Introductory Chemistry Modules [11 ed.] 1506698565, 9781506698564

Citation preview

Periodic Table of the Elements l

lA

VIIA

VIIIA

I .ll08

I.OOX

-Ul03

H

H

He

IIA

2

3 4

5 6

7

IIIA

6.9-1 1

9.012

10.811

IVA 12.0 I I

Li

Be

B

C

I

VA ).U)()7

Vii\ 15.999

I

2

IX.998

20.179

N

0

F

Ne 10

3

-I

5

6

7

8

9

22.990

2-U05

26.982

2lUJR6

30.97-1

32.06

39.948

Na

Mg

Al

Si

s

35.-153

p

Cl

Ar

11

12

1118

\VB

VB

VIB

VllB

13

14

15

16

17

18

W.098

40.08

44.956

47,90

50.942

51 ,996

54.938

K

Ca

Sc

Ti

V

Cr

Mn

i - Vlll - - , 55.847

Fe

58.933

Co

58.71

Ni

1B

JIB

63.5-16

65.37

Cu

Zn

69.72

Ga

72.59

Ge

74 ,922

78.96

79 .90-l

83.80

As

Se

Br

Kr

19

20

21

22

23

24

25

26

27

28

29

30

31

32

33

34

35

36

85.-l68

87.62

88.906

91.22

92.906

95.94

98.906

101.07

102.91

106.-1

107.87

112.41

114.82

I 18.69

121.7S

12760

126 9()

13 1.30

Rb .n

Sr

y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

38

39

-W

➔I

42

43

44

45

46

-47

48

49

50

51

52

53

54

l .'2.91

1:l.7.3~

Cs

Ba

55

56

(223)

138.91 ~ 178.-19

180.95

IK:'-85

186.2

190.2

192.22

195.()1)

196.97

200.59

20-1.37

207.2

208.98

(209)

(2 10}

(222 )

La

Hf

Ta

w

Re

Os

Ir

Pt

Au

Hg

Tl

Pb

Bi

Po

At

Rn

72

73

74

75

76

77

78

79

80

81

82

83

84

85

86

226.03

57 (227)

(261)

(262)

(263)

(262)

(265)

l266J

(269)

(272)

(277)

Fr

Ra

Ac

Rf

Db

Sg

Bh

Hs

Mt

87

88

89

104

105

106

107

108

109

110

111

11 2

1-10, 12

1-10.91

1-14.24

( 1-15)

150.-l

151.96

157.'.lS

158.93

162.50

IM.9J

167.'.lo

168.93

173.0-1

174.97

Ce

Pr

Nd

Pm

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

j~

59

60

61

62

63

65

66

67

70

231.03

238.03

237.05

(244}

(2-13)

(247)

(2S-1)

(256)

(255)

71 (257)

Th

Pa

u

(25 1>

68 (257)

69

2.12.04

64 (247)

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

91

92

93

94

Lr

90

95

96

97

98

99

100

IOI

102

103

Uun Uuu Uub

I §

ai) aii)

----I ll)

aj

INTRODUCTORY CHEMISTRY MODULES

ail

* ELEVENTH EDITION *

a ii

mi ~

mit ~ ~

Anthony J. Pappas Marta E. Goicoechea-Pappas

XanEdu

• •

.

This document has been printed directly from the authors' electronic file.

Copyright©2018, 2016, 2015, 2008, 2003, 2000, 1997, 1996, 1993, 1992, 1990 by Anthony J. Pappas and Marta E. Goicoechea-Pappas Eleventh Edition All rights reserved. No part of this book may reproduced or distributed in any form or by any means. or stored in a database retrieval system without the written permission of the author. For information, contact the authors at Miami Dade College, Chemistry Department. 1101 1 S.W. 104 St.; Miami, FL 33176.

Printed in the United States of America

10 9 8 7 6 5 4 3 2 l

ISBN 978-1-50669-856-4

Xan Edu 530 Great Road Acton, MA 01720 www.xanedu.com

• • • • • •' ' ' '• '•

•

•

1 PREFACE AND ACKNOWLEDGMENTS The study of chemistry will provide you, the student, with the opportunity to study certain abstract concepts and to learn new problem-solving techniques necessary in understanding chemical phenomena. This modular study guide was created with several purposes in mind: 1. To provide students with a brief summary of those topics that are emphasized both in lecture and examjnations. 2. To provide exercises that are sirrular in scope to those used in examinations. The exercises are also intended to provide you - the student - with an opporturuty to master the topics that are covered in each module and that constitute the course competencies. 3. To emphasize the importance of problem solving. No matter what your career goals are, learning how to solve problems is indispensable; with this in mind , many of the problems provided have been designed to stimulate your intellectual curiosity and to encourage analytical reasoning and thinking. We are more than willing to give consideration to any feedback and thus welcome your comments as to the usefulness of this modular study guide , to any suggestions that you may have, and to any errors or ambiguities that have been inadvertently overlooked. The free and open source program - Jmol - was used to construct the molecular geometries used in this study guide. We would gratefully like to acknowledge the support of a Wolfson Minigrant in the preparation of this manuscript. We would also like to especially thank several of our colleagues (Larry Bray, Ana A. Ciereszko, Eileen Delgado Johann , Georgina C. Hart, Michael McGauley, Barry Moss, Maria E. Tarafa, Joan Lee, and Dionne Dickson) and students at Miami Dade College for providing corrections, consultations, and/or suggestions for the improvement of this manuscript. We would also like to thank Sally Jacobson , who input the initial manuscript into the computer, for her patience and perseverance.

Anthony J. Pappas Marta E. Goicoechea-Pappas

--

---------------------------------LIST OF MODULES Module

Page

l

Rounding Off Numbers , Scientific Notation, Significant Figures and Powers of I0

1-1 to 1-10

2

Mathematical Operations Involving Powers of Ten

2-1 to 2-7

3

Measurements and Conversions - Part I

3-1 to 3-10

4

Measurements and Conversions - Part II

4-1 to 4-14

5

Introduction to Matter: Elements and Compounds

5-1 to5-25

6

Electronic Structure and Chemical Periodicity

6-1 to 6-20

7

Chemical Bonding and Molecular Geometry

7-1 to 7-22

8

Writing and Predicting Chemical Formulas, Oxidation Numbers and Nomenclature

8-1 to 8-20

9

Calculations Involving Elements and Compounds Composition Stoichiometry

9-1 to 9-21

JO

Balancing Chemical Equations and Reaction Stoichiometry

I0- I to I0- I8

11

Chemical Reactions

ll -l tol l-18

12

Solutions and Solution Stoichiometry

12-1

to

12-14

LIST OF APPENDICES page

A,ppen 1x

Al

Metric Prefixes and Conversions

Al -I

A2

Sofobility Rules and Electromotive Series

A2-1

A3

Supplemental Problems for Modules 2, 3 and 4

A3-l toA3-2

A4

Exceptions to the Octet Rule

A4-l to A4-2

AS

Practicing Algebraic Manipulations

AS-I toAS-3

A6

Practice Questions for Final Exam

A6-l to A6-8

A7

Practice Multiple Choice Questions

A7-1 toA7-29

A8

Module Summary

A8-I to A8-26

A9

Mastery Exercises

A9-l to A9-42

II

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a

C C

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•

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SUPPLEMENTARY PODCASTS To access the free supplementary podcasts that have been developed for some of the topics discussed in this course visit iTunesU. In the SEARCH FIELD in iTunesU type

"Miami Pappas or Goicoechea-Pappas" and a list of all podcasts developed for this course will appear. You may either download the podcasts to your computer, iPhone , and/or iPod or you may watch them directly from iTunes.

lll

r

I i i i

II

II

II I I I I

' ' I I I I

'' I ''4

• • •

•4

•4

PRELUDE Chemistry is that branch of science (systematic study of the structure and behavior of the natural and physical world by way of

performing experiments and making observations) that deals the composi tion , structure and properties of

matter (an ything that has mass and occupies space).

The ancient premise was that everything was composed of Earth, W ater, Air, and Fire (Greek Classical Elements). Th.is premise has been superceded by the contemporary view that all matter is made up of one or more ~~'if~~@{J~~

than one of the more than 110 elements (substances that can't be broken down into simpler substances which are the p1imary constituents of all matter) that are found in the Periodic Table (a table which displays

-

• .... . ,

.,

n.r-.c~ . ... < ...

-

...-

..

·"'

lr"•.f'

,..

•

...

*.

.....

-'-c-•ar.r.~ •

chemical elements).

·•

Chemists, as well as all scientists, use a systematic approach (i.e., scientific method) in their investigations. It is through this approach that the many amazing advances in chemistry have been made . Basically. the scientific

method e ntails making systematic observations and performing experiments so that the formulation, testing, and/or modification of a hyporhesis (i.e .. proposed reasoned ex.planation) can be made. The Latin expression Nanos gigantum humeris insidentes basically conveys that intellectual gain s are built upon the work / research

performed by notable thinkers of the past . One s uch notable thinker from the past, w hi ch is called the father of modern c hemistry , is Antoine Lavoisier (1743 - 1794). Lavoisier stated the first version of the law of conservation of mass, recognized and named oxygen and hydrogen , introduced the metric system, wrote the first

extensive list of elements, and helped reform chemical nomenclature. In thi s and other chemistry courses you will further be exposed to the contribution that other scientists have made in advancing the discipline of chemistry , which has been traditionally grouped by the type of matter being studied or 11

the general field of material being studied. The five 11 main disciplines / branches of chemistry are: •

Organic Chemistry - deals with carbon containing substances.

•

Inorganic chemistry - deals with substances not containing carbon.

•

Physical Chemistry - deals with the description of the theoretical basis of the behavior of matter and energy.

•

Analytical Chemistry - deals with the development and use of techniques that help to identify and quantify the amount of material present in a sample.

•

Biochemistry - deals with the composition and changes in living species.

At first you may think that you are in a mathematics course. This is because for o ne to truly become proficient in chemistry, one needs to first master basic mathematical concepts. Once you have had the opportunity to master several mathematical concepts, units of measurement and convers ions - you will be guided through the rest of the competencies (e.g., introduction to matter, electronic and molecular structure , composition and reaction stoichiometry, chemical periodicity, nomenclature, reactions, solutions and solution stoichiometry) that make up this course. T he following puzzle is provided to help you review some of the main ideas discussed in this prelude.

II

.• .•

..•-"" "" --.. ..-.. ..• •.. ..• ..•• •• .• ill!

•• •• •• •• •• • ~

•

2 3

4

5

6

7

8

9

Across 7. a systematic approach that scientists use in their investigations that entails making observations, performing experiments and formulating , testing and/or modifying a hypothesis (2 words) 9. substance that can't be broken down into substances that are the primary constituents of all matter

Down l. a proposed reasoned explanation 2. branch of science that deals with the composition, structure and properties of matter 3. father of modern chemistry 4. branch of chemistry that deals with carbon containing substances 5. _ __ _ , Water, Air and Fire are the four Classical Greek Elements

6. branch of chemistry that deals with non-carbon containing substances 8. anything that has mass and occupies space Ill

ii ii

p II"

:; • p•

!(] •

•

=~ :i ., ., ., • • t

••

I - Rounding Off, Sig Figs. and Powers of 10

Introqyctorx Chemistry

ROUNDING OFF NUMBERS, SIGNIFICANT FIGURES, AND POWERS OF TEN

I.

The Number Line Many students have a difficult time working with negative numbers. The number line shown below provides a great visual aid to help you understand the magnitude (i.e., numerical value) of numbers. N u mbers are getting smaller. The more negative the number, the smaller the numbe r.

-4

-3

-2

- I

0

2

4

3

Numbers are getting bigger.

II.

Rounding Off Numbers The following are the rules that need to be followed when rounding off a number.

Examples of each rule are provided below showing how to round each example to the hundredth place. Examples Rules 1. If the digit after cut-off point is less than 5;

then, the digit before the cut-off point remains the same.

2. If the digit after the cut-off point is greater than 5 or 5 followed by digits that are not all zeros , then the digit before the cut-off point is increased by one.

Number

I .2014 0. 99j46

Answer 1.20 0.99

0.99j9 1 .29~5 1 1.29j501

1.00 1. 3 0 1.30

a) If the digit before the cut-off point is odd, then round this digit up.

1. 99 ~ 2. 17 j500

2.00 2.1 8

b) If the digit before the cut-off point is even, then retain this digit.

I. 98 !5 2 .00 [5 00

3. If the digit after the cut-off point is only a 5 or 5 followed by ONLY zeros , then:

A. J. Pappas and M.E. Goicoechea-Pappas

i

1.98 2 .00

1-1

lntroduc,to~ Chemistry

I - Rounding Off, Sig Eigs. and Powers of lO

Problem 1: Round off each of the following numbers to the first decimal a) d) g) j)

place (i.e., to the tenth place): b) 25.151 e) 6.7501

0.125 7.150 3.95 99.051

c) 3.250 t) 7.755 i) 17 .05

h) 3.95 1 k) 199.95

I) 199.949

III. Scientific Notation (SN), Exponential Notation, and Powers of Ten When numbers are very small or large, it is more convenient to write those numbers in exponential notation or more specifically in scientific notation. Regardless of a number's magnitude, all numbers can be expressed in scientific notation. A number written specifically in scientific notation has the following general form: Exponent

Any number from± I - 9

(Power o f Ten)

Any non-decimal number

'-... [Bl.

X

lOX

/

Coefficient

For example , J.2 1 x 10 3 , 2.9x 10·4 , and-1. x 10°. In exponential notation the number before the power of ten (i.e., the coefficient) does not have to be from 1 to 9 (e.g., 12.31 x 1056) . The following are examples that illustrate what a number rai sed to a power of ten means:

10° = 1 Positive Exponents 101

102

=

Negative Exponents 10-1 = -1

IO

10

= 10 X 10 = 100

=

0.1

10-2 = 2.. x 210

10

1 = -100 - = o.o1

To convert numbers into scientific notation , use the following guidelines: a) As you move the decimal place to the left (i .e., make the coefficient smaller), the power of ten (i.e., exponent) must increase by the same amount. For example, Making coefficient smaller by 3 powers of ten

"

Exponent must get iarRer by 3 powers of ten

1 7 5 0 .Q = 1~. 0 3 2

X

JQO =

l.75QQ

X

/

103

I

A. J. Pappas and M.E. Goicoechea-Pappas

l-2

Introductory Chemioo

I - Rounding Off._Sig Fi.g,s. and Powers....oU.Q

b) As you move the decimal place to the right (i .e., make the coefficient larger), the power of ten (i .e., exponent) must decrease by the same amount. For example, Exponent must get smaller by 2 powers of ten

Making coefficient larger by 2 powers of ten

"

0. 050 =0.0 50

X

10 O = 5.0

/2

10 ·

X

\....A.JI 2

Proble m 2: Write each of the fo llowing numbers in scientific notation (SN): a) 22 , 400.

b) 0.007

d) 22.4

e)

g) 0.00905

h) 0 .273

i) 0 .000000029

j)

k) 892. x 105

I) 5.72x 10-2

12.3

m) 0 .00072

X } Q-8

p) 0.02010

X

)0-9

n)

c) 8,000.

1.9

f) 0.00035

1982. X 10-9

q) -39.152

X

0)

108

0 .00200

106

X

r) -0.00 198

X

10-lO

IV . Inputting Numbers Having Exponential Notation Into a Calculator

Depending on your calculator one of the followinl keyl is used to input numbers written in exponential notation ⇒ EXP EE x l0x

l

I

I

I.

