 3’

5’

[ From Essent ial Biochem ist ry; Prat t , C.W. and Cornely, K.; Copyright © ( 2004) by John Wiley & Sons, I nc. Reprint ed wit h perm ission of John Wiley & Sons, I nc.]

Critical Thinking Question: 9. RNA is used as t he “ go- bet ween” t o carry inform at ion from DNA and incorporat e it int o prot eins. Explain why m olecules like t ransfer RNA ( Figure 6) would be necessary in such a process. ( Hint : What are t he building blocks of prot eins?)

10. Consider t he t ransfer RNA in Figure 6b, in which t he hydrogen at om s are not shown: a. Describe where in t he t ert iary st ruct ure you would expect t he ribose and phosphat e backbone t o be.

b. Ribose is a 5- m em bered ring. I dent ify ( and circle) at least t hree ribose rings in Figure 6b. Does t heir locat ion m at ch your predict ion in Exercise 10a?

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c. Describe where in t he t ert iary st ruct ure you would expect t he base pairs t o be.

d. Draw an arrow t o at least t wo locat ions in Figure 6b where hydrogen bonds bet ween base pairs would be.

Exercises: 1. I dent ify each species as a nucleoside or nucleot ide. a. adenine b. cyt idine c. deoxyt hym idine d. UDP 2. I dent ify each base as a purine or a pyrim idine. a. cyt osine OH

N HN

b. hypoxant hine,

N N

3. Read t he assigned pages in your t ext book, and work t he assigned problem s.

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Ch e m Act ivit y 4 0

Glycolyis ( How is glucose convert ed t o pyruvat e?)

Information: Cat abolism is t he breakdown of food m olecules for energy. This process involves digest ion and hydrolysis of carbohydrat es, lipids, and prot eins; degradat ion of t hese m olecules int o sm aller ones in t he cyt osol of cells; and finally, oxidat ion of t hese m olecules t o produce energy in various m et abolic pat hways. I nit ial st ages of cat abolism occur in t he cyt osol ( cyt oplasm ) , while lat er st ages ( and m ost of t he oxidat ion react ions) occur in m it ochondria. These pat hways m ay be of t hree t ypes: line a r , in which t he product of one st ep is t he subst rat e for t he next ; cyclic, which is like linear except t hat one subst rat e is regenerat ed during t he cycle so t hat t he st eps repeat over again; and spir a l, in which t he st eps repeat , but t he subst rat e m olecules change slight ly each t im e. We begin our exam inat ion of m et abolic pat hways wit h t he oxidat ion of glucose t o produce energy. Glucose is m et abolized t o CO2 and H2 O via t hree consecut ive pat hways: glycolysis ( in t he cyt osol) ; t he cit r ic a cid cycle —or Krebs’ cycle—( in t he m it och on dr ia l m a t r ix ) ; and t he e le ct r on t r a n spor t cha in ( in t he in n e r m it och on dr ia l m e m br a ne ) . We will exam ine each of t hese t hree pat hways in t he next t hree Chem Act ivit ies.

Definitions: aerobic process – a cellular process which requires oxygen anaerobic process – a cellular process which does not require oxygen

Critical Thinking Questions: Refer t o Figure 1 ( on t he following page) t o help you answer CTQs 1- 11. 1. Which st eps of glycolysis are referred t o as t he “ energy invest m ent phase?”

2. W hy are t hese st eps referred t o as t he “ energy invest m ent phase?”

3. I n st ep 4 of glycolysis, t he 6- carbon m olecule fruct ose- 1,6- bisphosphat e is convert ed int o t wo 3- carbon m olecules—glyceraldehyde- 3- phosphat e and dihydroxyacet one phosphat e. What m ust happen t o t he dihydroxyacet one phosphat e in order for it t o cont inue proceeding t hrough t he glycolyt ic pat hway?

4. Consider st eps 6- 10 of glycolysis. a. Why is t here a “ 2” in front of each subst rat e in st eps 6- 10 ( e. g., 2 1,3- Bisphosphoglycerat e, 2 3- Phosphoglycerat e, et c.)

b. Why are t hese st eps referred t o as t he “ energy payoff phase?”

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Figure 1: In glycolysis, the six-carbon glucose molecule is converted to two three-carbon pyruvate molecules, producing 2 ATP and 2 NADH molecules. CH2OH O

H

Glucose

H OH

ATP

H

H

HO

1

OH H

ADP

CH2OPO3-2

OH

O

H H OH

Glucose-6-phosphate

H

HO

OH H

2 -2

O3PO CH2

Fructose-6-phosphate

Energy Investment Phase

HO

H

H OH

3

OH CH2OH

O

H

ATP

H

H

ADP -2

O3POCH2

Fructose-1,6-bisphosphate

CH2OPO3-2

O

H

HO

H

4

H

H

O C H

H OH

C

OH

CH2

Glyceraldehyde3-phosphate

+

CH2 OPO3-2

Dihydroxyacetone phosphate

2 NADH + 2 H+

6

OPO3-2

O C

2 1,3-Bisphosphoglycerate 2 ADP

H

OH

C

CH2 OPO3-2

7

O–

O

2 ATP

C

2 3-Phosphoglycerate

H

OH

C

CH2 OPO3-2

8

O–

O C

2 2-Phosphoglycerate 9 2 H2O

H

OPO3-2

C

Energy Payoff Phase

CH2 OH O–

O C

2 2-Phosphoenolpyruvate 2 ADP

O

CH2 OPO3-2

5

2 Pi + 2 NAD+

C

OH

CH2

10

2 ATP

OPO3-2

C

O–

O C

2 Pyruvate

C

O

CH3

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5. ATP is used in t wo st eps ( 1 and 3) and generat ed in t wo st eps ( 7 and 10) . Explain, t hen, how glycolysis gives a net yield of 2 m olecules of ATP.

6. Only one of t he t en st eps of glycolysis is an oxidat ion. Which one? How can you t ell?

7. Using only t he inform at ion in Figure 1, it is possible t o det erm ine t he class of enzym e used for each of t he t en st eps of glycolysis. Which of t he six classes of enzym es would be required for t he following st eps? a. St ep 3 b. St ep 6 c. St ep 2 d. St ep 8 8. I s oxygen required as a subst rat e for any of t he t en st eps of glycolysis? 9. Based on your answer t o CTQ 8, is glycolysis an aerobic or anaerobic pat hway? 10. Eryt hrocyt es ( red blood cells) cont ain no m it ochondria. Can eryt hrocyt es obt ain energy from glycolysis? Why or why not ? 11. The net chem ical equat ion for t he first st ep of glycolysis is: Glucose + ATP qwe glucose- 6- phosphat e + ADP Writ e a sim ilar equat ion for t he following st eps of glycolysis: a. St ep 7: b. St ep 4: c. St ep 10: d. St ep 6:

Information: Glycolysis does not require oxygen, but it does require an oxidizing agent , NAD + , in st ep 6. Under aerobic condit ions, t he NADH produced during glycolysis is convert ed back t o NAD + in t he m it ochondria, and glycolysis can cont inue. But under anaerobic condit ions, t he cell needs anot her m et hod t o regenerat e t he NAD + , or else energy product ion would cease. This m et hod in m ost organism s is t o produce lact at e ( lact ic acid) , as shown in Figure 2.

Critical Thinking Questions: 12. Lact ic acid is, of course, acidic. When m uscles cont ract , t hey squeeze t he blood out of t he vessels, and t herefore are operat ing anaerobically. What would happen t o t he pH of m uscle cells during prolonged cont ract ion? How m ight t his help explain m uscle soreness from a workout ?