The following is an example of how to input numbers written in exponential notation into several types of typical calculator. Number

V.

Keys to Press

Calculator Display

1.2

X

JQ- 3

OJ □ [I]

] .2

X

10-3

1.2

X

10- 3

[I) □ [TI [ill ~ [I] □ [TI ~ [TI] [I]

I

EXP 1 1 I II]

w

I I I

1.2 -03

I

l .2"03 1 1.2 ,.,o-031

Significant Figures, Accuracy and Precision Significant figures , abbreviated as SF or sig figs, indicate how precisely measurements have been made. In a measurement, it provides the number of digits that are reported which are believed to be correct by the person pe rforming the measurement. The last number in the measurement is estimated.

A. J . Pappas and M.E. Goicoechea-Pappas

1-3

•

Introductory Chemistry

J-

4

RQuruling Off, Sig Figs..._aud_Fowers of IO

Precision is how closely individual measurements agree with each other and provides an indication of the level of detail of a measurement. A measurement in which more significant figures can be procured is more precise. A( I I

0 cm 8

( 0

cm

11[ Ii

I"

C

i

ii

Ii I ,

Not Drawn

'f ') I 1J

Length of Paper Clip Using Ruler A and B II II

I

2.45 cm (3 SF) More ,,.,;,,.

t

3

measurement.

Estimated Digit

I

3

lo Scale

Accuracy is how closely a measured value agrees with

neither accurate or precise

set

2~cm 12sF] ~;:::: : : ,

precise,

•• ••

precise and accura1e

but not accurate

To find how many significant figures (SF) a number has: 1) locate the first nonzero digit (start from left

value.

~

right)

First non-zero

dit

0.002500

2) then starting with the first nonzero digit -

'J1ox ). multiply the exponent by the reciprocal of the root (i .e., 1/>).

Solution: Problem 1: a) 10-9

X

Calculate the following: 10- 12

d) 101/10 -9 g)

II.

Juli

Multiplying and Dividing Numbers Having Powers of Ten 1) Place the powers of ten together and the coeffic ients together.

2) The final answer must have the same number of significant figures as the coefficient with the least num ber of sign ificant figures (SF). 3) Round off correctly and preferably report the answer in scientific notation (SN). A. J . Pappas and M.E. Goicoechea-Pappas

2 -1

ln.trJ ductoi;y Cbe.mis=truY- - ~~=~ = -Z--~M~a~th,_,Operations Involving Powers of JO

Example E:

( I .76 x 10200) x (7 .650 x 10-8) = ? 3 Sf

4 SF

Solution: ( 1.76 x 7 .650) 13 .464

J .35

X

X

(

(1 0 200 x 10-8)

x

10200 + (-8))

\01 93 . - -

=

13 .464

X

10 192

FinalanswerwiththccorrectnumbcrofSF ( i.e .. 3), rounded off correctly and wrmcn m SN.

Example F: Divide 1.760 x 10 125 by 35 .20 x 10- 20 4SF

1760xlo125

4SF

Note that 1.760 xlQl2S / 35.20 ,

10 125

1 .760

x (10 125·(-ZO))

Solution: - - x - -2-0 = 0.05000 35.20

10-

- 5 000 -

Problem 2: a) (2 .0 b) C)

•

X

l0

143

.--

~

x 10-20 = -·- - - 35.20 X 10- 20

= 0.05000

x

10 145

Final answerwiththecorrectnumberofSF(i.e .. 4). rounded off correctly and written in SN.

Perform the following calculations. Report each answer in scientific notation with the correct number of significant figures.

.,

r

10-225 ) (6.Q X 10 102) =

X

4.80 X I Q-107 X 5.00 x 10- l l = (4.40 X to-l0?)/(2.2

d) (2 .00

X

1025)/(8 .00

X

X

10-8) = 1o-&O) =

a a

e) Multiply 1.50 x 10- 5 by 5 .5 x 10· 3

f)

~ • vr

••

Divide 5.00 x l OlOO by 12.5

g) 2.QQ X }Q-I02

6.00 X 10-I0? =

X

h) -2 .00 x 10- 102 x 6.00 x 10 109 = i)

5.00

j)

-V16.xl0 200

X

}Q· lO?

2.5

X

X

X

}010 l =

1099

k)

1)

Vo.064 0 x 1 0 300

••• n•

-4.00 x l0200= =

NOTE: Unless otherwise infonned. when no coefficient is wrinen in front of a POT. assume that the coefficient is exactly I.

■ b b

•►

-►

0.00800

------- = (20 0. X 5.0 X lQlOO)

A. J. Pappas and M.E. Goicoechea-Pappas

.■

2-2

il

I

~

Introductory Chemistry

2 Math Operations Involvinj! Powers of JO

III. Adding and Subtracting Numbers Having Powers of Ten

1) The exponents must have the same power of ten before addition or s ubtraction is performed. If the exponents don't have the power of ten, it is usually best if you manipulate the coefficient of the number having the smaller exponent in such a manner so that its power of ten is made the same as that of the other number.

2) Once the powers of ten are the same, the coefficients can be added or subtracted. T he exponents are neither added or subtracted. 3) The final answer must have the same number of decimal places as the coefficient (after any necessary exponent manipulations have taken place) with the f ewest decimal places . 4) Round off correctly and preferably report the answer in scientific notation (SN). Example G : 1.86 x 10- 15 - 9.27 x 10- 16 = ? Solution: 1.86 x 10-15 - 0.927 x 10- 15 =

1.86

X 1o -15

In this case, the number with the smaller POT ~~-e., 10· was converted to a larger POT (i .e .. 10· ).

- 0.92 7

X

10-JS

0.93 3

X

10-lS ____. 0.93

2SF

- 9.2 7

9.3 3

X

X X

)0-l 6 ]0-l 6

10-lS = 9.3

X

10- 16

N OTE: Upon exponent manipulation. the number of SF must remain the same.

OR

18.6

2 SF

X

16 )

In this case. the number with the larger POT (i .e .. 10· 15) was converted to a smaller POT (i.e., 10· 16).

10- !6 _ . 9.3

X

10- 16

Note that the number of significant figures (i .e., 2 SF) obtained is the same regardless of which number was manipulated. Example H: 9 .76 x 10 101 +

8.65 x 10 100 =

Solution: Since the exponents are not the same, manipulate one of the numbers (e.g., 8.65) so that it has the same exponent (i.e., 10 1)

as the other number. 9.76 + 0.86 5

X

10 101 10101

10.62 5

X

10 101 --+ 10.62 X 10 102 = 1.062

X

Used Rule 3a for Rounding Off

4SF

A. J. Pappas and M.E. Goicoechea-Pappas

Answer written in SN .

4SF

X

10 102 2-3

• a p!ii • I"""". :z)

Intmduc.tory C.b.emistry,

Problem 3:

2 Math Operations Involving Powers of 10

Perform the following calculations. Report each answer in scientific notation with the correct number of significant figures.

a) (9.800x 1Q· l04) + (9.200x 10· 103) = b) (1.20 x 10475 ) - (0.180 x 10476) =

,...-::

:t

c) Add 92.l an.d 9.9 d) Subtract 0.10 1 from 1.25 x 10-2

e) 4.7 1 X 105 + 9.92 X 106 =

f) 3.69 X 105 + 8 .1 2 X }05 = g) 9.999 X 108

+ 9

107 =

X

wp

h) 5.123 x 10·9 + 6.21 x 10- 10 i) 4.70 X

105

j) 9.69 X

105

9 .920 X

106 =

9.1 20 X

105

• µ-

• r•

=

k) 2.73xlo- 101 + 9.70xt0· 102 + 9.880x10- IOO

=

IV. Mixing Mathematical Operations

In these problems first take care of addition and/or subtraction within parentheses - properly round off the answer and make sure that it has the correct number of significant figures. Then proceed to do any exponentiation. multiplication and/or division. If the answer needs to be reported in scientific notation (SN) do this manipulation at the end. Make POTs the same.

Example I:

(9950. + 0.765 X 10 2 )

9 9 . 50

X

102

Q. 7 6 5 X 10 2

+

( 1.751 X 10- 3 ) 5 X 6.800 X 10 22 0

l OO . 2 6 5 x l 0

100.26 X 10 2

2

Round off to 2 decimal places; however. number has 5 SF

1 02

100.26

Solution: - - ----=------ --- - x - -3 -= 5 220 5 22 3 5 (1.751 X 10-

_ _1_0_0._26_ _ 16.46 X 6 .800

X

)

X

6.800 X 10

-

102 - (-15) - 220 :

(1.751)

X

6.800

0 .8958 X 10·203 =

NOTE: For addition and subtraction one needs to keep track of decimal places (when the numbers have the same POT) .

(10- ) X 1 0

0

8.958 X 10·204

Final answer written in scientific notation (SN) with the correct number of significanl figures (SF).

For multiplicatio n and division one needs to keep track of S F.

A. J. P appas and M.E. Goicoechea-Pappas

:I :;; .• • .-•= -

2-4

·= •

• •

5 ..

• i .. ■ b◄

•• •

2 Math Operatjgns Involving Powers of 10

lmm..ductory Chemistry

Problem 4:

a)

b)

c)

d)

e)

Perform the following calculations.

Report each answer in scientific notation with the correct number of significant figures.

(0.165 X 10 1

+ 9.35)

=

5.500 X 10 1

(1.75 X 10 2 ) (2.20 X 10- 3 ) (625 - 6.20

(7.48

X

X

=

10 2 )

10- I02 + 639.

X

10-I0 4)

(32.00)( 4.690 X 10- 99 )

=

144 X 10 3 00

=

4.00

1/(64.0

X

10 102 )

(20.70 - 20.20)

-

f)

3.5581 X 10 8 4.99x 10-9 X 3. 11 X 10-7

g)

(0.17200 - 0.16700) 0.002250 =

h)

0.00000172 X 0.572 0 .0000000000090 X 0 .0755112 =

i)

(3.21 xlo- 101

j)

0 .00000000033910 0.0000000000131 x 273 . =

Problem 5:

-

=

3.3xlo- 102)7.799xI0 100 =

Perfonn the following calculations.

Report each answer in

scientific notation with the correct number of significant figures. a) 6.00x 10- 112 x 2.00 x 10-115 = A. J. Pappas and M .E. Goicoechea-Pappas

2-5

lntroductory_C,...b..,.,em .....i..... st~tY' - - - - - - - - -- 2~~ti.onsmolYi..oJt.Powers of 10 b)

8 .00 x 10- 114 x -4 .00 x 10 112 =

c) -8.0x 10- 114

-4.0x10 112 =

x

d) 0.00 120 x 0 .00500 x 103 = (270 .1 - 273 .l) (2. 19 1 x 10-5) 6.712

e)

f)

104

=

(2.5 1 X 10-2 )2 (0.007 11 2)2 (6.7 1 X IQ-3) 2 = 2.51

g)

X

X

lQ-S X 7 .21 X l Q-8

1.oo x 1.oo x 6 .2 x 1o-s

=

.....-

(3 .11 x 1o-8)3 (7 .121 x 1o-6)3 h) (5.29 x 10 6 ) 3 (2.1 1 x 10-2)2 =

5.123 X ] 9 - 6.21 X 10-IO =

j)

7.88

X

10-l l7 - 5 .2 11

4.J (256 X k)

•

o-

i)

X

10-116

.•" .-! ·= •• •• ---= ..-.. ... = -..

M,

=

lQZOO)

(1 0. 70 - 10.50) -

Ill II

1)

(2. 000 X 10

5

)

(115.700 - 115.20 0)

II q

=

A. J. Pappas and M.E. Goicoechea-Pappas

..-:t

2-6

~

.....

-....... ....-... ..

· .!t..P~ers..of

~Chemi

Q

ANSWERS 1.

2.

a) 10-21

b) 103

c) 103

f) 1020

g) 106

h) 10 3

a) l.2xlo-l22

e) 8.2 X 10-& g) 1.20 X 10-208

h) -1.20 x 108

i) 1.2 X 10- 5

j) -l.6 xI0301

k) 2.50x 10- 1

I) 8.0 X 10-!06

a) 1.01 80 X 10- lOl

b) -6.0 x 10474

C) 1.020 X }02

d) -8.8x l0-2

e) 1.039 X ]07

f) 1 .18 1 X 106

g) 1.09 X 109

h) 5.744 x 10-9

i) -9 .450 X 106

j) 5.7 X 104

X

e) 10·4

b) 2.40 x 10· 117 d) 2.50 x 10 104 f) 4.00 X 1099

C) 2.0

3.

1o-99

d) 10 16

k) 1 .0250 x 1o -99 4.

5. ~

_, _, _,

~

.., ~ ~ ~

~ lll'J)

a) 2.000 x 10-l

b) 8 x 10-2

C) 9.242 X 10-S

e) 8.0 X 1034

d) 6.00 x 10 150 f) 2.29 X 1023

g) 1.12 X 10-S

h) l.4 x 106

i) 2.25

j) 9.48

a) 1.20 X 10-226

b) -3.20 x 10- 1

c) 3.2 X 10-l

d) 6.00x 10-3

e) -9.8

t) 7 .08 X 10-4

X

10-JO

X

10-2

g) 2.9 X 10-S

h) 1.65 X 10-55

i) 4.502 X 10· 9

j) -4.423 X 10-l 16

k) 2.0 x 10 51

1) 4.00

A. J . Pappas and M.E. Goicoechea-Pappas

X

105 2-7

l

-~ •• IA

3 - Measurements and Conversions · ,Par.tJ

Introductory Cbemistry

MEASUREMENTS AND CONVERSIONS - PART I I.

Use of Conversion Factors in Calculations

For a commonly known relationship (i.e., equality):

1 ft = 12 in The respective conversion factors to above equality are: 1 ft 12 in

or

12 in 1 ft given quanrity { wi1h uni1

The conversion factor that is used in

calculations is the one that allows for unit cancellation. The example to your right shows how 24 inches can be converted into feet by the use of one of the above conversion factors.

? ft what is desired

=

24)6' (1~¥) ~

=

2 ft

un"vante d units

cancell ed o ut

Example A: How many cents are there in 5 dollars? Solution:

Known equality

⇒

1 dollar= 100 cents

cents) = 500 cents

? cents = 5 dol)drs ( 1OO 1 do)rar

Problem 1: Given the following time equalities, perform the following one or two step conversions: 1 min= 60

sec

1 wk= 7 days

1 hr= 60 min 1 month = 4.35* wk

1 day= 24 hr

1 yr= 12 month

* The values, 4.35 has been rounded off to 3 SF. The other values are exact.

a) 25 hr= _ _ mm

b) 44 hr= _ _ days

c) 2.5 months= __ days

d) 0.50 wk = _ _ hr

II. The Metric System and International System (SI) of Units The metric system was developed with the intent of simplifying conversions among units of measure. The metric system is a decimal system in which units are related to each other by powers of ten~ prefixes

are used to indicate either fractions and multiples of ten. A. J. Pappas and M.E. Goicoechea-Pappas

3- l

- ~meots and ConY.ersioos - Ea

The International System of Units (abbreviated as SI Units from the French Le Systeme international d'unites) is the modern form of the metric system that has been adopted by scientists to provide additional uniformity to units used in science. This system of measurements is nearly universally employed and most countries do not even maintain official definitions of any other units. A notable exception is the United States of America, which still uses the "English System" of units in addition to metric, and SI units. The following table show the seven SI base units.