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Figur e 2 : Unde r a n a e r obic con dit ion s, pyr u va t e is conve r t e d t o la ct a t e . The pr odu ct ion of a ce t yl- SCoA u nde r a e r obic con dit ion s is oft e n ca lle d t h e “br idge ” or “t r a n sit ion” st e p t o t h e cit r ic a cid cycle . O O aerobic conditions anaerobic conditions CH3 C C O NAD+ NADH + H+ pyruvate CoA-SH NADH + H+

NAD+ O

OH O CH3

CH C lactate

CH3 C S CoA acetyl-SCoA

O

+ CO2

citric acid cycle

Exercises: 1. Lact at e event ually m akes it s way t o t he liver, where it is convert ed back int o pyruvat e. Explain why lact at e product ion is called a “ dead- end” pat hway.

2. I n t he absence of oxygen, yeast cells regenerat e NAD+ by convert ing pyruvat e int o et hanol and carbon dioxide. Writ e a balanced chem ical equat ion for t his react ion.

3. Using only t he inform at ion in Figure 1, it is possible t o det erm ine t he class of enzym e used for each of t he t en st eps of glycolysis. Which of t he six classes of enzym es would be required for t he following st eps? a. St ep 1 b. St ep 4 c. St ep 5 d. St ep 9 4. The net chem ical equat ion for t he first st ep of glycolysis is: Glucose + ATP qwe glucose- 6- phosphat e + ADP Writ e a sim ilar equat ion for t he following st eps of glycolysis: a. St ep 3: b. St ep 5: c. St ep 8: d. St ep 9: 5. Read t he assigned pages in your t ext book and work t he assigned problem s. - 169 -

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Ch e m Act ivit y 4 1

Citric Acid Cycle ( How is pyruvat e oxidized t o CO2 and H2 O?)

Information: Following glycolysis, t he next st age of aerobic glucose cat abolism is t he oxidat ion of acet ylCoA t o CO2 and H2 O in t he m it ochondria, t he cit r ic a cid cycle —also known as t he cit rat e cycle, Krebs’ cycle, or TCA ( for t ricarboxylic acid) cycle. This Chem Act ivit y will focus on t he cit ric acid cycle ( CAC) . Following t he CAC is t he pat hway including e le ct r on t r a n spor t and ox ida t ive ph osphor yla t ion ( Chem Act ivit y 42) . Recall t hat t he end product of aerobic glycolysis is pyruvat e, which is t hen convert ed t o acet yl- CoA ( See Figure 2 of Chem Act ivit y 40) . The CAC t akes t he carbons of acet yl- CoA and oxidizes t hem t o CO2 and H2 O, producing t he reduced cofact ors NADH and FADH2 , and a sm all am ount of ATP. Figur e 1 : Th e ch e m ica l st r u ct u r e of Coe nzym e A. N ot e t h e r e a ct ive t h iol gr ou p. NH2 N

N

O N

N

CH2O

O

O

P

O

P

OH HO

CH3 OH O

OH

OPO3-2 phosphorylated ADP

CH C NH CH2 CH2 C NH

CH2 C CH3

CH2 CH2SH

O

O

pantothenic acid

aminoethanethiol

Critical Thinking Question: 1. The part of Coenzym e A t hat originat ed in pant ot henic acid is act ually only an acid residue. I t is t he result of t wo condensat ion react ions—wit h am inoet hanet hiol on one end, and wit h t he phosphorylat ed ADP on t he ot her end. Draw t he st ruct ure of pant ot henic acid. ( Recall t hat condensat ion react ions norm ally j oin t wo m olecules and release a w a t e r m olecule.)

Figur e 2 : Th e r e a ct ive t h iol gr ou p of Coe nzym e A ( CoA or CoA- SH ) ca n ca r r y a n a ce t yl gr ou p. The r e su lt in g m ole cu le is a bbr e via t e d “a ce t yl- CoA” or “a ce t yl- SCoA” O

O CoA

SH

CH3 C

CH3

R

coenzyme A

C

S

CoA

R

acetyl-CoA

Under aerobic condit ions, t he NADH and FADH2 produced in t he CAC are oxidized back t o NAD + and FAD ( We will see t his in Chem Act ivit y 42) . While oxygen is not direct ly required for t he CAC, a lack of oxygen would cause NAD + and FAD t o be unavailable, and so t he CAC would slow t o a halt . Seven of t he eight CAC enzym es are found in t he m it ochondrial m at rix. The eight h is bound in t he inner m it ochondrial m em brane. CA41

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Figur e 2 : Th e cit r ic a cid cycle ox idize s a ce t yl- CoA t o CO 2 a n d H 2 O

CO2

FADH2

FAD CO2

[ From Essent ial Biochem ist ry; Prat t , C.W. and Cornely, K.; Copyright © ( 2004) by John Wiley & Sons, I nc. Reprint ed wit h perm ission of John Wiley & Sons, I nc.]

Critical Thinking Questions: Refer t o Figure 2 t o help you answer CTQs 1- 8. 2. I n st ep 1 of t he CAC t he acet yl group is t ransferred from acet yl- CoA t o oxaloacet at e. St ep 8 produces oxaloacet at e. Explain why t his pat hway is said t o be a cyclic pat hway.

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3. I n each t urn of t he CAC, t wo m olecules of CO2 are released. I n which st eps is CO2 released? 4. I n which st ep of t he CAC do t wo carbons ent er t he cycle?

5. What m et abolit e ( subst rat e) provides t he source of carbons for t he CAC t o oxidize?

6. Four of t he eight st eps in t he CAC are oxidat ions. Which four? How can you t ell?

7. I t is difficult t o oxidize a t ert iary alcohol since t he carbon wit h t he hydroxyl group has no hydrogen t o rem ove. What do you t hink is t he purpose of St ep 2?

8. No ATP is produced direct ly by t he CAC, but one GTP ( guanosine t riphosphat e) is produced in St ep 5. Ot her “ energy- currency” cofact ors produced in t he CAC are NADH and FADH2 . For each t urn ( 8 st eps) of t he CAC, circle t he num ber of each t hat are produced. Num ber of NADH produced in one t urn of t he CAC ( circle one) –

1

2

3

Num ber of FADH2 produced in one t urn of t he CAC ( circle one) –

1

2

3

9. Adding your result s from CTQ 8 t o t he cofact ors produced in glycolysis and t he bridge st ep ( see Chem Act ivit y 40, Figure 2) , t ot al up t he num ber of ATP m olecules and ot her cofact ors produced in t he ent ire oxidat ion of one m olecule of glucose t o CO2 and H2 O via glycolysis and t he CAC. ATP ____ GTP ____ NADH ____ FADH2 ____ 10. Using only t he inform at ion in Figure 2, it is possible t o det erm ine t he class of enzym e used for t he eight st eps of t he CAC. Which of t he six classes of enzym es would be required for t he following st eps? a. St ep 2 b. St ep 4 c. St ep 5 d. St ep 7 CA41

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Exercises: 1. The chem ical equat ion for t he first st ep of t he CAC is: Oxaloacet at e + acet yl- SCoA + H2 O qwe cit rat e + CoA- SH Writ e a sim ilar equat ion for t he following st eps of t he CAC: a. St ep 2: b. St ep 4: c. St ep 5: d. St ep 7: 2. Describe what cellular respirat ion is, in general ( including when it occurs) .

3. Using your t ext book as a reference, sket ch and label a diagram of a m it ochondria ( in a hum an m uscle cell, for exam ple) .

4. Read t he assigned pages in your t ext , and work t he assigned problem s.

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Ch e m Act ivit y 4 2

Electron Transport/Oxidative Phosphorylation ( How can energy from t he reduced cofact ors be used t o m ake ATP?) Following t he CAC ( Chem Act ivit y 41) in t he aerobic oxidat ion of glucose is t he final oxidat ive pat hway, which includes e le ct r on t r a n spor t and ox ida t ive ph osphor yla t ion, known also as t he e le ct r on t r a n spor t ch a in ( ETC) . At t he end of t he cit ric acid cycle, all six carbons of glucose have been oxidized t o CO2 , and a few nucleot ide t riphosphat es ( ATP and GTP) have been produced. However, m ost of t he energy has been saved in t he reduced cofact ors NADH and FADH2 . The ETC t akes t he elect rons from NADH and FADH2 and passes t hem t hrough a “ chain” of m any ot her cofact ors and enzym e com plexes, unt il t hey finally end up adding t o O2 t o m ake H2 O.

Information: The Electron Carriers in the ETC The prot eins and cofact ors shown in Figures 1- 4 carry elect rons wit hin and bet ween t he ETC enzym e com plexes, eit her one or t wo at a t im e. Figur e 1 : Th e st r u ct u r e of FM N is sim ila r t o FAD ( Ch e m Act ivit y 3 7 , Figu r e 5 ) , but w it h a ph ospha t e gr ou p ( a t bot t om ) in pla ce of AD P. O

H

N

O

N NH

NH

+ 2 H+ + 2 e– N

O

N

N

CH2

N

CH2

H

C

OH

HO

C

H

H

C

OH

H2C

H

H

C

OH

HO

C

H

H

C

OH

OPO3-2

O

OPO3-2

H2C

FMNH2

FMN (Flavin Mononucleotide)

Figur e 2 : I r on - su lfu r pr ot e ins a r e pr ot e in s con t a in ing ir on - su lfu r clu st e r s—in w h ich t h e ir on is a t le a st pa r t ia lly coor dina t e d by su lfu r . A t ypica l ir on - su lfur clu st e r is sh ow n ; cu r ve d lin e s in dica t e con t in ua t ion of t h e pr ot e in ba ck bon e . O HN

CH2

CH2 S

HN + e

S CH2

S

S

S 2+

NH

Fe

Fe

O CH2

S

O CH2

NH

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HN

Fe3+

O –

3+

O

CH2 S

S

S

S NH

CH2

S Fe3+

O

CA 42

O HN

S

O CH2

NH

Figur e 3 : Ox idize d a n d r e du ce d for m s of Coe n zym e Q ( ubiqu inon e ) O

OH

CH3O

CH3

CH3O + 2 H+ + 2 e–

CH3 CH3O

(CH2 CH

C

CH3

CH2)10H

CH3 CH3O

(CH2 CH

O

C

CH2)10H

OH

Coenzyme Q

Coenzyme QH2

Figur e 4 : Ox idize d a n d r e du ce d for m s of t h e h e m e gr ou p of a b cyt ochr om e . Cyt ochr om e s a r e he m e - con t a ining pr ot e ins. O-

O-

O N

N Fe+3

N

O N

+ e–

Fe+2

ON

N

N

ON

O

O

Critical Thinking Questions: Refer t o Figures 1- 4 t o help you answer CTQ’s 1- 6. 1. Which of t he four t ypes of elect ron carriers in t he ETC are on e - e le ct r on carriers?

2. Which of t he four t ypes of elect ron carriers in t he ETC are t w o- e le ct r on carriers?

3. Which t wo elect ron carriers in t he ETC are pr ot e in s? What m et al ion cofact or is required for t hese carriers?

4. Which t wo elect ron carriers in t he ETC are coe nzym e s?

5. Cyt ochrom es are hem e- cont aining prot eins. What ot her com m on blood prot ein cont ains hem e?

6. What is t he im port ant m et al ion in hem e?

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Information: The Enzyme Complexes The four enzym e com plexes of t he ETC ( brilliant ly nam ed Com plexes I – I V) are int egral prot eins in t he inner m it ochondrial m em brane. They serve t o t ake t he elect rons from t he reduced cofact ors NADH and FADH2 and t ransfer t hem t o O2 t o m ake H2 O. Figur e 5 : Th e e le ct r on - t r a n spor t cha in , con sist ing of fou r e n zym e com ple x e s a n d ot h e r cofa ct or s, ope r a t e s in t h e in n e r m it och on dr ia l m e m br a ne path of electrons outside (intermembrane space)

complex I

complex III

e–

complex IV

cyt c e–

FeS FeS

inner mitochondrial membrane

inside (matrix)

cyt b

e–

e– NADH + H+

cyt a3

e–

FMN

NAD+

FADH2

cyt a

e–

Q

e–

complex II FAD

e– ½ O 2 + 2 H+

2 H 2O

Critical Thinking Questions: Refer t o Figure 5 t o help you answer CTQ’s 7- 13. The pat h of elect rons t hrough t he various com plexes and cofact ors follows t he hollow arrows ( Ÿ) . 7. Which com plex accept s elect rons from NADH? 8. Which com plex accept s elect rons from FADH2 ? 9. Which com plexes cont ain iron- sulfur clust ers? 10. Which com plexes cont ain cyt ochrom es? 11. What m olecule carries elect rons from Com plex I t o Com plex I I I ? 12. What m olecule carries elect rons from Com plex I I I t o Com plex I V? 13. Which com plex t ransfers elect rons t o oxygen? 14. For elect rons ent ering t he ETC from NADH, which of t he com plexes I – I V do t he elect rons pass t hrough? List t hem .

15. For elect rons ent ering t he ETC from NADH, which of t he com plexes I – I V do t he elect rons pass t hrough? List t hem .

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Information: Oxidative Phosphorylation All t he react ions of t he ETC are energet ically favorable, and “ use up” t he energy st ored in t he reduced cofact ors. This energy m ust be saved som ehow t o m ake ATP. The energy is saved by using t he favorable elect ron- t ransport react ions t o “ pum p” prot ons ( H+ ions) out of t he m at rix int o t he int erm em brane space. This “ prot on gradient ” ( also called “ pH gradient ” ) is a high- energy st at e, because t he nat ural t endency is for ions t o diffuse from areas of higher concent rat ion t o areas of lower concent rat ion. The only way for t he prot ons t o reent er t he m at rix is t hrough an enzym e com plex called ATP synt hase. So, t he prot ons are pum ped out of t he m at rix, and as t hey ret urn, providing energy like wat er t urning a wat erwheel, t hey are used t o at t ach a phosphat e t o ADP t o m ake ATP. This process is called ox ida t ive ph osphor yla t ion, because t he energy needed t o phosphorylat e ADP t o m ake ATP com es from t he oxidat ive st eps of t he ETC. Figur e 6 : W h e n e le ct r on s t r a ve r se Com ple x e s I , I I I , a n d I V of t h e ETC, Pr ot on s a r e pum pe d out of t he m a t r ix . Pr ot ons r e e nt e r t hr ough t he ATP synt ha se com ple x ( V) , pr ovidin g e n e r gy t o m a k e ATP. Com ple x I I ( n ot sh ow n ) pum ps n o pr ot on s.

[ From Biochem ist ry, 2E; Voet , D. and Voet , J.G.; Copyright © ( 1995) , John Wiley & Sons, I nc. Reprint ed wit h perm ission of John Wiley & Sons, I nc.]