Unit

Symbol

Measured Quantity

meter

m

length

kilogram

kg

second

s

mass time

Kelvin

K

temperature

mole

mol

amount of substance

ampere candela

A

electric current

cd

luminous intensity

There are other units called SI derived units that can be obtained from the seven SI base units [i.e., m3 ⇒ unit of volume (e.g., dm3 = L, cm3 = mL)]. A. Basic Units in the Metric System Mass

Length

Volume

grams (g)

meter (m)

liter (L) -4L

"'1 g

It's a

,trl /J

-3 kg

.. 1mm

... soµm

A. J. Pappas and M.E. Goicoechea-Pappas

I fl ot

1 ml = 1 cc =1 cm3 1 L= 1 dm3

3-2

3 - Measurements and Conversions -~J

Introductory Chemjstey

B. Prefixes to Know

Prefix

Symbol

tera

T

1012

( 1,000 ,000 ,000 ,000)~

giga

G

109

( l ,000 ,000 ,000)

mega

M

106

( I ,000 ,000)

u.SA: billion

kilo

k

103

(1,000)

EI sewhere: thousand million

deci

d

10- 1

(0.1)

centi

C

10-2

(0 .01)

milli

m

10-3

(0 .001)

micro

µ

10-6

(0.00000 I)

nano

n

10-9

(0 .00000000 I)

pico

p

10- 12 (0.000000000001)

Numerical Value

V

USA: trillion Elsewhere: billion

~

The above prefixes are synonymous with their numerical value and the following equalities can be written: 1 ML _

=

1 km =

6

10 10

3

-2

L

m

1 £g = -=cl--=-0- g Problem 2:

M k

= 106

= 10 3

1 msec 1 gm

= 10-3

=

-6 10 -9

d = I o- I

1 ng = 10

c =l0-2

lgs = 10

- 12

sec m

g

s

3

m = 1oµ = 10-6

n = 10-9 p = 10-12

Predict the following:

a) The length of a closet (3 km, 3 dm, or 3 m) b) The length of a pencil (18 cm, 18 dm , or 18 µm)

c) The width of a paper clip (10 mm, lO cm, or 10 Mm)

d) The distance from someone's home to school (4 km, 4 m , or 4 cm) e) The mass of a pair of socks (20 kg, 20 cg, or 20 g)

f) The mass of a cat (1.4 g, 1.4 mg, or 1.4 kg) g) The mass of a foo tball player (0 .120 kg, 0.120 dg , or 0.120 Mg)

h) The volume of liquid that can fit in a spoon (4 L, 4 mL, or 4 kL) i) The volume of paint in a large paint can (4 ML, 4 L, or 4 dL)

j) The volume of water in a cup (2.5 dL, 2.5 GL, or 2.5 L) A. J. Pappas and M.E. Goicoechea-Pappas

3-3

3 - Measurement~ andJConversions. . :~EanJ

Introductory Chemistry

To interconvert between metric units, one or two steps are needed. One step is needed when converting from a prefixed unit to one of the basic (base) units or from a basic (base) unit to a prefixed unit. Two steps are needed when converting from one prefixed unit to another. The process is as follows: p refix unit 1

step a

step b

base unit

dg

prefix unit 2

Mg

g

Example B: Convert 5 .63 m to cm.

Solution:

? cm= 5.63 m

( 1cm ) =

5.63 x 102 cm

1 0- 2 m

Example C : How many liters are in 274 mL?

Solution:

10 3 ? L = = 274 mL ( 1ml

L)

= 2.74 x 10- 1 L

Example D: How many kg are in 2 .94 g?

Solution:

? kg

= 2.94 g (

1

=

~g ) 10 g

2.94 x 10 -3 kg

Example E: Convert 35.4 dm to mm. Solution:

? mm= 35.4 dm (

10

1

m) (110-mmm ) = 3.54 x103 mm

1 dm

3

Example F: Convert 5 .63 x 10-3 m to ~tm.

Solution: ? cm= 5.63 x 10- 3 m (

1

µm ) = 5.63 x 10 3 µm 6 m

10-

Example G : How many kg are in 5 .32 mg?

Solution: ? kg

=

3

g) ( 1 kg ) = 5.32 x 10-6 kg

105.32 mg ( - -1 mg

10 3 g

Example H: How many pg are in 2.94 x 10-4 Gg? 4

10

9

g) ( 1_pg g ) = 2.94 x 10

Solution: ? pg = 2.94 x 10- Gg ( Gg 1 A. J. Pappas and M.E. Goicoechea-Pappas

10

12

17

pg

3-4

3 - Meas.urements and C,o.oyersiQJ.l - a.cLJ

lntmd11ctory Ch~_try

Problem 3: Do the follow ing conversion problems . a)

b) c) d)

e) f)

g) h)

15 dm to meters 5 .2 kg to grams

15 µm to cm 4.33 mg to kg j) k) 17.5 g to mg 64.9 kg to mg I) m) 2 .0 mL to µL n) 160. L to kL o) 60 .l cg to µg p) 0 .00526 mm to cm i)

2 rnL to liters

75 cm to mm 540. mm to km 50. kg to cg 45 dg to kg 14 g to µg

III. The English, American or Non-Metric System The English , American , Non-Metric or SI system is the system that is commonly used in the United States. Compared to the metric system, it is a mess! Commonly used equalities in the English , American or Non-Metric System.

Volume

Mass

1 pt= 16 fl oz

1 lb= 16 oz

1 ft= 12 in

1 qt= 2 pt

1 to n = 2000 lb

1 yd= 3 ft

1 gal= 4 qt

Length

1 mi= 5280 ft

The values for each of the above equalities are exact (i.e., each has an infinite number of significant figures). The process of interconverting among English-English units may involve several steps. IV. Metric

~

English Conversions

In order to interconvert between metric and English units and vice versa, you must memorize the following conversions which bridge the gap between these two systems, as well as, the metric and English conversions given above.

Volume I qt

=

0.946* L

Mass I lb= 454.* g

Length 1 in= 2.54cm

* The values, 0.946 and 454., have been rounded off to three significant figu res. The other values are exact.

A. J . Pappas and M .E. Goicoechea-Pappas

3-5

is)

3 - Measure.ments and Co_nyersions -

Int1'09JLC1,2ry Chemistry

Example I: Convert 5.00 qt to liters.

Solution: ? L

0 .946

= 5.00 qt ( - --

1 qt

L) = 4.73 L

Example J: How many megagrams are there in 500. tons? 2000

lb) (454 g) (1 Mg) = 454 Mg ---

Solution: ? Mo= 500. tons ( - - 0 1 ton

10 6 9

1 lb

Example K: Convert 0.75 pints to microliters .

qt)

1 Solution: ? µL = 0.75 pt ( - 2 pt

L) ( 1 _µL ) = 3 .5 x 10s µL

(0.946 ---1 qt

10

6

L

Example L: How many mare in 5 ft 2 in? We chose to convert feet into inches because it takes one less conversion step to get into the metric system m doing it this way.

Solution: a) Convert 5 ft to inches .

?m

= 5 ft

(1 2

in) = 60 .

l ft

b) Add the inches that there are in the 5 ft to the 2 inches

60 in + 2in

= 62in

c) Convert the total number of inches to meters

cm) ( -10-- -m)= 2

2.54 ? m = 62 in ( . 1 in Example M:

1 cm

1.6 m

A bottle of smartwater® says that its content of wate r is 1 pt 7.7 fl oz. How many mL of water are in the bottle?

Solution: a) Convert 7 .7 fl oz to pints (pt) ? pt = 7 .7 fl

oz

(161:itoz) = 0.48 pt

b) Add the number of pt that there are in 7. 7 fl oz to the 1 pt 1 pt+ 0.48 pt= 1.48 pt (assume th at the 1 pt is exact) c) Convert the total number of pt to mL

? mL = 1.48 pt

(1 qt) (0.946 L) ( 1mL ) pt qt 2

A. J. Pappas and M.E. Goicoechea-Pappas

1

10- 3 L

=

700. ml

3-6

., •F

•:

•• ••

4

3 - Meas.urements and Conversions - Part I

lntroducrocy Chemisrry

Example N:

A 6 lb 12 oz baby has a mass of how many decigrams?

Solution : a) Convert 12 oz into lb

? lb= 12 oz ( l lb ) 16 oz

= 0.75 lb

b) Add the number of oz that there are in 12 oz to the 6 lb

6 lb+ 0.75 lb = 6.75 lb (assume that the 6 pt is exact) c) Convert the total number of lb to dg ?dg = 6.75lb (

4549 1 lb

) ( ld: )= 3.06x l 04 dg 10-

g

Problem 4: Do the following conversion problems. a) 4.75 L to q t b) 5.00 quarts to liters c) 20.0 gal to rnL d) 2 1.0 qt to µL e) 40. g to tons f) 1.00 gal to mL g) 5.00 lb to kg h) 36.0 in to cm i) 16 in tomm j) 20.0 mm to in k) 100. m to ft 1) 100. yd tom m) 9.48 ft to cm n) 100.0 m to yd o) 1.1 2 kg to lb p) 7 lb 3 oz to kg

V. Using Percent / Percentages as Conversion Factors Fundamentally, a percent tells you how many of a particular item are present in a group of items that has exactly 100 items. For example, if 12% of the coins in a wallet are pennies, one could say that if there were 100 coins in a wallet, 12 would be pennies. The following equation can be used to obtain the percent of a particular "item" in a group of items. [NOTE: The sum of . . # a f items of interest all percentages must % item of interest = total# of items x 100 add up to 100.]

To solve unit conversion problems in which the percent of a particular 11 item11 is given, it is advantageous to use the given percent as a conversion factor. For example, given that 20% of the students in a class are females , this implies that 80% of the class are males and depending on what is being given and what is being asked for, one of the following conversion factors can be used: 20 females

100 students

80 males

100 students

100 students

20 / emales

100 students

80 males

A. J . Pappas and M.E. Goicoechea-Pappas

20 females 80 males

80 males 20 females

3-7

3 - MeaSJ.1re.ments and Conversions - Part]

lutroductotY, Chemistry

Example 0 : The composition of a 14-karat gold ring was found to be 11.66 g gold and 8.63 g copper. What is the percent (by mass) of copper in the ring? 8.63 g copper

m

Solution: w copper = ______;____ • 100 = 42.5% copper . ) (.mnng

(11.66 + 8.63) g r ing

[NOTE: If 42.5 % of the ring was made out of copper, then the remaining 57.5% (100 - 42.S) was gold]. Example P: How many girl students are in a class of 46 students in which 37% are girls?

37% girls ⇒ 100 students= 37 girls= 63 boys Solution: ? girls = 46 students (

37

girls

100 students

) = 17 girls

~

[NOTE: If there were 17 girls in a class of 46 students, then it is implied that 29 students (46 - 17) are boys.] Example Q: A 500. mL bottle of booze only contains water and alcohol. If the label reads: Alcohol Content 30% by volume, how many mL of water are in the bottle of booze? 30% alcohol =- 100 ml booze

= 30 ml alcohol = 70 mL water

(by volume)

[NOTE: If there were 350 mL of water in the bottle of booze then there are 150 mL of alcohol (500 - 350) Example R: If your score on a multiple choice exam was 75% and you got 93 questions correct, how many questions did you get wrong? ~

57

••

•• • •

~ •

s

I

70 mL water ) 2 Solution : ? mL water= 500. mL booze ( - - - - - = 3.5 x 10 mL 100 mL booze

75%

•• •• r ••

JOOquestions= 75 correct= 25 wrong

(% of questions

that were correct)

Solution: ? wrong = 93 correct

2 5 wrong) ( 75 correct =

31 wrong

[NOTE: If you got 93 questions correct and 31 questions wrong, the exam must have had 124 questions (93 + 31).] A. J. Pappas and M .E. Goicoechea-Pappas

3-8

•

•• •

:a

•• ••

...... .... -a

;; ~

lntroductor:v Chemistcy

3 -,Measurements and CoOYe.rsions - Part l

Problem 5: a)

b) c) d) e)

t) g)

A 25 .0 g mixture containing salt and water is 2.2% salt by mass. What is the mass (in g) o f salt in the mixture? A class of 48 students has 36 girls, what percent of the class are boys? An assortment of coins contains 20 pennies, 2 nickels, and 6 dimes. What percentage of the coi ns are nickels? There are 20 male students taking a class. If 40% of the students are males, then how many students are in the class? 5% of the volume of a nameless brand of beer is alcohol. ln a 12 fl. oz. can of this same beer, how many mL are NOT alcohol? A sam ple of air was found to contain 2 1% oxygen by volume. How many mL of oxygen are present in 2.0 L of air? The mass composition of a rock is: 20% spodumene, 32% quartz, 27% albite, 14% microcline, 6% muscovite, and small traces of other minerals. If the rock was found to contain 1.6 g of quartz, what is the mass (in g) of the rock?

Problem 6: A Hodgepodge of Conversion Problems a) How many feet are there in 200. cm? b) Convert 2.36 kilograms to milligrams. c) How many milliliters are there in 3.25 liters? ct) How many millimeters are in 9 .46 centimeters?

e) Calculate the distance in feet covered by the participants in a l 00. meter swimming race.

f)

What is the mass in mg of 64.9 kg?

g) What is the volume in mL of 6.00 quarts? h) How many pints of gin will a 3 .70 liter flask hold?

i) Calculate the length in yards of a wire which is 15.0 cm long. j) What is the mass in kg of an object that weighs 425 .6 cg? k ) The 10 ,000. meter run is one of the events in the Olympic Games. What is the distance in miles? 1) A 2.00 liter bottle of soda is equivalent to how many quarts? m ) Convert 2.5 gallons to liters. n) A byte (B) is a unit of digital information that most commonly consists of eight bits . How m any bits are there in 1 TB ? o) Using the time equalities provided on pg 3 -1 and given that 1 jubilee is equal to 50 yr and that 1 fortnight is equal to 2 wk, how many fortnights are there in 5¾ jubilees? (Use 3 SF to report your answer)

A. J. Pappas and M.E. Goicoechea-Pappas

3-9

l

ANSWERS 1.

a)

l.5

X

10 3 min

b) 1.8 days

2. a) 3m c) e)

10mm 20 g g) 0.120 Mg

i)

3. a) c) e)

4L 1.5 m 2 x 10-3 L

5.40 x 10-4 km g) 4.5 x 10-3 kg i) 1.5 x 10- 3 cm k) 1.75 X 10 4 mg m) 2.0 x 10 3 µL o) 6.01 X 105 µg

j

I:

I

c) 0.21 yr b) 18 cm d) 4km f) 1.4 kg h) 4mL j) 2.5 dL

b) 5.2 X 103 g d) 7.5 x 10 2 mm f) 5.0 X 106 cg h) 1.4 X 107 µg j) 4.33 x 1o- 6 kg I) 6.49 X 107 mg n) 1.60 x 10-l kL p) 5 .26 x 10-4 cm

4. a) 5.02 qt c) 7.57 x 10 4 mL e) 4.4 x 10-5 tons g) 2.27 kg i) 4.1 x 10 2 mm k) 3.28 X 102 ft m) 2.89 x 10 2 cm o) 2.47 lb

b) d) f) h) j) 1) n) p)

5. a) 0.55 g salt c) 7% nickels e) 337 mL g) 5.0 g

b) 25 boys d) 50 students f) 420 mL oxygen

6. a) c) e) g) i) k) m) o)

b) d) f) h) j) I) n)

6.56 ft 3.25 X 103 mL 3.28x 10 2 ft 5.68 X 103 mL 1.64 x 10- 1 yd 6.2137 mi 9.5 L 7 .50 x 10 3 fortnights

d) 84 hr

4.73 L 1.99 x 107 µL 3 .78 x 10 3 mL 9.14 x 10 1 cm 7 .87 X 10-I in 9.14xl0 1 m 1.094 x 10 2 yd 3.26 kg

2.36 x 106 mg 9.46xI0 1 mm 6.49 X 107 mg 7.82 pt 4.256 x 10-3 kg 2.11 qt 8 x 10- 12 bits

!

I·

A. J. Pappas and M.E. Goicoechea-Pappa

3-10

'

' ~

I

Introductory Chemistry

MEASUREMENTS AND CONVERSIONS - PART II

if

., iJ

4 -Meas.urements and CoUYersions - Part II

I.

Square and Cube Units of Length

it ii ii

When conversions require a square unit of length (area) or a cubic unit of length (volume), you must first write the conversion fac tors as you normally would for conversions not involving the square or cubic unit, then square or cube the numbers and units in those factors that are squared or cubed , and finally multi ply or divide accordingly.

it

Example A: How many square feet (ft2) are in 6.12 x 105 cm2 ?

-• ...... ........

Solution: First write conversions without squaring them; then square them in the next step . 1 in 1 ft 2 2 5 ? ft = 6.12 x 10 cm x 2.54 cm x 12 in

= 6.12 X [05 cm2

....

...

5

X

= 6.59

X

r

2.;~nCffi

x (2 .5 4)2 cm2

= 6.12 X 105X

~

~ ~

= 6.12

~

= 6.59

~

~

!Ill!) ~

"" ~

--~

X

(2_~4 i'2)2 X

105 X (3 .28 X 10-2) 2

= 6.12 X 105

~

~

X

x

l._f!_12 ( 12 in) (1)2 ft2 (12)2 in2

}02 ft2

~ ~

X

(1)2 in2

2

= 6.12 x 10 cm

111111

(

1.07

X

X J0-3

Instead of squaring each of the above numbers individually , it is much easier to multiply and divide all the numbers that are to be squared and then to square the resulting number.

102

Example B: How many ft 3 are in 0.500 km3? Solution:

ft

? 3 = 0500 km 3

i::) (1~~m)\ (2~1:mr

1 X (

3

x

X

(i~~tnr

(103)3 m3 (1)3 cm3 (1)3 in3 (1 )3 ft3 = 0.500 km3 x (1 )3km3 x oo-2)3 m3 x (2.54)3 cm3 x (12)3 in3

= 1.77

X

10 10 ft3

A. J. Pappas and M .E. Goicoechea-Pappas

4-1

Instead of cubinv each of the above numbers individua11y , it is much eas ier to multipl; and divide all the numbers that are to be cubed and then cube the resulting number. For example, 1Q3 1 l 1X 10·2 X 2.54 (

= 0 .500 X = 0.500

= 0.500

(3.28

X

3.53

X

X

X

1)3

X

}2

103)3

1Q10

= 1.77 X l 0lO Example C: The volume of water inside a car's radiator is 6.0 liters. Calculate thi s volume in cubic inches . (1 mL = 1 cm3)

Solution: ,,

l mL

-' - 60L -· m - · · x 10- 3 L

?.

(1 in) 3

1 cm 3

-x 1 mL x (2.54 cm) 3

_ . 3 2 10 37 10 · x ·

Problem I: Convert the follow ing a)

25. cm3 to mm3

b) 450 . m2 to ctm2

c)

0.0723 krn 2 to m2

d)

1 .5 f t3 to L

e)

3.4 in2 to cm2

t)

8 .00 in 3 to mL

III. Fractional Conversions

As before, the conversion fac tors that are used are the ones that lead to the cancellation of unwanted units . The follow ing is an example o f how 939. mlhr can be converted to km/min by using the appropriate conversion factors.

i

?

Example D:

Sol"•ttL·on .·

?.

:nt

i

= 939. ;

(160~~)(66~in) t

= 0.0 156

km min

Convert 8.0 lb/gal to g/L

_g_

L --

8.0 lb

454 g

~

l qt

- L == 9.6 x 102 1 gal x 1 lb x 4 qt x 0 .946

A. J. Pappas and M.E. Goicoechea-Pappas

L 0

4-2

...-.. ..... ...... ..,.-. .....,.-...... .......• ....... ....... ....... .... -.... ...... ...... ..

.....-. ----... .-.. .--.. -

~ ~ aj aj

lntrQductory Chemism

Example E: Convert 0.075 g/L to mg/dL

S l .

mg 0.075 g 1 mg 10- l L o utwn: ? dL = 1 L x 10-3 g x 1 dL

~

a ll) ~

cm

cm

3 .25 m I cm 1 min x -- x · · sec - 1 m in 10-2 m 60 sec

Solution·

? -

Example G:

-

--

-

S.42 sec

If a car is going 60.0 miles per hour, then how many cm/sec is the car traveling?

Solution:

cm

?- = . sec

60 mi 1 hr

I hr 1 min 5280 ft x 60 min x 60 sec x l mi

12 in

X

1ft

2.54 cm X

1 in

cm = 2.68 x 103 sec

Problem 2: Do the following conversion problems . a)

8 .8 rn/sec to km/hr

b) 2.3 lb/qtto g/mL

c) 10.0 lb/gal to kg/L d) 25 cm/s to m/min e) 60 cents per doz oranges to cents per orange

f)

$3 .98/lb to cents per gram

~

g) 3 .0 cups/lb to L/kg

.-it

h) 6.30 mg/µL to g/L

lllll!t

i)

..... ..... ...

mg

= 7.5 dL

Example F: Convert 3.25 m/min to emfs.

~

a ii)

4 - Measur,:ements and Conversions - Part II

(Note: 1 pt = 2 cups)

65. mi/hr tom/sec

1111!1

.....

111111

...

IV. Density (d) and Specific Gravity (sp gr) .

density

=

mass volume or

d

m

't.v Um·ts C ommon Dens1 solids (s)

=V

liquids (l)

r./ cc = g • cc-• r./ml= g • mL- 1

gases (g)

r./L = g. L

3

spgr =

d substance dwater@ 4°C

=

d substance (

9

fmL)

cc= cm = mL dm3 =L

1.00 (9/ml )

A. J. Pappas and M.E. Goicoechea-Pappas

4-3

I

Introductory Chemist[)' _ _ 4 - Me_asurernt.s_and Conversions - Part II In a medicinal setting, the specific gravity of urine is usually part of a routine urinalysis that can be used to determine how efficiently the kidneys are working. Density and specific gravity are numerically equivalent; however , specific gravity is a unitless quantity. To obtain the density of a substance in the laboratory, one needs to measure the substance's mass and volume. Empty Beaker

I,

mL

Density and Specific Gravity Determination of an Unknown Liquid mass of liquid 31.000 g - 20.000 g = 12.000 g

, io.ooo

volume of liquid

~ l:J~ ~- - - 10.0 mL d uquid

= ~

!::o::,~

Sp gr liqui d : Electronic Balance

= 1.20 g/ mL

1.20 g/ mL = I .l0 1.00 g / mL

Graduated Cylinder

Density is temperature dependent (i.e., the density of a substance changes with a change in temperature). The density of a substance also differs depending on its physical state (i.e., when a substance changes from a solid, to a liquid, to a gas). See the example for water in the

table below. Substance

density (glee)

Platinum

21.5

Water

1.0

Gold

19.3

lee

0.92

Substance

density (glee)

Mercury

13.6

Steam

5.7 X 104

Lead

11.3

Air

1.3 X I0·3

Aluminum

2.7

Helium

1.8 X 104

Urine

1.001 -1.035

Ethanol (booze)

A. J. Pappas and M.E. Goicoechea-Pappas

}

Density is temperature dependent.

0.789

4-4

... ...

lillii lilil aii

-

Introductory Chemistry 4 - Measurements and Conversion~ Part II The volume of certain solids can be obtained in one of the following ways: water

water +

aw

solid

lilliiif

=

a,g- 20 ::;; 19 ;::;;_ 18 17 ;;;; 16

Volume of Solids via Water Displacement:

=-- ,s

aiJ i!lil aiJ

..:; 14 ;;;;; 13

~

=- 12 -l·t · }· ·· · VTot - 10 ~9 Vsolid 4.0

=I

='- 8

·· :~:J;

- ..

"C2

·····

1.0 mL {Vwater + Vsolid) mL (VTot - Vwater)

= 7.0 mL

Vwater

s

-3 0 2

1111,

Volume of Solids via Mathematical Formula V(cube)

=i

h (i = w = h) X W X

cuboid or V(rectangular prism)

-

t x wxh

NOTE: 1 mL = 1 cc= 1 cm3 A substance's density can be used as a conversion factor to interconvert between its mass and volume. mass

(g) •

density

~ volume

(ml)

Example H: Calculate the density of a piece of wood that has a mass of 7 .35 g and occupies a volume of 8.20 mL.

Solution:

dcwood)

7.35 g _g__ = 8.20 mL = 0.896 mL

A. J. Pappas and M.E. Goicoechea-Pappas

4-5

lntrodttctocy_,Chemi stzy

4 - Measurements and Conversions - Part II

Example I: How many grams are in 63 .2 mL of alcohol, if the density is 0.79 g/mL? Density (0.79 g/mL) is used to convert volume (ml) ~ mass (g)

Solution:

Example J:

• 0

63.2mL - 1ml

-

50 . g

Calculate the volume, in µL , of 10.0 cg of alcohol, if its density is 0 .79 g/mL. metric-metric conversions volume

Solution :

(0.79g) =

?o~

mass

? L = 10.0 co0 µ~

....______...,

,1.

(10-2 9) (1mL) (10-3 L) ( 1µL) = 1 cg

__.

0.79 g

10- 6 l

1 ml

.

127 L µ

Before density (0.79 g/mL) can be used as a conversion factor to convert a mass into a volume, the mass in cg needs to be converted into g

Example K: A cube of solid gold measures 2.000 cm on one of its sides and has a mass of 154.4 g. What is the density of gold in g/mL?

Solution.:

-e x w x

h

= side

3

because -e

=w = h

V (cube) = (2.000 cm)3 = 8.000 cm3 = 8.000 mL

Then, d

m

154.4 g

v

8.000 mL

= - = --- =

g 19.3 0 mL

Since the mass unit asked for is in grams and the one given is in lb we need to convert 0.255 lb into grams: mL mL

g)

. g = 0.255 lb (454 l lb

?

= 116. g

data:

? mL = 33.0 mL - 20.0 mL = 13.0 mL Finally, the density can be obtained:

=~ = V

116. g = 8.92 _Jl_ 13.0mL

A. J. Pappas and M .E. Goicoechea-Pappas

• '•

't

I

t

t

t

Next the volume of the cylinder, in ml, can be obtained from the water displacement

d

'

t

NOTE:

1 cm 3 = 1 mL

Example L: When a 0.255 lb cylinder was dropped into 20.0 mL of water , the level of the water rose to 33.0 mL. Given this information, what is the density of the cylinder in g/mL?

Solution:

' '

t

First calculate the volume of the cube: V (cube) =

t

mL

4-6

lntmductor,: Chemistry

4 - Measurements ~ad Conversions - Part II

Example M: Given that the specific gravity of a liquid is 1.12, how much volume (in in3) will 2 .00 kg of this liquid occupy? sp gr and density are numerically equi valent , they differ in that sp gr is unitless while its corresponding density has units of g/mL. Since sp gr= 1.12 => d = 1.12 g/mL volume

~

mass

(103 g) (1.12 1 ml ) (1cm3) ( 1 in )3 = 109 in3 ~g 1 1 2.54

Solution: ? in3 = 2.00 k

kg

Problem 3:

g

mL

cm

Before density (1.12 g/mL) can be used to convert a mass into a volume, the initial mass in kg needs to be converted into g.

a) If a sample of a liquid occupies a volume of 17 .45 mL and has a mass of 16.3 g, calculate its density (in g/mL). b) Concentrated sulfuric acid has a density of 1.84 g/mL. Calculate the mass (in grams) of 1.00 liter of this acid. c) What is the volume (in mL) of a liquid (density = 2.0 7 g/mL) weighing 130. grams? d) 30.0 mL of ethyl alcohol (density = 0.790 g/mL) are added to a graduated cylinder that weighs 44.28 grams. What will be the mass of the graduated cylinder plus the alcohol (in grams)? e) The density of glycerine is 1.26 g/mL. Calculate the mass (in g) of 4 70 . mL of glycerine.

f) The density of water is 62.4 lb/ft3 . What mass of water (in kg) will a rectangular can hold if it measures 2.00 ft by 3.00 ft by 6.00 in? g) A brass cylinder weighing 100. g was placed into a graduated cylinder containing 20.0 mL of water; as a result, the water level rose to 32.6 mL. Given this information, what is the density (in glee) of the brass cylinder? V . Temperature Conversion Problems Temperature provides an indicator of the degree of heat present in a substance or object. Hotter substances or objects, which have a greater amount of kinetic or thermal energy, move faster than colder objects. OF oc K '18.6

T

37

180 • F

77

I

32

l

-459

Fahrenheit

373

100

.2 12

25 0

T 100

·c

l

-273

Celsius I Centigrade

A . .J. Pappas and M.E. Goicoechea-Pappas

310

298

273

0

i

Boil ing

Pt of H 2 0

Body Temp.

100 K

l

Room Temp.

Freezi ng Pt. of H20

Absolute Zero

Kelvin

4-7

lnrroductory Chemistry 4 - Measure~Conyersions - Part II Interconversion among the above temperature scales can be accomplished by using the following formulas:

I

0

I

I

P = ( 1.8 • °C) + 32

K = °C + 273

NOTE: The degree sign for the Kelvin scale is not used ; hence, K is used instead of °K. Example N: Convert -40°C to P ; and then to K . 0

Solution:

°C to K

K = -40 + 273

° F = (1.8 • -40) + 32 = -72 + 32 = -4Qop

= 233K

Problem 4: Convert the following: a) 150°F to

0

c

b) 25°C to °F

c) 350 Kto °C

d) 53°C to K

e) 95°F to K

f) - 183°C to °F

VI. Calorimetry and Heat Transfer (OPTIONAL) Calorimetry is defined as the measurement of heat changes. Chemical reactions and physical changes usually occur with either the simultaneous evolution (exothermic process) or the absorption (endothermic process) of heat. The amount of heat energy transferred in a process is usu ally expressed in calories (cal) or in the SI derived unit of Joules (J) = (ko • m\ 2 ).

1 cal= 4.1 84 J

A calorie (cal) is defined as the amount of energy required to raise the temperature of 1 g of water by l °C. [NOTE: In nutrition labels a Cal= kcal] Specific heat (aka, specific heat capacity) given the symbol s, is defined as the amount of heat necessary to raise the temperature of 1 g of a substance by l °C. Substances with high specific heats require large amounts of heat be transferred for their temperature to change. Water is an example of such a substance. Metals are examples of substances with low specific heats (i.e., they require a small amount of heat be transferred to change their temperature). A. J. Pappas and M.E. Goicoechea-Pappas

Substance

specifi c heat (Cll/g • •c)

Water (C)

1.00

Ice (s)

0.485

Ethanol (C)

0.583

Aluminum (s)

0.214

Mercury (s)

0.033

Gold (s)

0.03 1

4 -8

...... ...... ..... lillii Slit

llilliii lilii

lotroductory Chemistry

4 - Measurements and Co.nyersions - PaaJI

If the specific heat of a substance is known, the heat (q) that is absorbed or released in a given process can be calculated by using the following equation . q (he at energy)

lilli)

l

m (mass)

q

= m • s • ~T

⇒

⇒

cal or

J

g

s (specific heat)

⇒

cal g

•oc

J

or

g

•oc

c hange in te mperature

Example O: How much energy (in cal) is required to raise the temperature of 352. g of water (s = 1.0 0 cal/g . °C) from 32°C to 95°C?