Critical Thinking Questions: 16. Suppose t hat as t wo elect rons proceed t hrough t he ETC as shown in Figure 6, Com plexes I , I I I , and I V each pum p t w o prot ons out of t he m at rix. Considering your answers t o CTQs 14 and 15, how m any t ot al prot ons are pum ped: a. For t he t wo elect rons originat ing in NADH? b. For t he t wo elect rons originat ing in FADH2 ? 17. Considering your answer t o CTQ 16, if t he oxidat ion of one NADH t o NAD+ via t he ETC leads t o product ion of 3 ATP m olecules via ATP synt hase ( Com plex V) , how m any ATP could be produced from oxidat ion of one FADH2 ? Explain your answer.

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Exercises: 1. Com plet e t he t able below wit h t he num bers of product s t hat would be obt ained in t he various st eps from t he aerobic oxidat ion of glucose. Then t ot al t he ATP t hat one glucose is “ wort h.” Glycolysis ATP ( or GTP)

2

“ Bridge t o CAC” 0

CAC

Tot al

2

4

# of ATP “ wort h” 4

NADH FADH2 2. Biochem ist s m easure pH t o det erm ine t he values given in CTQ 16 for t he num ber of prot ons pum ped out by each ETC enzym e com plex. Explain how variat ions in t hese m easurem ent s would lead t o a different answer t o Exercise 1.

3. I n Figure 2, t he iron at om s are coordinat ed t o four am ino- acid residues in a prot ein. All four am ino- acid residues are t he sam e. I dent ify t he am ino acid.

4. Read t he assigned pages in your t ext , and work t he assigned problem s.

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Ch e m Act ivit y 4 3

Fatty Acid Oxidation ( How is energy produced from fat s?)

Information: Fat s ( t riacylglycerols, Figure 1) are degraded during digest ion t o fat t y acids and m onoacylglycerols, but are reform ed int o t riacylglycerols ( “ t riglycerides” ) and packaged wit h prot ein int o chylom icrons for t ransport in t he bloodst ream . Aft er fat s ent er a cell, t hey again are broken down int o fat t y acids. Fat t y acids are degraded in m it ochondria t o acet yl- CoA, which yields energy via t he cit ric acid cycle and elect ron t ransport , in t he sam e m anner as we have seen previously. Figur e 2 : Ox ida t ion of fa t t y a cids be gin s a t t h e E ca r bon .

Figur e 1 : St e a r in , a t r ia cylglyce r ol O

O H2C

C O

(CH2)16CH3

HC

C O

(CH2)16CH3

H2C

C

(CH2)16CH3

R

CH2

E

CH2

C

OH

D

Figur e 3 : Fa t t y a cid a ct iva t ion t a k e s pla ce in t h e cyt osol. ATP

O OH

+

CoA-SH

fatty acid

AMP + 2 Pi

O SCoA

fatty acyl-CoA

Fat t y acid oxidat ion is usually called E- oxidat ion ( bet a- oxidat ion) , since t he Ecarbon of t he fat t y acid is oxidized in t he process ( see Figure 2) . The free fat t y acid is first act ivat ed for oxidat ion by at t aching it t o coenzym e A in a process t hat requires ATP. Then, t he act ivat ed fat t y- acyl- CoA is t ransport ed int o t he m it ochondrial m at rix.

Critical Thinking Questions: Refer t o Figure 3 ( above) and Figure 4 ( following page) t o help you answer CTQs 1- 6. 1. How m any “ high- energy phosphat e bonds” are hydrolyzed during fat t y acid act ivat ion ( Figure 3) ? How can you t ell?

2. Which of t he four st eps in t he E- oxidat ion pat hway are oxidat ions? How can you t ell?

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Figur e 4 : E- ox ida t ion r e m ove s 2 - ca r bon unit s fr om fa t t y- a cyl- CoA t o pr odu ce a ce t yl- CoA. H H O D E CH3(CH2)14 C C C SCoA fatty-acyl CoA H

H FAD FADH2

CH3(CH2)14

C

H

O

C

C

SCoA

trans-'2-enoyl-CoA

SCoA

E-hydroxyacyl-CoA

H H2O

OH CH3(CH2)14

C

O CH2

C

H NAD+ NADH + H+ O CH3(CH2)14

C

O CH2

C

E-ketoacyl-CoA

SCoA

H CoASH

O CH3(CH2)14 C SCoA fatty acyl-CoA (2 C atoms shorter)

O +

CH3 C SCoA acetyl-CoA

Critical Thinking Questions: 3. Which of t he six classes of enzym es would be used in: a. St ep 1 b. St ep 2 c. St ep 3 d. St ep 4

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4. Suppose t hat st ep 1 of E- oxidat ion begins wit h an 18- carbon fat t y- acyl- CoA, as in Figure 4. St ep 4, t hen, produces acet yl- CoA and a 16- carbon fat t y- acyl- CoA. a. What would happen next in order t o oxidize t he 16- carbon fat t y- acyl- CoA?

b. Explain why E- oxidat ion is said t o be a spiral rat her t han a cyclic pat hway.

5. Consider t he E- oxidat ion of a C14 fat t y- acyl- CoA. a. How m any spirals of E- oxidat ion would be needed t o com plet ely degrade t he C14 fat t y- acyl- CoA t o acet yl- CoA? b. Explain why t he answer t o part ( a) is not 7.

c. How m any FADH2 m olecules would be produced? ____ d. How m any NADH m olecules would be produced? ____ e. How m any acet yl- CoA m olecules would be produced? ____ f.

Rem em bering t hat each acet yl- CoA can be oxidized in t he cit ric acid cycle t o produce 3 NADH, 1 FADH2 , and 1 GTP, t ot al t he num ber of ATP produced from t he C14 fat t y acid aft er com plet e aerobic oxidat ion, including elect ron t ransport . Don’t forget t o subt ract t he num ber t hat you gave in t he answer t o CTQ 1. Show your work.

Information: Ketone bodies Acet yl- CoA m ust ent er t he cit ric acid cycle ( CAC) in order t o be used for energy. I n order for t his t o happen, oxaloacet at e ( a carbohydrat e) m ust be present t o react wit h acet yl- CoA in st ep 1 of t he CAC. During fast ing, or if t he diet is high in fat and low in carbohydrat es, t here is not sufficient oxaloacet at e t o react wit h, and so t he concent rat ion of acet yl- CoA builds up. When t his happens, ot her enzym es cat alyze a condensat ion of acet yl- CoA wit h it self t o form blood- soluble m olecules called ket one bodies ( see Figure 5) . These react ions are reversible, so ket one bodies m ay revert t o acet yl- CoA and be used for energy when oxaloacet at e is available. I f t he concent rat ion of ket one bodies becom es t oo high, as can happen in diabet ics or on a very low carbohydrat e diet , a condit ion called k e t oa cidosis can result . Since t he ket one bodies are not com plet ely m et abolized, acet one can diffuse out of t he bloodst ream int o air in t he lungs. Also, since t he ket one bodies acet oacet at e and E- hydroxybut yrat e are acidic, t he pH of t he blood decreases. This affect s t he abilit y of t he hem oglobin t o carry oxygen, and breat hing can becom e difficult . Unt reat ed, t his condit ion can lead t o com a, or even deat h. - 181 -

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Figur e 5 : An e x ce ss of a ce t yl- CoA le a ds t o t h e for m a t ion of t h e “k e t one bodie s” a ce t oa ce t a t e , E- h ydr ox ybu t yr a t e , a n d a ce t on e

O

O

CH3 C SCoA + CH3 C SCoA

2 CoA-SH

O

O

N AD H + +H

OH

+

NA D

O

CH3 CHCH2 C OE-hydroxybutyrate

CH3 C CH2 C Oacetoacetate

O CO2

CH3 C CH3 acetone

Critical Thinking Questions: 6. Ket one bodies can be rem oved by t he kidneys and excret ed in t he urine. Would t his increase or decrease t he effect iveness of a low carbohydrat e diet ? Explain your answer.