Solution:

q = m • s • LiT

q = (352. g)

(1.00 ~~c)

(95°C - 32°C) = 2.2 x 104 cal

g

Example P: How many grams of iron (s = 0.444 J/g-°C ) will absorb 411 . cal of heat when it is heated from 25.0°C to 155.0°C?

Solution: The unit fo r q (heat) and that part of the heat unit in s (specific heat) must be the same. Therefore, we need to con vert the 411. cal into J. ?

_

. J - 411. cal

84 J) l(4.11 cal

= l.72xl0 3 J

Now we can use the equation:

=m• s•

q

3

l .72 x 10 J

=m•

~T

(o.444 -g •1oc-) 1.72

m=

1

X

• (1 55.0 - 25.0)°C

103 J

0 .444 g ··c x ( l 55 .0 - 25.0)°C

=

29 .8 g

Example Q : If 450 cal of heat are added to 37 g of ethyl alcohol (s = 0.59 cal/g • °C) at 20°C , what would its final temperature be?

Solution: q = m • s • 6-T A. J . Pappas and M.E. Goicoechea-Pappas

4-9

4 -Measurements and Conversions - Part II

Introductory Chemjstey

x L\T 450 cal = 37 g x 0.59 g~ ,oc

~0.59

4

AT=

~?ir

g·oc X

37 g

= 2l ' C

Since heat was added, the final temperature must be greater than the initial temperature.

L\T = T fina l -

T initiat

21 °C = T final

-

T final

20°c

= 2 1°C + 20°C = 41°C

Problem 5 (OPTIONAL): a) How much heat (in cal) is necessary to raise the temperature of 27 .0 rnL of octane (s = 0.526 cal/g . °C ; d = 0.703 g/mL) from 50 .6°C to 67.2°C? b) How many grams of water (s = L.00 cal/ g • °C) will release 1367. J of heat when cooled from 45.2°C to 36.2°C? c) What will the final temperature be, if 82.0 cal of heat are added to 32.0 g of carbon tetrachloride (s = 0.210 cal/g • c) at 33.0°C? 0

d) If 1794 J of heat are added to 23 .67 g of a liquid (s = 0.590 cal/g • °C) at 25.3°C , by how many °C will its temperature rise?

VI. More on Conversion Factors Thus far, you have been exposed to basically two type of conversion factors in problems: a) those that help you interconvert between the same type of unit [e.g ., metric-metric (1 km = 103 m) and metric-English (1 in = 2.54 cm) conversions involving mass, length, volume and time]. b) those that help you convert between two "unrelated" units (e.g., conversions involving density and percentages). In this section the intent is to continue to establish how two unrelated "fractional units" (e.g., miles/hr, m/gal , Cal/g, mg/tablet, females/class) can be used to solve unit conversion problems . Example R: If a doctor ordered that you take 0 .050 g of Benadry 13 every 4 hours, how many Benadryl tablets would you be taking in one day from the stock - 25 mg tablets - that you had at your disposal? A. J. Pappas and M .E. Goicoechea-Pappas

4-10

---

== .; • ••

•• • •• • • =•• = •= re

... -•• •• •. • ■I

re

11

or 7

"a •

•

..• ◄

..

!II

=

.., -» .., Iii

Iii) Iii)

Iii)

llit llii

...-. .-.... lllii

.......

Two "u nrelated" units w hose relationship is established by the stock that is availab le (25 mg tablets)

~

~

-!t 1111 111!1 IIIIJ IIIIJ

Dr.'s orders

?

~

11!1

.•

ll

l!l1 1111 Ill;

time conversion

10- 2 m

1

2.54

12

5280 ft

59 mi

Length (distance) conversions writte n in such a man ner that cause unwanted units to be cancelled.

l

time conversion

Before you can convert the given d istance (km) to the desired time (mi n) you must convert from km to mi because the car's speedometer has units of mi per hr (mph). It is this reading that establishes the relationship between distance and time.

Example T: If the cost of gasoline is $3 .25 per gal, how much will 3.25 L of gasoline cost (in $)? Solution: Two "unrelated" units [ .e., money($) and volume (L)J whose relationship is established by the price per gal (i.e., $3.25/gal)

~ ( 1qt ) (1gal ) ($:.:5) = ==

?$

32.5 L --"-0.946 L

-

-

-1 gal

4 qt

$27 .91

Volume conversions wrinen in such a manner tha1 cause unwanted units to be cancelled.

Example U : If a car's fuel tank holds l.80 ft3 of gasoline. how many km will you be able to drive this car before its fuel tank is empty given that thjs car's fuel economy is 12 .5 L per 100 km?

Solution:

~

11!!1

available stock

day

3m) ( 1cm )( lin )( 1/t) ( 1 )> (10 ( 1hr)(60min) -km -- --. - -. - hr = 35. min• cm in

. (",

. mm ==7.5

~

~

metric-metric conversion

tablets

12

Two "unrelated" units [i.e., distance (km) and iime (m in )] whose relationship is established by the car's speedometer. (i.e., 59 mi/hr)

.., ~

hr)

E xample S: As you are drivi ng on a Canadian highway, you see a sign telling you that the distance to the nex t Rest Area is 5 .5 km If you maintain a speed of 59 mph (mi/hr), how many minutes will it take for you to make it to the Rest Area? Solution :

~

111!9 11!1

~

? tablets == 0.05 g ( 1 mg ) ( 1 tablet) ( 24 day 4 hr 10- 3 g 25 mg 1 day

lllil

... ...

4 Measurements and Conversions - PanJ.I

Intrnd,uctory Chemistry Solution:

?

Two "unrelated" units [i.e., di stance (km) and volume (fl')) whose relationship is established by the car' s fuel economy (i.e., 12.5 L per JOO. km)

k!:~::t(12 in)3(2.54. cm)3(1mL )(10-3L) (~ km) 3

1 ft

1 m

1

cm 3

1 mL

12.s L

= 408

km

Volume con,·ersions written in such a manner that cause unwanted uni ts to be cancelled .

A. J. Pappas and M.E. Goicoechea-Pappas

4- 11

lntr.o

Problem 6: a)

To relieve her men trual cramp a girl took 2 1/ 2 Ibuprofen tablet • Given that each tablet contain 200 mg of Ibuprofen, how many µg of Ibuprofen did he take?

b)

An anesthesiologi t placed 2.5 cc of Diprivan (Propofol) in a patient' IV. If the stock that was adminj tered contained 10 mg Propofol /mL, how many grams of Propofol wer administered?

c)

You are travelling by car through Europe. The cost of ga oline at a peculiar station is I .50 Euro/L. When you fill the car's tank you find that the pump deliver the ga oline in gallon . If you put in 7 .6 gal of ga oline into the car' tank and the exchange rate i 1.20 Euro/Dollar, what will your bill (in ) be?

d)

There are 110 Calories in 1 serving (8 fl oz) of Red Bull 1K•. If you drank 450 mL, how many Calories did you ingest?

Problem 7: An Assortment of Problems

a) A speed ign in England read 30. km/hr, over what peed in mph (mile /hr) would you get a ticket? b) Mercury melt at -38.87°C. What i it melting point in °F? c) A piece of ilver (density = 10.5 g/mL) dropped into wat r di places 21.56 mL of water. What is the mas (in grams) of thi piece of silver? d) 30.00 mL of carbon tetrachloride (den ity = 1.595 g/mL) i added to a graduated cylinder that weighs 47 .98 g. What will be the ma (in grams) of the cylind r plus the carbon tetrachloride? e) What i the total ma (in g) of the following: 0.002000 kg, 200. cg, and 1.00 dg? f)

405. mg, 0.500 g,

A shoe box measures 12.0 11 by 5.5 11 by 3.5 11 • Calculate it volume in cubic centimeters. [ OTE: =inch] 11

g) A European dairy farm has a land area of 0.425 km 2 . Determine this area in acres. 1.000 acre = 43,560 ft2 . h) Calculate the volume (in mL) of a ample of CCl4 having a ma s of 80 .0 g if the density of CCl4 is 1.60 g/mL. i)

Find the density of ethyl alcohol (in glee) if 80.0 cm 3 "weighs" 63 .3 g.

j)

A rectangular block of the element barium weighs 52.5 g and has the following dimen ions: 5.00 cm by 3.00 cm by 1.00 cm. Given this information , what is the density (in g/cm 3) of barium?

A. J. Pappas and M.E. Goicoechea-Pappas

4-12

Introductory Chemistry 4 - Measurements and Conversions - Part II k) Vitamin C ha a density of I .65 grams per cm3. What is the volume (in cm 3) of a 500 mg tablet of vitamin C? 1)

What i the density (in g/ mL) of a medication if the contents in ide a syringe filled to 3 .00 mL weigh 3 .85 g?

m) Given that 140. mL of chlorine gas weigh 0.450 g; find its density in grams per liter (g • L"1). n) The recommended IV dose to edate a patient with Rocuronium is

2 mg/kg of body weight. If the patient weighs I00 lb, how many mL of this sedative should be injected into the IV, if the stock that is on hand contains I 00 mg Rocuronium per 10 mL? o) If 1.25 cc of Ketamine HCl, a drug used for the induction and maintenance of general anesthesia, were given to a patient through their IV line, how many grams of Ketamine HCl did the patient receive, if the stock that is on hand contains 500 mg per 10 mL? p) The equalities listed below use units of measurements that in all likeliliood you are not familiar with. Consider these equalities, as well as, other equalities that have been provided in the previous module when answering the ensuing questions. Use 3 significant.figures to report each answer. 1.94 x 10- 2 baktun = 200. fortnights

700. days= 4.86 x 10-3 baktun 1 femtosecond (fs) = 10-15 second (s) 300. bushels = 1200. pecks

400. cords = 1.65 x 1o5 pecks 120. in 3 = 5.42 x 10-4 cords 10- 11 Darcy= 9.87 x 10 4 Barn 220. Barn= 5.44 x 10-30 Acre 100. Acre= 4.25 x 10- 28 square parsec (pc 2) 500. pc 2 = 4.76 x 10 53 nm2 1. 11.

How many femtoseconds (fs) are there in 7 .12 fortnight ? How many GL are there in 243 bushels?

iii. How many m2 are there in 450. Darcy? 1v. How many µL/Ts are there in 300. cord /baktun?

A. J. Pappas and M.E. Goicoechea-Pappas

4-13

ANSWERS 1. a) 2.5 x 104 mm3 c) 7.23 x 104 m2 e) 2.2 x 101 cm2

b) 4.50 x 104 dm 2

d) 4 . 2 x 10 1 L f) 1.31 x 102 mL

2. a) 3.2 x 101 km/hr c) 1.20 kg/L e) 5 ¢/orange g) 1.6 L/kg i) 2.9 x 10 1 m/sec

b) d) f) h)

3. a) 9.34 X 10-l g/mL

b) 1.84 x 103 g

C)

6.28

X

10 1 mL

e) 5.92 X 10 2 g g) 7 .94 glee

4 . a) 65.6°C c) 77°C

1.1 g/mL 1.5 x 10 1 m/min 8 .77 X lO- l ¢/g 6 .30 x 10 3 g/L

d) 6.80 x 10 1 g f) 8.50x10 1 kg

b) 77°F d) 326 K f) -297°F

e) 308 K

5. a) 166 cal c) 45.2°C

d) 30.7°C

6. a) 5.0 x 105 µg c) $35.95

b) 2.5 X 10·2 g d) 209 Cal

7. a) 1.9 x 10 1 mph

b) -37 .97°F

c) 2.26

10 2 g

b) 36. g

m) 3.21 g/L

d) f) h) j) 1) n)

o) 0.0625 g p) ii. 8.55 x 10-6 GL

1021 fs p) iii. 4.44x 10-10m2

X

e) 5.00 g

g) 1.05 x I 0 2 acres i) 7.9IxI0- 1 g/cm3 k) 3 .03 x 10- 1 cm 3

9.583 x I0 lg 3.8 x 103 cm 3 5.00 x 10 1 mL 3.50 g/cm 3 1.28 g/mL 9.1 mL

p) i. 8.61

X

p) iv. 8.75 x 10 13 µL/T

A. J. Pappas and M.E. Goicoechea-Pappa

4-14

.... lilil

--------....-.. lili)

INTRODUCTION TO MATTER : ELEMENTS AND COMPOUNDS

lilliii)

aii)

a .ii

...... .... ~ ~

~ ~

~ ~

I. Matter Matter is defined as anything that has mass and occupies space. In fact, chemistry is that branch of science which studies matter and the changes that it unde rgoes.

A. Classifications of Matter T he followin g concept map shows the classifications of matter. Matter

Anything that has mass and occupies space I

The symbol for all known elements is found 1n the Periodic Table.

I

I Pure Substance

physical process

-'

Mixture ~

r

Contains more than one pure subscane

NOTE: Gaseous mixtures are homogeneous

I

I Element

I Compound

chemical_ Pure substance made up Pure substance that r can't be broken down ' process of more than one into simipler substances element Na or 0 2 or 0 3

I

I H omogeneous Mix:ture

Heterogeneous Mixture

Mixture chat is uniform throughout

Mixture that is not uniform throughout

H20 or NaCl or (NH4)z$ apple Juice or NaCl, H20

salt

oil and mter or cheesburger homogeneous mixture = solution

All known elements are found in the periodic table (see the back side of the front cover of this text). A language analogy can be applied to matter. The letters of the

alphabet are like the symbols used for an element; however, elements may have a 1 - 3 letter symbol. The first letter is always capitalized while the other letters (if

~

any others are present) are in lower case (e.g., C, Ag or Uuu). There could also be a subscripted whole number - not including 0 , l and negative numbers - after an

~

element s symbol (e.g., H2 or P4) .

9ll!t

Just like more than one letter makes a word; a compound is made up of more than one element. The chemical formula of a compound is used to express its

.....--....... ...... ....... ~

~

1

elemental composition (i.e., the proportion of each particular e lement that it is composed ot). There are no spaces in a word; the same concept is applied to a compound. However, the chemical formula of a compound can have letters. subscripted num bers and parentheses; furthermore, more than one letter will be capitalized [e.g., H2O, CO2, C(>lf12O6, Fe3(AsO4)2 or (Hg2)3(PO4)2]. If there is

more than one of the same element in a compound, a number conveying how many are present is subscripted right after its symbol. If parentheses are used, the subscripted number outside the parentheses is multiplied by each

subscripted number inside the parentheses in order to ascertain the total number A. J. Pappas and M .E. Goicoechea-Pappas

5- 1

5 Jntroduction to Matter

Introductory Chemistry

of that particular element that are present in the compound. used as a subscript and is understood to be there.

NOTE: 1 is not

The elemental composition (i.e., which elements are present and in what proportion) of several compounds is illustrated in the following table.