7. A pat ient on a “ low- carb” diet is being m onit ored by a physician. Why m ight t he physician sniff t he breat h of t he pat ient t o check his healt h?

Exercises: 1. Consider st earic acid, an 18- carbon sat urat ed fat t y acid. a. Calculat e t he t ot al num ber of ATP t hat could be produced by t he com plet e oxidat ion of st earic acid.

b. Calculat e t he rat io of t he num ber of ATP form ed per carbon at om in st earic acid.

c. Considering t he com plet e oxidat ion of glucose t o yield 38 ATP, calculat e t he rat io of t he num ber of ATP form ed per carbon at om in glucose.

2. Carbons t hat are in m ore reduced oxidat ion st at es can release m ore energy when t hey are oxidized. On average, are t he carbons in a fat t y acid m ore reduced or m ore oxidized t han t hose in a carbohydrat e? Explain.

3. Read t he assigned pages in your t ext , and work t he assigned problem s. CA43

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Ch e m Act ivit y 4 4

Other Metabolic Pathways ( I n what ot her ways is energy produced or used?)

Information: Catabolism of Proteins We have seen how carbohydrat es ( specifically, glucose) and fat s can be used for energy. Prot eins can be hydrolyzed int o t heir am ino acid com ponent s, which can be t ransform ed int o ot her energy- producing m olecules as shown in Figure 1. Figur e 1 : Ca r bon s fr om a m ino a cids a r e de gr a de d t o CAC in t e r m e dia t e s a n d ot he r r e la t e d m e t a bolit e s

[ From Biochem ist ry, 2E; Voet , D. and Voet , J.G.; Copyright © ( 1995) , John Wiley & Sons, I nc. Reprint ed wit h perm ission of John Wiley & Sons, I nc.]

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CA44

Critical Thinking Questions: 1. Assum ing t hat no ATP are used or form ed in t he t ransform at ion of t he am ino acid serine int o pyruvat e, how m any ATP can be produced using serine for energy? Show your work. ( Hint : Consider a ll t he cofact ors produced during it s oxidat ion, - see Chem Act ivit y 41, Figure 2- if necessary.)

2. Assum ing t hat no ATP are used or form ed in t he t ransform at ion of t he am ino acid lysine int o acet oacet yl- CoA, how m any ATP can be produced using lysine for energy? Show your work. ( See also Chem Act ivit y 43, Figure 5.)

Information: Glycogen metabolism Recall t hat glycogen ( also called anim al st arch) is a polym er of D- D- glucose connect ed by D1,4 linkages wit h D- 1,6 branches. When energy is plent iful, excess glucose is convert ed int o glycogen, m uch of which is st ored in t he liver. When energy is needed, t he glycogen can undergo phosphorolysis t o produce glucose- 6- phosphat e. These processes are sum m arized in Figure 2. Figur e 2 : Glu cose is st or e d a s glycoge n ( syn t he sis) , a n d r e le a se d a s glucose - 1 ph ospha t e ( phosph or olysis) OPO3 - 2 OPO3 - 2

glycolysis

glucose- 1- phosphat e

glucose- 6phosphat e

phosphorolysis

UTP

Pi

2 Pi UDP

+ glycogen

UDP- glucose

UDP

glycogen wit h one glucose added CA44

- 184 -

Critical Thinking Questions: 3. How m any “ high- energy phosphat e bonds” are hydrolyzed in order t o at t ach one glucose t o glycogen? Explain how you can t ell.

4. Explain why glycolysis st art ing from glycogen yields one m ore ATP t han when st art ing wit h glucose it self. You m ay need t o refer t o Figure 1 of Chem Act ivit y 40.

Information: Gluconeogenesis The brain norm ally requires glucose as an energy source ( alt hough it can survive on ket one bodies when fast ing, if required) . When glycogen supplies run low, liver enzym es can perform a process called gluconeogenesis, which lit erally m eans “ new birt h of glucose.” Figur e 3 : Glu con e oge n e sis pr odu ce s glu cose w h e n glycoge n st or e s a r e de ple t e d. Pyr u va t e a n d ox a loa ce t a t e ( fr om t h e CAC) a r e st a r t in g m a t e r ia ls, bu t a ce t yl- CoA is n ot . 6 ATP 6 ADP amino acids

2 pyruvate

4 ADP 4 ATP glucose

2 oxaloacetate

amino acids

X fatty acids

acetyl-CoA

Critical Thinking Questions: 5. Recall t hat som e am ino acids break down t o m ake cit ric acid cycle int erm ediat es. Persons on very low calorie diet s oft en lose m uscle m ass along wit h body fat . Explain why t his is probably unavoidable.

6. Hum ans can survive poorly on a diet of prot ein wit h very lit t le fat or carbohydrat es, but not at all on a diet of fat wit h lit t le prot ein or carbohydrat es. Explain.

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Information: Anabolism—Pentose phosphate pathway Cat abolic pat hways produce energy, and are considered oxidat ive, but anabolic ( biosynt het ic) pat hways require energy and are considered reduct ive. While oxidat ive pat hways require oxidized cofact ors such as NAD + and FAD, reduct ive pat hways require cofact ors in t heir reduced form . The elect ron t ransport chain keeps m ost of t he cofact ors NADH and FADH2 in t heir oxidized form s, so t hey are not available in t heir reduced form s for biosynt hesis. Therefore, reduct ive pat hways m ake use of different cofact ors, such as NADPH. ( By at t aching a phosphat e group t o NADH, it becom es NADPH.) The pe n t ose phospha t e pa t hw a y ( also called t he phosphogluconat e pat hway or hexose m onophosphat e shunt ) produces NADPH for what is called “ reducing power” —t he abilit y t o perform biosynt hesis. This pat hway can also be used nonoxidat ively t o produce ribose- 5phosphat e, which is needed for t he backbone st ruct ure of t he nucleic acids RNA and DNA. Figur e 4 : Th e pe n t ose phosph a t e pa t hw a y pr odu ce s r ibose - 5 - ph ospha t e a n d N AD PH nonoxidative phase 3 H2O 3 H+ + 3 CO2 3 ribulose-5P

3 glucose-6P +

6 NADP

nucleic acids

ribose-5P

6 NADPH

Figur e 5 : Th e t h r e e m a j or u se s of glu cose

glyco

lysis

pyruvate

storage glycogen

glucose pent

o se p h p a th w o s p h a te ay

ribose-5-phosphate

Critical Thinking Questions: 7. What are t he t wo m ain product s of t he pent ose phosphat e pat hway? For what purpose are t he product s used?

8. Of t he t hree fat es of glucose in Figure 4 ( st orage, PPP, or glycolysis) , t wo occur in m ost cells at all t im es. The ot her one alt ernat es on and off. Explain.

CA44

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9. Which of t he t hree pat hways would increase in rat e m ainly during cell replicat ion or division? Explain.

Information: Fatty acid synthesis Fat t y acid synt hesis is a spiral pat hway sim ilar t o t he reverse of E- oxidat ion. Main differences are t he locat ion ( synt hesis t akes place in t he cyt osol) , and t he coenzym es and cofact ors used. Also, t he fat t y- acyl groups are bonded t o an acyl carrier prot ein inst ead of t o coenzym e A. Figur e 6 : Fa t t y a cid syn t h e sis a dds 2 - ca r bon un it s fr om m a lonyl- ACP t o t he gr ow ing fa t t y- a cyl- ACP. ( ACP = Acyl Ca r r ie r Pr ot e in ) . O

O

CH3 C S–ACP + acetyl-ACP or fatty-acyl-ACP

-

CH2 C S–ACP malonyl-ACP

O2C

CO2 O CH3

C

O CH2

C

S—ACP

NADPH + H

E-ketoacyl-ACP

+

NADP+ OH CH3

C

O CH2

C

S—ACP

E-hydroxyacyl-ACP

H H 2O

CH3

C

H

O

C

C

S—ACP

trans-'2-enoyl-ACP

H NADPH + H+ NADP+ O CH3 CH2 CH2 C

S—ACP

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butanoyl-CoA or fatty-acyl CoA

CA44

Critical Thinking Questions: 10. Which of t he six classes of enzym es would cat alyze each of t he react ions shown in Figure 6 for fat t y acid synt hesis? a. St ep 1 b. St ep 2 c. St ep 3 d. St ep 4 11. A high blood glucose level st im ulat es t he release of t he horm one insulin, which act ivat es fat t y acid biosynt hesis. Why would t his be an appropriat e response of t he organism ?

Exercises: 1. Propose som e reasons t hat fat t y acid synt hesis requires NADPH rat her t han NADH as a cofact or.

2. Glycolysis convert s glucose t o pyruvat e, and gluconeogenesis convert s pyruvat e t o glucose. The conversion of glucose o pyruvat e o glucose by t his pat h is called a fut ile cycle. a. Calculat e t he net num ber of ATP m olecules t hat would be produced or required for t his fut ile cycle.

b. High blood glucose levels bring about t he secret ion of t he horm one insulin. I nsulin act ivat es glycolysis and inact ivat es gluconeogenesis. Explain why it is beneficial for t he organism t o have bot h effect s occur sim ult aneously.

c. Would you expect insulin t o act ivat e or inact ivat e glycogen synt hesis? Explain.

d. Would you expect insulin t o act ivat e or inact ivat e glycogen phosphorolysis? Explain.

3.

Read t he assigned pages in your t ext , and work t he assigned problem s.

CA44

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Ch e m W or k sh e e t 1

Stoichiometry (Mole Relationships): Practice Worksheet 1 1. Given t he equat ion for t he com bust ion of but ane: 2 C4 H10 + 13 O2 o 8 CO2 + 10 H2 O Show what t he following m olar rat ios should be. a. C4 H10 / O2 b. O2 / CO2 c. O2 / H2 O d. C4 H10 / CO2 e. C4 H10 / H2 O 2. Given t he react ion: 4 NH3 ( g) + 5 O2 ( g) o 4 NO ( g) + 6 H2 O ( l) When 1.20 m ol of am m onia react s, t he t ot al num ber of m oles of product s form ed is: a. 1.20

b. 1.50

c. 1.80

d. 3.00

e. 12.0

3. Silver sulfide ( Ag 2 S) is t he com m on t arnish on silver obj ect s. a. What weight of silver sulfide can be m ade from 1.23 m g of hydrogen sulfide ( H2 S) obt ained from a rot t en egg? b. I t 5.7 m g of silver sulfide are obt ained, what is t he percent yield? The react ion of form at ion of silver sulfide is: Ag( s) +

H2 S( g) +

O2 ( g) o

Ag 2 S( s) +

H2 O( l)

( Equat ion m ust first be balanced.)

4. Given t he following balanced equat ion: 2 KClO3 o 2 KCl + 3 O2 How m any m oles of O2 can be produced by let t ing 12.00 m oles of KClO3 react ?

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CW01

5. A som ewhat ant iquat ed m et hod for preparing chlorine gas involves heat ing hydrochloric acid wit h pyrolusit e ( m anganese dioxide) , a com m on m anganese ore. ( React ion given below) . How m any kg of HCl react wit h 5.69 kg of m anganese dioxide? HCl( aq) + MnO2 ( s) o balanced.)

H2 O( l) +

MnCl 2 ( aq) +

Cl 2 ( g)

( Equat ion m ust first be

6. Given t he following equat ion: 2 K + Cl 2 o 2 KCl a. How m any gram s of KCl are produced from 2.50 g of K and 3.00 g Cl2 ?

b. How m any gram s of KCl are produced from 1.00 g of Cl2 and excess K?

7. Given t he following equat ion: Na2 O + H2 O o 2 NaOH a. How m any gram s of NaOH are produced from 1.20 x 10 2 gram s of Na 2 O?

b. How m any gram s of Na2 O are required t o produce 1.60 x 10 2 gram s of NaOH?

CW01

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8. Given t he following equat ion: 8 Fe + S8 o 8 FeS a. What m ass of iron is needed t o react wit h 16.0 gram s of sulfur?

b. How m any gram s of FeS are produced?

9. Given t he following equat ion: 2 NaClO3 o 2 NaCl + 3 O2 a. 12.00 m oles of NaClO3 will produce how m any gram s of O2 ?

b. How m any gram s of NaCl are produced when 80.0 gram s of O2 are produced?

10. Given t he following equat ion: Cu + 2 AgNO3 o Cu( NO3 ) 2 + 2 Ag a. How m any m oles of Cu are needed t o react wit h 3.50 m oles of AgNO3 ?

b. I f 89.5 gram s of Ag were produced, how m any gram s of Cu react ed?

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CW01

11. Molt en iron and carbon m onoxide are produced in a blast furnace by t he react ion of iron( I I I ) oxide and coke ( pure carbon) . I f 25.0 kilogram s of pure Fe2 O3 and 100.0 kilogram s of pure carbon are used, how m any kilogram s of iron can be produced? The react ion is: Fe 2 O3 + 3 C o 2 Fe + 3 CO

12. The “ average hum an” requires 120.0 gram s of glucose ( C6 H12 O6 ) per day. How m any gram s of CO2 ( in t he phot osynt hesis react ion) are required for t his am ount of glucose? The phot osynt het ic react ion is: 6 CO2 + 6 H2 O o C6 H12 O6 + 6 O2

CW01

- 192 -

Answers to Stoichiometry Practice: 1. a. 2/ 13 b. 13/ 8 c. 13/ 10 d. 2/ 8 e. 2/ 10 2. The correct answer is d. NH3 o ( NO + H2 O) = 4 o 10 4 / 10 = 1.20 / x x = 3.00 m ol 3. a. 8.95 m g Ag 2 S b. 64% 4. 18.00 m ol O2 5. 9.54 kg HCl 6. a. 4.77 g KCl b. 2.10 g KCl 7. a. 155 g NaOH b. 124 g Na2 O 8. a. 27.9 g Fe b. 43.9 g FeS 9. a. 576.0 g O2 b. 97.4 g NaCl 10. a. 1.75 m ol Cu b. 26.4 g Cu 11. 17.5 kg Fe 12. 175.9 g CO2

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CW01

Ch e m W or k sh e e t 2

Gases: Practice Worksheet Information: R

0.0821

L ˜ atm K ˜ mol

1 at m = 760 t orr = 760 m m Hg

Suggested demonstration: How to solve gas law problems A. Sim ple subst it u t ion

B. Ch a nging condit ion s

1. A t em perat ure of 0°C and a pressure of 1 at m are defined as t he St andard Tem perat ure and Pressure, or STP, for calculat ions involving gases. Use t he ideal gas law t o calculat e t he volum e occupied by one m ole of air at STP.