Compound

Elemental Composition

a) H20

2 Hand I 0

co

b)

1 C and 1 0

c) C0Cl2 d)

1 Co and 2 Cl

(NH4)3As04

In Exampled, multiply the subscript outside the parentheses with each subscript inside the parentheses. Recall that when there is no subscript that a 1 is understood to be there.

3 N, 12 H , I As, and 4 0

A mixture is like a group of words. Mixtures are made up of more than one pure substance (either several elements and/or compounds). The vast majority of matter that you have encountered are mixtures composed of more than one pure substance [e.g., air (mixture of gases); consumer products - vinegar, wine, Gatorade®, diamond ring, 14-kt gold; your body- urine, blood, saliva, sweat].

Problem 1: Classify each of the following as either an: A. element B . compound C. homogeneous mixture D. heterogeneous mixture b) Co(P03)2

a) K

c) "normal" urine

d) air [N2, 0 2, Ar, CO2, and other gases]

e)

h) IF

i) Ss

j) Sterile Saline (i.e., 0.9% NaCl) k) 160-ProofRum [80% C2HsOH, ~20% H2O, < 1% other compounds] I) testosterone [chemical formula: C19H2gO2~ o H As an FYI. its molecular structure - which specifies how. the e leme nts are •bonded to A tru c . s ctura I ,ormu 1a 1s each other - 1s shown to your n ght. o ; very useful in organic

Problem 2: Complete the following table. Compound a)

Na3803

b)

C6HsCOOH

c)

Al (SeO4)3

d)

(NH4)2Te04

A. J. Pappas and M.E. Goicoechea-Pappas

~

.,--= ::; ._.-.. .........--= --...-.... -------.-=.. • •

f) 18-kt gold [75 %Au and 25% ocher metals] g) tap water

~

chemistry (study of carbon containing compounds).

Elemental Composition

5-2

-.:

-= --a --=

~

9111!1 .-!I

...... .....

...... -..

5 - Introduction to Matter

Introductory Chemistry

B. States of Matter and Interconversion Among States of Matter Depending on its temperature (and pressure) , a sample of matter can exist in any one of the foliowing common physical states: solid, liquid or gas. There is another state of matter (i.e., plasma) that we won't be discussing. The following diagram shows several properties for each state. solid

(s)

(t) eool -== =::!:

definite shape & volume

dee. pressure

~

or

L·

heat

~

gas (g)

liq u id heat cool or in c . pressure

0

indefin it e shape (assumes shape of container) defininte volume

Physical

Changes

i nde finite shape & volume (assumes shape and volume of container)

attractiv e force between particles decr eases compressibility inc .-e ases because distance between pariticlcs is increasing density ( d

= V)

decreases (as a result of a volume increase)

The diagram below illustrates how temperature and pressure affect the interconversion among states of matter and the terminology that is employed in describing these changes in state (i.e., phase transi tions).

~

(c-o ~o

• • decrease

boiling point (bp) of a substance is the same its conden~acion poim (cp) .

~t1. ')

~,.">

0 ">

J

~=heat

Uquld(t)

~~ ~-I

'o.,.;s-1l'J.0

,--t- = - 1-nc_re_as _e ~ 0

The tempcra1ure (T) at which vaporization (i.e., boiling) occurs is the same as that at which condensation occurs. Thus. the

.r~

-====:;v;;;:a=p=o:;:rt=z=at=l::o;:n==::::!.• Condensation (cool or Pt)

~

T he temperature (f) at which melting occurs

IG~.00. .. ~ •

is the same as that at which freazi ng occurs. Thus , 1he melting point (mp) of a IIIA and VA> VIA). Example M : Arrange the following atoms - N, Mg, Al, Si, K - i n order of increasing ionization .

--

VIIA VIIIA

IA

Solution:

IIA

I

IIIA IVA

Mg

3

Al

...., ~

N

2

◄

VA

VIA

~

V

Si

.5

K

~ K ~✓-.,. ~ found more towards the left side of the periodic -0 table; while non-metals are found towards the decreases right. Thus, it wou ld stand to reason that going across a period (from left =;> right) the metallic character of an element decreases. The crude sketch of the periodic tabl e to above shows what happens to the metallic character going across a period and up a family.

~o~

(,)

4)

Examp le 0 : Wh ich of the following elements - Al or Mg - has the least metallic character?

Solution: AI, because goi ng across a periodic (from left

⇒ right) the

metallic character of an element decreases . A. J . Pappas and M. E. Goicoechea-Pappas

6- 14

ln.troouctocy Chemista

6

An Assortment of Problems: 17 . List two types of EMR that: a) have less energy than IR rays b) are shorter than uv-rays c) have a greater fre quency than green light.

18. W rite the spectroscop ic electronic configuration for each of the fo llowing . Which are the core electrons and which are the valence electrons? a) rubidium

b) gallium

c) krypton

19. How many unpaired electrons are contained in one atom of chromium? 20 . Upo n examining the electronic configuration of Kr and Cr , which would be p aramagnetic? Which would be diamagnetic? 2 1. Iden tify the electronic sublevel that elements in the indicated section (A D) of each row are filling .

-IA

I

-

I/HA VIIIA

-

IIA

IIIA

IVA VA VIA

2

l 4

s 6

7

111B

\

IV8

vg

V18

\'118

,-- VI II -

'8

1B

'

~ I I I I I In I I I I I I ,-.

I

l

t

('

-

I

~

~

22. W hy do the elements in a g iven group resemble each other chemically , in terms of electronic structure ? 23 . Complete the following table pertaining to elements. Atomic Symbol No.

Shorthand Electronic Configuration

Name

Group No.

# of Metal. Valence Non-metal, Electrons Metalloid

5 Sn 9

Mg [Ne] 3s2 3p 1 [Ne] 3s2 3p5

Sb

82

24. W ha t is the ele ctronic configuration for each of the following species? a) K +

b)

s2-

A . J. Pappas and M. E. Goicoechea-Pappas

c) Ar

d) Cr 6-15

w

6 Electronic Structure & Chemical Periodlc~

Introductory Chemistry

25. What is the relationship of all the four species in problem 24? 26. Using spectroscopic notation, write the electro nic configuration for: 2 a) P b) V c) Cu d) Mg +

e) p 3-

f) Zn2+

4

g) Po

h) Ge +

use lines to 27. For each of the following, draw the orbital diagram (i.e., represent orbitals and a rrows to represent electrons). d) N 3c) sc 3+ b) Si a) S h) Fe2 + g) Mn2+ f) Cr e) Ni

28. Identify the element with each of the following shorthand electronic configuration. c) [Ar] 4s2 3ct10 4p3 b) [Kr] Ssl 4d10 a) [Xe] 6s2 29. Given the following list: As, Cl, Rb , Ga ~I IIA IIIA ; a) which is the smallest? l .:,: b) which is the largest? • u s 0 .g c) which has the largest ionization energy? 6 ' "0 d) which has the smallest ionization energy? e) which has the tendency to form a cation the most readily?

VIIA VIIIA

IA

IVA

VA

VIA

.,=. •

;..: !& UI

•

V

r

r

--

·!• Iii

7

f) g) h) i) j)

which has the tendency to form a cation the least readily? which loses an electron the most readily? which lo ses an electron the least readily? which has the most metallic character? which has the least metallic character?

..,

;,-[OPTIONAL] [OPTIONAL]

30. Though the element whose atomic number is 1 15 has been synthesized , it has not been placed in the periodic table in this tex t. Ans wer the following questions about the element whose atomic number is 115. a) In what group is it in? b) How many valence electrons does it have? c) How many unpaired electrons does it have? d) Is it paramagnetic or diamagnetic? e) Provide its shorthand electronic configuration?

zj 7

...... ......,... ■7 & 7

C

IIJ D•

f) Is it classified as metal, non-metal or metalloid? g) Is an electrical conductor or insulator? h) Is its atomic radius greater than or less than that of Rf (Z = 104)? A. J. Pappas and M. E. Goicoechea-Pappas

=-=

•►

>

6- 16

~

-~

...

lilllll

liillllll

6 Electron ic Stmcture & Che.mica] Periodicity

Introductory Chemistry

i) Is its ionization energy greater than or less than that of Rf (Z = 104)?

j) Is it more or less likely to lose an electron than Rf (Z = 104)? k) Is its atomic radius greater than or less than that of Bi (Z = 83)?

I) Is its ionization energy greater than or less than that of Bi (Z = 83)? m) Is it more or less likely to lose an electron than Bi (Z = 83)? n) Is it more or less likely to become a cation than Bi (Z = 83)?

~ ~ ~

-

3 1. a) Identify the sublevel in which each depicted orbital can be found in.

A.

B.

C.

aill

D.

CF)

=llliil

::lliiil :lliit ~ ~ ~

~ ~

~

b) Which energy leve ls (n) can each orbital be in found in? c) Consider the orbitals (A - D) depicted above. Which orbital can be used

to represent the statistical probability of the location where the last electron in an atom that each of the following are likely to be found in?

i. Mg

ii. Fe

iii . Pb

32. Which of the following represents a(n): A.

B.

C.

.., ~

D.

0

~

~ ~

-~

---

"" ~ ~

~

a) l s orbital

b) 2s orbital

c) 4p orbital

Calculations Involving Electromagnetic Radiation [OPTIONAL] 33. If an EM wave has a wavelength of 3 .70 x 10-6 m, what is its frequency (in

s- 1 = Hz)? 34 . What is the wavelength (in m) of an EM wave having a frequency of 5.60 x 10 15 sec-1?

35. If the frequency of a photon of violet light is 7 .37 x 10 14 s-1 , calculate its energy (in J) .

~

36. If the wavelength of an EM wave is 3.70 x 10-6 m, calculate its energy (in J).

~

37. What is the wavelength (in m) of a photon having an energy of 6.16 x 10-19 J ?

~ ~

~ ~ ~

A. J. Pappas and M. E. Goicoechea-Pappas

6-17

_ctr.onic Struct

ANSWERS Radio and TV waves

1. a)

Visible

Infrared (IR) rays

microwaves

Ultraviolet (UV) rays

Light

x-rays

gamma

('Is') rays

>

ROYGB I V

E

't

All travel at the same speed (i.e ., the speed of light. 3 x l0 3 m /-)

b) i and ii.

v ( hotter)

,1, (longer)

iii. A (

v (greater)

hotter)

E (lower)

c (same)

E (greater)

c (same)

..

c) 8.33 min

2.

A. excitation F. 656nm

C. photon H. 3

B. relaxation G. 410nm

D. emitted I. 6

1s2 2 2 2p6 3s2 3p4

= [ e] 3s2 3p4

b)

1s2 2 2 2p6 3s2

= [ e] 3s2

c)

1s2 2 2 2p6 3 2 3p6 4s2 3ct10 = [Ar] 4s2 3dl0

3. a)

E. pnsm

1 2 2s2 2p6 3s2 3p6 = [ e] 3 2 3p6 e) 1s2 2s2 2p6 3s2 3p6 4s2 3dl0 4p6 5 2 4ctl0 Sp6 6s2 4fl4 Sdl0 6p2 =

d)

[Xe] 6s2 4fl4 SdlO 6p2 1s2 2s2 2p6 3s2 3p6 4s2 3d 10 4p6 5s2 4d IO 5p6 6s2 4f I4 5d 10 6p6 7 2 Sf 14 6d2 =

f)

[Rn] 7s2 Sfl4 6d2

4. a) 6

b) 1 b) 2p 3

5. a) 3d6

6. a)

fl

T!

ls

2S

b) t1 ls

c)

fl ls

c) 2p6

T!T!T! - --

2p

H T! 2s

8.

a) [Kr] Ssl 4d5

d) 3s2 U

3s

H

2p

3s

H T! H 2p

T!

b) 3

a) 4

T1

Tl H H

- --

2s

7.

d) 1

c) 5

H

i!

3s

c) 0

tl H 3p

T! 4s

Tl Tl

H

3p

4s

i

i

i

3d

T! H 3p

d) 0

b) [Xe] 6

I

4f14 SdlO

c) [Kr] Ssl 4ctl0 9.

a) 0, diamagnetic c)

10.

0, diamagnetic

a) ls 2

c) ls2 2s2 2p6 3s2 3p6 4s0 3dl0

b) 2, paramagnetic d) 3, paramagnetic b) 1s2 2s2 2p6 3 2 3p6 4s0 3d7 d) ls2 2 2 2p6 3 2 3p6 4s0 3d10

11. a) c1- or s 2- or p 3- or Si4- or K+ or Ca2+ or Sc 3 + or Ti4+

b) 1- or Te2- or cs+ or Ba2+ or La3+ A. J. Pappa and M. E. Goicoechea-Pappas

6-18

...... ..• .... ..•.... .... .... -

Introductory Chemistry

12 . a) He

b) He

13. a) 6

b) 8

6 Electronic Structure & Chemical PerioQicitY c) Ne c) 4

14. a) Group IHA , b) Group VIIIA,

d) Xe d) 1

e) Kr e) 2

[Rn] 7s2 Sfl 4 6dl0 7pl ,

3

[Rn] 7s2 5fl4 6ct10 7p6,

8

15 . a) S < Se < Te < Po c) 0 < N < C < B 16. a) c) 17. a) c)

f) 1

b) He < F < S < P

S > Se > Te > Po b) He >F > P > S N > 0 > C > B radio waves, TV waves, microwaves b) x-rays and y-rays blue light, indigo light, violet light, uv-rays, x-rays and y-rays

18. a) 1s2 2s2 2p6 3s2 3p6 4s2 3d IO 4p6 5s 1

(core e-: not bold, valence c-: bold)

b) 1s22s2 2p6 3s2 3p6 4s2 3d 10 4p 1

(core e-: not bold , valence e-: bold)

c) 1s2 2s2 2p6 3s2 3p6 4s2 3ct10 4p6

(core e-: not bold, valence e-: bold)

19. 6

(careful, this is an exception)

20. Kr - diamagnetic 21. A. 4s

Cr - paramagnetic

B . Sd

C. 6p

D. Sf

22. They all have the same number of valence electrons (i.e., they have the same number of outermost s and p electrons). 23 . Atomic Symbol No .

Name

Shorthand Electronic Configuration

Group No.

#of Valence Electrons

Metal , Non-metal, Metalloid

[He] 2s2 2p 1

IIIA

3

metalloid

IV

4

metal

5

B

boron

50

Sn

tin

9

F

fluorine

[He] 2s2 2p5

VIIA

7

non-metal

12

Mg

magnesium

[Ne] 3s2

IIA

2

metal

13

Al

aluminum

[Ne] 3s2 3p 1

3

metal

17

Cl

chlorine

[Ne] 3s2 3p5

7

non-metal

51

Sb

antimony

[Kr] Ss2 4d 10 Sp3

IIIA VIIA VA

5

metalloid

82

Pb

lead

[Xe] 6s 2 4f14 5d 10 6p 2

IVA

4

metal

[Kr]

ss2 4ct IO sp2

24 . a) 1s2 2s2 2p6 3s2 3p 6

b) 1s2 2s2 2p6 3s2 3p6

c) 1s2 2s2 2p6 3s 2 3p6

d) 1s2 2s2 2p6 3s2 3p6

25 . They are isoelectronic (i.e., they have the same electronic configuration). A. J. Pappas and M. E. Goicoechea-Pappas

6- 19

·c_illeriodici

b) 1 2 2s2 2p6 3s2 3p6 4 2 3d3 c) 1 2 2s2 2p6 3s2 3p6 4 1 3ct10

d) 1 2 2s2 2p6

e) 1 2 2 2 2p6 3s2 3p6

f) 1s2 2 2 2p6 3s2 3p6 4s0 3ct 10

g) 1s2 2s2 2p6 3 2 3p6 4s2 3d 10 4p6 5s2 4d 10 Sp6 6s2 4f 14 5d 10 6p4

h) 1s2 2 2 2p6 3 82 3p6 4 0 3ctl0 4p0

27. a)

i!

b)

T! 1s

ls

c) d)

e) f) g) h)

u. u.

f.J,

2s

2p

3s

i!

iJ. i! TJ.