2. A balloon filled wit h air at room t em perat ure ( 22°C) cont ains 5.0 L of air. What volum e would t he balloon occupy if it were cooled t o liquid nit rogen t em perat ure ( –196°C) ?

3. What is t he densit y of m et hane gas ( nat ural gas) , CH4 , at 25°C and 0.947 at m ? Express your answer in gr a m s pe r lit e r . ( Hint : d= m / V; assum e any V you like, and calculat e it s m ass) .

CW02

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4. When a 2.0- L bot t le of concent rat ed HCl was spilled, 1.2 kg of CaCO3 was required t o neut ralize t he spill. Considering t he react ion equat ion shown, what volum e of CO2 was released by t he neut ralizat ion at 735 m m Hg and 20° C? ( Hint : besides V, what variable in t he ideal gas law is not given?) CaCO3 ( s) + 2 HCl( aq)

ssd

CaCl 2 ( aq) + H2 O( l) + CO2 ( g)

5. The part ial pressure of air in t he alveoli, t he air sacs in t he lungs, was m easured as follows: nit rogen, 570.0 m m Hg; carbon dioxide, 40.0 m m Hg; and wat er vapor, 47.0 m m Hg. I f t he barom et ric pressure was 0.973 at m , what was t he pO2 in t he alveoli? Express your answer in m illim et ers of m ercury ( t orr) .

Answers: 1. 22.41 L

2. 1.3 L

3. 0.621 g/ L

4. 3.0× 10 2 L

5. 82.5 m m Hg or 82.5 t orr

Additional Exercises: 1. The densit y of dry air at 25°C is about 1.2 kg/ m 3 . I f you open a nat ural gas ( m et hane) spigot in t he chem ist ry laborat ory, will t he m et hane sink t o t he floor or rise t oward t he ceiling? ( Convert unit s and com pare wit h # 3 above) . 2. Read t he assigned pages in your t ext book, and work t he assigned problem s. - 195 -

CW02

Ch e m W or k sh e e t 3

Stoichiometry (Mole Relationships): Practice Worksheet 2 Information: Recall t hat an acid and a base react t o form a salt and wat er. This is called a neut ralizat ion react ion. I f we put a pH indicat or in a solut ion of acid and t hen add base unt il t he solut ion is neut ralized, t he indicat or changes color. This indicat es t hat we have added equal m oles of acid and base, and t he endpoint has been reached. This process is called a t it rat ion. Consider t he react ion of sulfuric acid and sodium hydroxide: H2 SO4 ( aq) + 2 NaOH ( aq) ssd 2 H2 O ( l) + Na 2 SO4 ( aq) We would have t o add 2 m oles of NaOH t o neut ralize 1 m ole of H2 SO4 , giving m ole conversion fact ors of: 2 m ol NaOH 1 m ol H2 SO 4

or

1 m ol H2 SO 4 2 m ol NaOH

To det erm ine t he m oles of base ( NaOH) , we need t o know bot h t he a m ou n t ( m L or L) of t he base and t he con ce n t r a t ion ( M or m ol/ L) of t he base t hat was added. We can t hen use t he m ole rat io t o det erm ine how m uch acid is present . The un it pla n would be: L NaOH o m ol NaOH o m ol H2 SO4 Once you calculat e t he m ol H2 SO4, t hen you m ay sim ply divide by t he volum e t hat was used t o get m ol/ L ( i. e., M) .

Exercises: 1. When 25.00 m L of a sulfuric acid solut ion of unknown concent rat ion was t it rat ed wit h sodium hydroxide solut ion, t he acid required 42.36 m L of 0.500 M NaOH t o reach an endpoint . a. Writ e a unit plan for t he t it rat ion ( st oichiom et ry) problem t o calculat e t he m oles of acid in t he sam ple. Then writ e conversion fact ors for each st ep.

b. How m any m oles of sulfuric acid were present ? Show work wit h all unit s.

c. What was t he concent rat ion ( in M) of t he sulfuric acid?

CW03

- 196 -

d. What was t he concent rat ion in % ( m : v) ? Make a unit plan first .

2. I n a t it rat ion, a 20.00- m L port ion of Lim e- a- Way cleaner cont aining t he m onoprot ic acid sulfam ic acid ( HSO3 NH2 ) required 30.26 m L of 0.102 M NaOH t o reach an endpoint . a. Writ e a balanced equat ion for t he react ion of sulfam ic acid and sodium hydroxide.

b. Writ e t he t wo m ole fact ors for t he t it rat ion.

c. Writ e a unit plan for t he t it rat ion ( st oichiom et ry) problem .

d. Calculat e t he concent rat ion of sulfam ic acid in Lim e- a- Way in m olarit y.

e. What is t he concent rat ion of sulfam ic acid in % ( m : v) ? Writ e a unit plan first .

Answers: 1. b. 1.06 × 10 - 2 m ol c. 0.424 M d. 4.15 % - 197 -

2. d. 0.154 M e. 1.50 % CW03

Ch e m W or k sh e e t 4

Functional Groups ( How are organic m olecules classified?)

Information: Organic m olecules can be t hought of in t erm s of fu n ct iona l gr oups—cert ain arrangem ent s of bonded at om s t hat have predict able propert ies and react ivit ies. The t ables below list som e funct ional groups t hat are im port ant in organic chem ist ry and biochem ist ry. Table 1 cont ains t hose which we will nam e by t he syst em at ic rules agreed upon by t he I nt ernat ional Union of Pure and Applied Chem ist ry ( I UPAC) . These rules assum e t hat t he m ost im port ant funct ional group is assigned a ba se na m e , and ot her funct ional groups are nam ed as su bst it u e n t s. Most funct ional groups have separat e rules depending on whet her it is t he base nam e ( nam ed last ) or t he subst it uent ( nam ed first ) ; t hese differences are indicat ed in t he colum ns “ base suffix” and “ nam e when subst it uent ” in t he t able. Table 2 cont ains som e funct ional groups which are easier t o nam e by com m on ( or t rivial) nam ing rules. Som et im es m ore t han one nam e is given for an exam ple because I UPAC has adopt ed som e com m on nam es as accept able alt ernat es. For exam ple, t he m olecule form aldehyde is syst em at ically nam ed m et hanal. The second ( syst em at ic) nam e is m ore descript ive, but alm ost nobody uses it . You should be fam iliar wit h bot h nam es.

Table 1: Organic Functional Groups (Systematic IUPAC Naming Rules) funct ional group

alkane

alkene

alkyne

generic st ruct ure

R

C C

C C

base suffix

- ane

- ene

nam e when subst it uent

- yl

- enyl

- yne

- ynyl

benzene

phenyl-

exam ples CH3 CH CH3 CH3 2- m et hylpropane

Br

t rans- 2- pent ene

4- brom ocyclopent ane

CH3 C C CHCH3 CH3 4- m et hyl- 2- pent yne

Cl

arom at ic

isopropylcyclohexane

Cl

1,4dichlorobenzene

2- phenylhept ane Br

haloalkane

CW04

R- X

—

fluoro- , chloroet c.