TJ.

2s

2p

3s

Ti

t!

T!

lS

2s

T! ls

t!

i! i! i!

2s

2p i!

u.

i

t!

f.J,

3S

iJ. ls

i!

i J, iJ. iJ.

i!

Zs

2p

3s

t!

t!

iJ. t! t!

tJ.

ls

2s

2p

3s

2s

2p

b) Rb h) Cl

g) Rb 30. a) VA

e) [Rn] 7s h) less 1) less

2

5f

14

33. 8.11 x 10 13 36. 5.38

X

- 1

JQ-20 J

iJ. 3s

iJ.

tJ. t!

TJ.

i! TJ. iJ.

3p

4s

3d

H

n

3p

t!

iJ,

t!

A. J. Pappas and M. E. Goicoechea-Pappas

X

l0-7

i

i

U.iii

i

i

i

3d

c) As e) Rb

c) 3 f) metal j) less n) more

34. 5.36x 10- 8 m

i

3d

4s

d) Rb j) Cl

i

i

i 4s

B. C. d B. n 2!:: 1 C. n 2!:: 3 either A or C iii. D smaller than a 2 orbital)

37. 3.23

i

i

3d

i J,

tJ. tJ.

i) Rb

i

4S

3p

b) Ag c) Cl

i

i

3p

b) 5 6d 10 7p 3 i) greater m) more

A. d A. n 2!:: 3 i. D ii. D (a 1s orbital i

u.

3p

2p

i J, tJ. f.J,

't.L

tJ.

i!

t!

i 3p

2s

H ls

i

3p

3s

t!

28. a) Ba 29. a) Cl

31. a) b) c) 32. a)

i!

2p

l!

T! ls

u.

i!

i

t!

f) Cl

d) paramagneti g) conductor k) greater

-.. ...... ..... ....• ......• .... ..•..• ..•.,. ..• ••

D. p D. n?!::2 b) A

35. 4.89

c) B X

10- 19 J

"

ID

6-20

•

Ii)

Ii)

Iii

e_cul.ar___G_eomelr'

Iii lit

.. .. -..... .. ...... --.. -.. .... ..-

CHEMICAL BONDING AND MOLECULAR GEOMETRY

Iii)

Iii

I.

Valence Electrons For the r pre entative (main group) elements (i.e., the "A" elements in the periodic table) the number of valence electrons equal the number of electrons that are in the outermost and p subshell. Also , the number of valence electron i equal to the column number for the representative ("A") elements.

Iii Iii

Iii

Electronic Confi uration for C 1s2

L___J

~

core electron

I

2s 2 2p 2I I

valence electrons

El ment that have the same number of valence electrons (i.e. , are in the ame column of the periodic table) have similar chemical properties . How many valence electrons do the following elements have? a) B

b) Rb

c) S

llllliit

Solution: a) three (B i in column IIIA) b) one (Rb is in column IA) c) ix (S is in column VIA) Problem 1:

11119

~

1119

11]1

8

II.

How many valence electrons do the following elements have?

a) Se

b) Ar

c) Si

d) H

e) 0

f) Na

Lewis Electron-Dot Structure of the Elements For element , the symbol for the element represents the nucleus and all but the outermost shell. The electrons in the outermost, valence, shell are repre ented by dots that are placed on four sides of an imaginary square (no more than 2 dots/side). Electrons should be spread to all four sides of the imaginary square before being paired.

i dots can be placed on - 0-any of the s ides of f the imaginary . quare. 4

Exam le B: Draw the Lewis electron-dot structure for the following atoms. c) 0 b) Al a) Na A. J. Pappas and M.E. Goicoechea-Pappas

7-1

Introdu

Solution: a) Sodium,

a, ha 1 valence electron o it Lewi structure i : Na·

b) Aluminium , Al , has 3 valence electrons so its Lewi structure is:

.

·Al · c) Oxygen , 0, has 6 valence electron

o it Lewis tructure is:

:O·

Problem 2:

Draw the Lewi

tructure for the following atoms .

a) K

b) B

c) P

d) Br

e)Mg

f)C

III. Lewis Electron-Dot Structure for Monatomic Ions For ion one tarts with the Lewi structure for the element and then adds or remove electrons as needed . Remember that po itive ions are formed when electron are lost by an atom while negative ion re ult from the gain of electrons. Metals become cations (lose e-). Metal in IA, IIA and Al tend to lo ea many electrons as is necessary to become iso l ctronic (i .. , same electronic configuration) with the noble gas in the previous row (i.e., they lose their valence electrons). Noble gases are very stable becau e they have a complete outer shell of electron . Non-metals become anions (gain e-). Non-metals gain a many lectrons a is necessary to become i oelectronic with the noble ga in the ame row (i.e., they gain as many electrons as necessary to have a full octet - 2 electrons at each side of the imaginary square) . oble gas s are stable and do not form ions. Exam le C: Draw the Lewis electron-dot tructure for the ion that each element in Example B is likely to form. Solution: a) For odium, Na, it one valence electron i lo t (become i oelectronic with Ne); the re ulting ion has a charge of+ 1. NOTE: Just like sub cript of 1 are not written, the l in a charge of+ 1 i not written. Na· A. J. Pappas and M.E. Goicoechea-Pappas

⇒

Na+ 7-2

.... ..• ...... .... .... .......

...... ..-.. -..... .....

Iii

~

al

.......

...... ... ...... ~

l.!Umuuctory Chemis.1.rY._ _~-------1- ·..,.Ce"'9~ he~@ ~·c"',a~LB~fillliig a.ruLM-oleculai:.,G~.metry b) For the metal aluminum, Al , its three valence electrons are lost (becomes isoe lectronic wi th Ne); the resulti ng ion has a charge of +3.

·Al· ⇒

c) For the no n-metal oxygen , 0 , si nce an oxygen atom contains 6 valence e lectrons, it will gain 2 electrons so that it has a fu ll octet (becomes isoelectronic w ith Ne); the resulting ion has a charge of -2.

:oProblem 3:

c) P

d) Br

e) Mg

f) S

IV . Electronegativity (EN) E lectronegativity refers to the relative tendency of an atom to attract electrons towards itself when it is chemi cally bonded to another atom. Commonly , electro negativities are expressed on a somewhat arbitrary scale, called the Pauli ng scale. A partial scale is shown below. Fluorine has the highest electronegativity (4 .0), and Cs and Fr the lowest (0.7) . NOTE: Non-metals h ave th e larger electronegativities and hydrogen has generall y a lower EN than other nonmetals, but higher than those of metals . IA

IIA

IIIA

IVA

2 .0

2 .S

~

Na

Mg

Al

Si

0.9

1.2

1.5

~

K

Ca

Ga

Li

B

VIlA

VIIIA

0 3 .S

1.8

N 3 .0 p 2. 1

2.5

Cl 3.0

Ge

As

Br 2.8

C

s

F

E

4 .0

N

n

0.8

1.0

1.6

1.8

2.0

Se 2.4

Rb 0.8

Sr

In

Sn

Sb

Te

1

1.0

1.7

1.8

(.C)

2 .1

2.5

a s

Cs 0.7 Fr

Ba 0.9 Ra

Tl 1.8

Pb

Bi

Po

1.9

I .9

2 .0

At 2.2

"s

0.7

0.9

1111!

1111!

VIA

,h Be I .S

~

VA

H 1.0

~

2

b) Ba

Ill§

~

••

:o: -

a) K

2 .1

"' "'

⇒

D raw the Lewis structure that most likely w ill form wh en each o f the fo llowing elements beornes an ion .

~

-"

[ or Ai+3 ]

AJ3+

EN increases

C

e

-

Given a list of elements, the element with the largest electronegativity (excluding Vil/A elements) will be the one that is at the most upper right hand corner of the periodic table.

Ill!

1111!

~

A. J. Pappas and M.E. Goicoechea-Pappas

7-3

• 7 - Chemical...fumding and Molecular Geometry

Introdu.ctory Chemistry

Without looking at their electronegativity values , arrang.e :11e following elements in order of increasing electronegativity: N , Mg, Al, K , Si

Exam le D:

Solution:

IA

VIIAVIIIA

I

IIA

IIIA

IVA

2

r,g

3 4

Al

VA N

VIA

I

J "'

:;

.

;

Si

K

. .,

a

i

~ K < Mg < Al < Si < N §

s

w

6

C

e

:,

7

- - - - 8ectronegativ1cy Increases -----i),a

t:., w

Problem 4: Arrange the following atoms in order of increasing electronegativity • a) S , Se,Te, Po

V.

b) Al , F , S, P

c) B , C ,N, O

How to Predict the Type of Bond Formed by Compounds A chemical bond is the force that holds 2 or more atoms together. There are two main type s of bonds: covalent and ionic. A covalent bond consists of a pair of electrons being shared between two atoms. This is the type of bond that exists between two non-metals. There are two types of covalent bonds: non-polar and polar. A non-polar covalent bond consists of the pair of electrons bei ng shared "equally" by both elements (i.e., when the difference in the electronegativity of the 2 elements is from 0.0 to 0.4). A polar covalent bond consists of a pair of electrons being shared unequally between the two elements (i.e., when the d ifference in electronegativity of the 2 elements is from 0.5 to 1.9).

A bond becomes more polar as the difference in EN between the two bonded atoms increases. The atom that is more electronegative has the electrons closer to itself and therefore develops a partially negative charge (6-) while the less electronegative atom develops a partially positive charge (o+). Non-Polar Covalent Bond (llEN =0 • 0.4)

Polar Covalent Bond (tiEN =0.5 . 1.9)

Elements in the bond are "equally" sharing electrons

Elements in the bond are unequally sharing electrons

X: Z Since X and Z have very similar EN, the electron pair, which is represented by : is '"equally" shared between X and Z.

A. J. Pappas and M .E. Goicoechea-Pappas

&+

&-

X : Z

+--X :Z

The more EN element pulls the electron pair towrds ( : ) itself and develops a partially necative charge (&--) while the less EN element is starting to lose its electrons and develops a partially positive charge (&+ ). A bond dipole also displays bond polarity (an arrow is used co point towards the more EN element and a "+" is placed on top of the less EN element ( ,----- ).

7-4

-..• •• -.... ..• ......• .... •• ...... ...... ....... •• •• ..•

•• •• •• •• ••

.... ....

aii)

111M

Iii)

7 - Chemical Bonding and Molecular Geometr)',

I.n.u:aductory Chemistry

Example E: While referring to electronegativity values, classify the bond formed between the foJlowing e lements as : non-polar covalent, polar covalent, or ionic . c) Li and 0

b) CandCl

a) Cl and Cl

Solution: a) non-polar covalent because the EN difference is 0. b) polar covalent because the EN difference is 0.5 (12.5 - 3.01) c) ionic because the EN difference is 2.5 (1 1.0 - 3.51). Also , the bond between a metal and non-metal is likely to be ionic. Example F: Depict the bond polarity that exists in H-F by using both partial charges (8+/8-) and a bond dipole ( I ., ). Solution: The electrons in the polar covalent bond are c loser to the fluorine (because it is more EN) than to the hydrogen; therefore, the fluorine atom becomes partially negative and the hydrogen atom becomes partially positive . 8+ 6H -F

I

H -

•

F

Example G: Indicate the polarity (8+/8-) of each atom in the molecule Cl2?

Solution:

Since both atoms forming the covalent bond are the same (there is no difference in electronegativity) , the electrons are shared equally. The bond is thus non-polar.

Example H: Which compound is more polar, BrF or BrCl? Also , indicate the polarity of each bond by using partial charges and a bond dipole. Solution: Electronegativity increases going up a column Relative to Br, since Fis more electronegative than Cl , Br-F is more polar than Br-CL The greater the distance between elements in the periodic table the greater the ~EN and the more polar the bond is going to be. 6+

&-

Br-F

I

•

Br-F

&+ Br-

&Cl

I

•

Br-

Cl

An ionic bond is formed when the difference in electronegativity between the two elements is greater than or equal to 2.0 (i.e., a metal (small EN) and nonmetal (large EN)). The more electronegative element (the non-metal) takes electrons away from the less electronegative element (the metal) - thus forming ions. The non-metal gains the electrons (becomes an anion) while the metal loses its electrons (becomes a cation). A. J . Pappas and M.E. Goicoechea-Pappas

7-5

7 - Chemical Bonding and Molecular Ge ~tr.¼

Introductory Chemistry Ionic Bond (llEN ~ 2.0)

Covalent Bond (b.EN < 2.0)

This type of bond typically occurs between a metal and a non-metal and is electrostatic in nature. The metal transfers its valence electrons to the non-metal.

This type of bond typically occurs between nonmetals which sh:are electrons.

X•

•Z

➔

[X+]

X·

[: z-1

.z

...

X: Z = X- Z

The non-metals Xand Zare sharing a pair of electrons (i.e.. 2 electrons)

An electron is being transfered from X (a metal) to Z (a non-metal)

Problem 5:

Without looking at the EN value fo r each atom, iden tify the type of bond (polar covalent, non-polar covalent or ionic) that each is likely to form . a) 0 and O b) S and O c) Na and N d) Ca and N

e) P and F Problem 6: H- F VI.

f) H and H

g) C and O

h) Pb and F

Without looking at EN value for each atom , which of the following bonds is the: a) most polar and b) least polar? H- H H- Cl H- Br H- I

Lewis Electron-Dot Structure of Binary Ionic Compounds The Lewis structure for binary ionic compound is obtained by simply combining the Lewis structure of each individual ion (the metal cation and non-metal anion) that are its components. The metals transfers all of its valence electrons to the non-metal. The ratio of ions is as such because the net charge must add up to zero. [See section III of this module to review how to write the formula of monatomic ions.] Example I: Write the Lewis structure for the following binary 1omc compounds : a) NaBr b) BaF2 c) Al20 3 Solution: a) For NaB r, ( Na+) (: Br: -)

F: .. -h For Al20 3, ( AJ3+ h (: O: 2 - )3

b) For BaF2, ( Ba2+ ) ( : c)

Enclose the Lewis sturcture o f each ion in parenth eses or brackets and any subscripts are placed outside the parentheses.

Problem 7: Draw the Lewis stmcture for each of the following compounds: a) MgF2 b) Ca3N2 c) SrO d) KI

e) Ba3N2

A. J. Pappas and M.E. Goicoechea-Pappas

f) AlF 3 7-6

...... ...... ...-. .--. ...... ...... -...... a ii a ii aii

......