- 198 -

CF2 Cl 2 dichlorodifluorom et hane

Br t rans- 1,2dibrom ocyclopent ane

Table 1 (continued): Organic Functional Groups (Systematic IUPAC Naming Rules) funct ional group

alcohol

generic st ruct ure

R- OH

base suffix

- ol

nam e when subst it uent

hydroxy-

exam ples O CH3 CH C OH OH 2- hydroxypropanoic acid

CH3 CH CH3 OH 2- propanol

OH

phenol

OH

phenol

—

Cl

Cl

2,5- dichlorophenol t hiol

aldehyde

R- SH

O R C H

- t hiol

- al

m ercapt o-

—

HS- CH2 CH2 CH3 1- propanet hiol

ket one

carboxylic acid

acid salt

am ide

O R C R

O R C OH

O R C O

O R C NR2

- one

- oic acid

- oat e

- am ide

—

—

—

—

- 199 -

O C

O H C H m et hanal ( form aldehyde)

H

O CH3 CH CH Cl 2- chloropropanal

benzaldehyde

O CH3 C CH3 2- propanone ( acet one)

O CH3 C OH et hanoic acid ( acet ic acid)

O CH3 C O Na sodium et hanoat e ( sodium acet at e)

O CH3 C NH2 et hanam ide ( acet am ide)

HSCH2 CH2 OH 2- m ercapt oet hanol

O CH3 C CHCH3 CH3 3- m et hyl- 2- but anone ( m et hyl isopropyl ket one)

O C

OH

benzoic acid O C

O

benzoat e O C

N CH3 CH3

N,Ndim et hylbenzam ide

CW04

Table 2: Organic Functional Groups (Common Naming Rules) funct ional group

est er

generic st ruct ure

nam ing rules

O R C OR'

nam e R’ as alkyl group; t hen like acid salt

exam ples O C

O CH3 CH2 C OCH3 m et hyl propanoat e

O

CH3 CHCH3

isopropyl benzoat e nam e each

acid anhydride

et her

O O R C O C R

R- O- R

O R C

as acid alphabet ically; t hen “ anhydride” nam e each R as alkyl group alphabet ically; t hen “ et her”

O O CH3 CH2 C O C CH2CH2CH3

O O CH3 C O C CH3

but anoic propanoic anhydride

et hanoic anhydride ( acet ic anhydride)

OCH3

CH3 CH2 OCH2 CH3 diet hyl et her

m et hyl phenyl et her

CH2OCH3

benzyl m et hyl et her

R- S- R

nam e each R as alkyl group alphabet ically; t hen “ sulfide”

CH3 CH2 SCH2 CH3 diet hyl sulfide

disulfide

R- S- S- R

nam e each R as alkyl group alphabet ically; t hen “ sulfide”

CH3 CH2 SSCH2 CH3 diet hyl disulfide

am ine

R N R R

nam e each R as alkyl group alphabet ically; t hen “ am ine”

CH3 NH2 m et hyl am ine

( CH3 ) 2 NCH2 CH3 et hyl dim et hyl am ine

am ino- prefix

O CH3 CH C OH NH2 2- am inopropanoic acid

HO CH2 CH CH3 NH2 2- am ino- 1propanol

sulfide

am ino group

- NH2

Notes: O C

x

Several funct ional groups cont ain a carbonyl group ( considered a funct ional group by it self.

x

When t wo or m ore funct ional groups are present , choose t he m ore im port ant one t o be t he ba se n a m e . I n general, t he m ore carbon- oxygen bonds cont ained by a funct ional group, t he higher it s im port ance.

CW04

- 200 -

) ; t he carbonyl is not

Exercises: 1. I dent ify t he funct ional group t hat corresponds t o t he following descript ions. a. A carbonyl group connect ed t o at least one hydrogen. b. An oxygen bonded t o t wo sat urat ed carbons. c. A benzene ring bonded t o a hydroxy group. d. A carbon bonded t o t wo ot her carbons and double- bonded t o an oxygen. 2. List all t he funct ional groups ( ot her t han alkane) in each of t he following st ruct ures.

a.

O CH3CH2 C CH2CH3

H3C

b.

H3C

c.

O Br CH2 C OH

d.

O CH3 H2N C CH2CHCH3

e.

f.

CH3 H3C C OH CH3

g.

OH O CH3 CH CH2 C O

3. Give a correct ( com m on or syst em at ic) nam e for each m olecule in Exercise 2. a. b. c. d. e. f. g. 4. Draw t he st ruct ure t hat corresponds t o each of t he following nam es. a. diisopropyl disulfide

- 201 -

CW04

b. but yl m et hyl et her

c. 2- m et hyl- 2- am inopropanal

d. benzoic anhydride

e. 2,4,6- t rim et hylphenol

f.

pent yl am ine

g. 2- m ercapt oacet ic acid

5. List all t he funct ional groups ( ot her t han alkane) in each m olecule nam ed in Exercise 4. a. b. c. d. e. f. g.

6. Read t he assigned sect ions in your t ext and work t he assigned problem s.

CW04

- 202 -

5

6

7

8

9

10

11

12

5

13

6

14

7

15

8

16

9

17

2

18

Kr 83.80 54

Br 79.90 53

Se 78.96 52

As 74.92 51

Ge 72.61 50

Ga 69.72 49

65.39 48

63.55 47

58.69 46

58.93 45

55.85 44

54.94 43

52.00 42

50.94 41

47.88 40

44.96 39

237.0

(244)

Pu

(243)

Am

(247)

Cm

(247)

Bk

(251)

Cf

(252)

Es

(257)

Fm

(258)

Md

(259)

No

(260)

Lr

U

Pa 231.0

Th

232.0

Np

175.0 103 173.0 102 168.9 101 167.3 100 164.9 99 162.5 98 158.9 97 157.3 96 152.0 95

150.4 94

(145) 93

144.2 92

140.9 91

140.1 90 238.0

71

Lu

70

Yb

69

Tm

68

Er

67

Ho

66

Dy

65

Tb

64

Gd

63

Eu

62

Sm

61

Pm

60

Nd

59

Pr

(289)

58

(283)

Ce

(277)

(272)

Ds (269)

Mt (268)

Hs (265)

Bh (262)

Sg (263)

Db (262)

Rf (261)

Ac

227.0

Ra

226.0

Fr

(223)

Rn (222)

At (210)

Po (209) 116

Bi 209.0

Pb 207.2 114

Tl 204.4

Hg 200.6 112

Au 197.0 111

Pt 195.1 110

Ir 192.2 109

Os 190.2 108

Re 186.2 107

W 183.9 106

Ta 180.9 105

Hf 178.5 104

La

138.9 89

Ba

137.3 88

Cs

132.9 87

Xe 131.3 86

I 126.9 85

Te 127.6 84

Sb 121.8 83

Sn 118.7 82

In 114.8 81

Cd 112.4 80

Ag 107.9 79

Pd 106.4 78

Rh 102.9 77

Ru 101.1 76

Tc (98) 75

Mo 95.94 74

Nb 92.91 73

Zr 91.22 72

Y

88.91 57

Sr

87.62 56

Rb

85.47 55

40.08 38

V

39.10 37

30

Zn

29

Cu

28

Ni

27

Co

26

Fe

25

Mn

24

Cr

23

22

Ti

21

Sc

Ca

K

Ar 39.95 36

Cl 35.45 35

S 32.07 34

P 30.97 33

Si 28.09 32

Al 26.98 31

Mg

Na

22.99 19

24.31 20

Ne 20.18 18

F 19.00 17

O 16.00 16

N 14.01 15

C 12.01 14

B 10.81 13

Be

9.012 12

Li

6.941 11

He

4

4.003 10

3

H

4

2

1.008 3

1

1

Periodic Table of the Elements