... ... ..._ ..... .. ~

VII. Lewis Electron-Dot Structure of Covalent Species When a covalent pecies (e.g., C12 - diatomic element, H20 - covalent compound, H30+ - polyatomic ion) is formed , electrons are shared between bonded atoms. A covalent bond is formed when non-m tals chemically combine. NOTE: The species whose Lewis structure you will be asked to draw will not break the octet I duet rule . Guidelines to Drawing Skeletal Structures: a) the central element (CE) is the element with the lowest numbered subscript. U ually the first element in the formula is the central elem nt , except H - it i never the central element. For those species having only one central lement (i.e. , the central element having no sub cripts), the remaining atom in the formula are placed surrounding the central element. b) In oxyacid (HxEOy) and oxyacids anions (HxEOy-), H is bonded to the O and not to the E. Eis a non-metal. c) oxygen atoms do not bond to each other except in: 02 , 03, and peroxide (e.g. , H202) . d) for species that have more than one central element, the mo t symmetrical keleton possible is used . Exam le J: U ing the above guideline , draw a skeletal tructure for each of the following: a) SOC12

b) CO2

c) H2S03

d) C2H4

Solution: 0

0 a)

Cl

s

Cl

b) 0

C

0

c) H 0 S

H H 0 H

d) H C C

H

al!9

.... ....

~ 111111

Guidelines to Drawing Lewis Structures: a) draw the skeletal Lewis structure b) draw the Lewis electron dot structure for each atom. For th sake of keeping the drawing as neat a po ible , direct single l ctron on adjacent atoms towards each other . c) draw a line from a ingle electron on the central element to a ingle electron on the surrounding atoms (the sharing of an electron pair constitute the formation a covalent bond) o that each of the atom end up with an octet around it (except hydrogen - it only ha two electron ). The pair(s) of electrons being hared mu t be placed between the two atoms forming the bond. A. J. Pappas and M.E. Goicoechea-Pappas

7-7

Iotroductory Chemistry

7 Chemical B_pnding and Motecu.I.a.r__Geometo:

OR: You may follow the procedure outlined as an Addendum at the end of this module. Example K : Draw the Lewis structure for CH4. Step I: Draw skeletal Lewis structure. C is the central element (it has the smallest subscript).

H

H C H H H

Step 2: Draw Lewis dot structure for each atom. [Notice has the electrons on H were drawn directed towards the central element.]

X

•

Hx • C . xH •

X

H

Step 3: Start forming covalent bonds by drawing a line from a single electron on one atom to a single electron on the other atom. When finished , C should have an octet (8 e- surrounding it) while H should only have a duet (2 esurrounding it). Step 4: Clean up the drawing (remove the shared electrons and leave the individual bonds in their place) .

H

I

H-C-H

I

H

Note, C (in CH4) has 4 sing le bonds . Example L: Draw the Lewis structure for H20 . Step 1: Draw skeletal Lewis structure. 0 is the central element (it has the smallest subscript).

H

Step 2: Draw Lewis dot structure for each atom . Place the two H with their one valence electron around O (place them at the sides of the imaginary square where oxygen's electrons are not p aired) .

Hx · .O. · xH

A . J. Pappas and M .E. Goicoechea-Pappas

0

H

..

7-8

.... ...-... ..• ...-, -""' ••"" ...... ...... --• -••

•• •• •• •• •• • II

9!!I

• el

----......

lillll

---

~

~ ~

~ ~ ~ ~

~

~ ~ ~

Step 3: Start forming covalent bonds by drawing a line from a single electron on one atom to a single electron on the other atom . When finished , 0 should have an octet (8 e- surrounding it) while H should only have a duet (2 e- surrounding it) .

~

~ ~

••

H- O - H ••

Note the O (in H20) has 2 single bond and 2 lone pair . Exam le M: Draw the Lewis structure for NCb. draw keletal and Lewis structure ~

..

xx

xx

Clx •N • xCl ~ xx xx • X ; Cl~

form bonds (share ingle e- on adjacent atoms)

..

xx

xx

; Cl-N-Cl;

t

XX

XX

clean-up drawing ~

..

xx

xx

Cl- N -

Cl~

XX

XX

I

~

~Cl~

xx

xx

Cl~ xx

Exam le N: Draw the Lewis structure for ethene, C2H4. draw keletal and Lewi structure

~

H

~ ~

..

H - - .. O ~x H

Step 4: Clean up the drawing (remove the shared electrons and leave the individual bond in their place).

~ ~

iar Geomruy

- Chemical Bondin

H X • • C . xH

X

•

.

.

Hx • C •

form bonds (share ingle e- on adjacent atom ) H

t

clean-up drawing

H

H

H

I

t

H-•CL-eCL.....tcH

I

H-C-== C-H

~ ~

~ ~ ~ ~ ~

~ ~

Exam le 0 : Draw the Lewis structure for hypoclorous ac id , HClO .

This is an oxyacid, H bonded to O and not to the central element, Cl. draw skeletal and Lewis structure

..

:O .. •

..

xx X

Qx xx

form bonds (share ingle e- on adjacent atoms)

·H

:a..

..

xx r---1[

0

xx

clean-up drawing

x--.- H

xx

:0 .. - 0 - H xx

~ ~

~

~

~

A. J. Pappas and M .E. Goicoechea-Pappas

7-9

""' ln.,.tr...,o=d~u ~ ~ - - ~====~==~~~~b,!:;!~olecular..Geometr;x

Problem 8: Draw Lewis structure for each of the following sub tance :

~ HF

b) N2

a) CO2

~ 02 h) CHClO

VIII. Lewis Electron-Dot Structures for Polyatomic Ions To draw the Lewis structures for polyatomic ions, one mu t take into account that electron have been added (if it i a negative ion) or subtracted (if it is a positive ion) from the molecule. If the ion has a: -1 charge, add 1 e· to the mo t electronegative (EN) atom; -2 charge, add 2 e·, one to each of the 2 most EN atoms; -3 charge, add 3 e·, one to each of the 3 most E

atoms;

+1 charge, ubtract 1 e· from the lea t EN atom. Exam le P: Write the Lewis Structure for draw skeletal and Lewis structure X

.

~O x - C· . xx X

xo ~ xx

add 1 e- to each of 2 most EN atoms

X

xQ~ X)
:'.

Introductory Chemistry d) Lewis Structure

Molecular Geometry (Shape)

Linear

I I

there are only 2 elements

I

•

Electronic Geometry

H-Q:

••

Linear Bond Angles

Undefined: Since there are only 2 atoms Problem 9: What is the molecular geometry (shape) for each of the following? c) NBr3 b) CS2 e) C032-

f) SiBr4

Problem 10: Provide both the electronic geometry and the bond angle(s) for each of the species in the previous problem? XI. Resonance [OPTIONAL] For many molecules and molecular ions , a single Lewis structure does not satifactorily account for the chemical and physical properties of the species. The true state of the species is better represented by a combination of two or more Lewis structures. These individual structures, are called resonance structures or contributors to the structure of that species. Resonance comes into play whenever there is/are multiple bonds present in a molecule or molecular ion . Resonance structures differ from each other only in the arrangement of electrons; only nonbonding electrons and electrons in a

multiple bond change locations in different resonance contributors. The nuclei of atoms in different resonance structures are in the same position.

A. Equivalent Resonance Structures If one of the atoms (let's call it X) around the central element has a double bond, and there is another atom of the same kind that is singly bonded to the central element , then equivalent resonance structures

exist. Equivalent resonance structures contribute equally to the real structure. The important thing to keep in mind, is that resonance structures are not true structures; the true structure is a composite or

hybrid of all the resonance structures. The element ozone, following equivalent resonance structures. A. J. Pappas and M.E . Goicoechea-Pappas

I I

o3 , has the 7-14

• •• •

• •• •• •• •• ••

•• •• •• •• •• •• •• •• •• 8111

..

Iii Iii Iii

.. .... -----

7 Chemical Bonding and Molecular Geometry

Introductory Chemistcy

A n arrow having two heads is used to indicate resonance structures

liiil

Ii)

Ozone really exists as a resonance hybrid (a combi nation) of the two equivalent resonance forms (each O - 0 bond is in-between being a single and a double bond).

Problem 11: Draw all possible equivalent resonance structures for each of the following species. b) SiTe2

m ii

&:lllii

~

[OPTIONAL] XII. Formal Charge (FC) A formal charge is basically a bookkeeping system that counts bonding electrons as though they were equally shared between atoms and its determination is useful in drawing "correct" Lewis structures. The following equation is used to assign a formal charge to bonded elements. FC =

atom's valence e- in non-bonded state

Group/

atom's valence ein bonded state

I

Half of the electrons in a bond are given to one atom and the other half are given to the other atom; having done so, then count the number of electrons around each atom (including its lone pairs).

The sum of all the formal charges (LFC) is equal to the charge on the species: a) for neutral molecules, LFC = 0~ b) for polyatomic ions, I.FC = charge on the ion.

The following provides an example of how to assign a formal charge to each atom in a covalently bonded species.

FC = 4 - 4 = 0 A. J. Pappas and M.E. Goicoechea-Pappas

7-15

r 7 - Chemical Bonding and Molecular_Geometry

Introductory Chemistry Problem 12:

Assign a formal charge to each atom in the following species.

b)

.. /··o:10 :o, [ ..

.· 0 ·.

c)

~

N

A Bricola e of Problems: 13. Draw the electron dot structure for each of the fo llowing:

a) Cs

b) Sn

c)

I

d) Se2-

e) Kr

t) Sr

g) Na+

h) Brl -

14. Complete the following table by ill ustrating the correct electron dot structure for each group. Let E represent any element within that group. IA

15.

IIA

IIIA

IVA

VA

VIA

VIIA

VIIIA

Which is the most electronegative element?

16. Which group of elements is the least reactive? 17 . The bond between which of the following has the least covalent character (i .e ., more ionic character, the atoms are sharing their electrons the least)? Fr and F

18.

H and F

Cl and F

I andF

F and F

Indicate the polarity of I-Fusing both partial charges and a bond dipole.

19. Draw the Lewis electron-dot structure for each of the following compounds. c) CsCI

f) Rb3P

20. What type of bonding involves sharing electrons between two atoms? 2 1. Draw the Lewis structure for each of the following.

a) c1022

b) Cl03-

c) S03 -

d) AsH3

e) N03-

f) SOC}i

g) CH20

h) SiH4

A. J. Pappas and M.E. Goicoechea-Pappas

t t

7- 16

........ --.. -.. --1W

;.,jj

7 - Chemical Bonding and Molecular Geometr..y

Imroductory Chemistry

22.

What is the molecular geometry (i.e., the molecular shape) for each species in problem 2 1?

23. What is the electronic geometry for each species in problem 21? 24. What are the bond angles for each species in problem 21?

25. Draw the Lewis structure for each of the following . a) SiCl4

b) MgO

c) NBr3

d) GaN

e) H2Se04

f) H3As04

aii ~

ail

~

--' ~

26. T ry this out for fun. Draw the Lewis structure for each of the following skeletal structures. 0 0 H a)

H C O H

b)

H

C

N

H

27. What is the formal charge on the central element in each of the species in problem 21? [OPTIONAL]

=-' e -'

mj

f!!IIIII r:1111

f!IIIII

,. SIi

----_, -•,,,

•,.

A. J. Pappas and M.E. Goicoechea-Pappas

7- 17

Addendum A Quick-N-Dirty Method for Drawing the Lewi Structure of Covalently Bonded Species This method will be illustrated by drawing the Lewis structure of SO 3.

l. Find the total number of valence electrons by adding the number of valence electron from each atom in the formula.

1S 30

2. Draw a single bond from the central element to each of the element mrounding it. [The fir t element in the formula i u ually the central element (except for H - it is never the central element).]

O - S-0

1(6) = 6 3(6) = .lli 24 [For polyatomic anion add the ab olute value of the charge to the total number of valence electrons. For polyatomic cations ubtract the charge from the total number of valence electron .]

3 . Subtract the electron used so far m the tructure from the total number of electrons. Each bond contains 2 electrons.

I

0 3 bonds x 2 e - = 6 e 24 - 6 = 18 e-

.. .. :o-s-o: •• I ..

4. The remaining electrons are spread, as pair , to the surrounding element first . Each surrounding element (except H) will receive nough electron pairs to have an octet. If any electron are left over, give them to the central element. 4. If the central element does not have an octet (except: H, which has a duet) , then lone pairs are converted into bonding pairs. If possible, make 2 double bonds before making one triple bond.

.. 0. .. .

make a bond into a lone pair ~o that will have an octet

-0

..

..

:o-s- o: q =o=s-o: ··

I

:o:

··

··

I

:o:

··

The table to your left Ii ·t the typical number of of bond and !>OT Ille 0.825 mol

CCI p 2 2

There i no need to do~nythingelse since whole numbers were obtained.

9-15

In tmdu Problem 15 : Provid d with elemental mass percent composition data, determine each compound's empirical fo1mul a . a) 86.41 % Sn and 13.59% b) A hydrocarbon (CxHy) is l l .84%H c) Tetrahydrocannabinol (THC) i compo ed of C , H and O and it ma s p rcent com po ition of carbon and hydrogen are: 80 .2 1% and 9.62% , re pectively. XVII. Molecular Formula (MF) Determination To obtain th molecular formul a of a compound , both its empirical formula and molar ma mu t be known. The followin g equation hows how the molecular formula can be obtained. (

Mo l ar M ass (MM) ) . . l F l ( Emp1nca ormu a Mass EFM )

x Emp irical Fo rmula (EF) = Molecu lar Formul a (MF)

Example F': If the empirical formula for Freon-12 is CCl2F2, what is it molecular formula given that is molar mas i 12 1.0?

Solution: The empirical formula mass for l mol of CCl2F2 is: C: 1 mol x 12.0 9 / mol = 12.0 g Cl: 2 mol x 35.5 9 / mol = 71.0 g F: 2 mol x 19.0 9 / m ol = 38.0 g 121.0 g

¢::

mass of 1 mol of CCl2F2

Since the molar mass is given a 121.0 g/mole , the MF is :

9

( 121.0 /mol) 121.0 9/ mol

X

CCI F

2 2

=

ICCI F I .

2 2.

Molecular Formula

Example G ': A 125.0 g sample the hydrocarbon (CxHy) limonene (MM = 136) was found to contain 110.2 g of C. Given thi information, detennine its empirical formula.

Solution: 1) Find g H, by using the Law of Conservation of Mass .

= 125.0 g limonene - 110 .2 g = 14 .8 g

110.2 g C + ? g H ?gH

= 125 .0

g

2) Determine the empirical formula (EF).

a. Convert g ⇒ mol (i.e., 7 by each element's MM) C

110.2

g

12.01 9 / mol

H

14.s

g

⇒

C9.1 76 mol H1 4_7 mol

1.01 9 lmol

A. J. Pappas and M .E. Goicoechea-Pappas

9-16

b. + by smallest number (of moles).

C9 .176 mol H 9.176 mol

c.

14.7 mol 9.176 mol

::::::> CH 1.60

Multiply each subscript by a whole number o each within ± 0 .08 of a whole number. x

CH1.60

5

) C5Hg ~ EF

2) Determine the empirical formula ma s (EFM).

C: 5mol x 12.0 9 /mol - 60 .0 g H: 8 molx 1.01 9 /mol = 8.08 68.1 g

~

EFM

3) Determine the molecular formula (MF).

MM) ( EFM

X

EF

⇒

9/

(136 nlol) 68.1

mol

2 .00

Molecu lar Formula

X

C5Hs

= IC10H16

I

multipl y each

ubscript by 2

Problem 16: Provided with the following data determine the molecular formula for each compound. a) The molar ma CH2Br.

of a compound is 188

g/moi

and its empirical fomiu la i

b) The empirical formula and molar ma s of a compound are C2H3Cl and 187.5 g/mol, re pectively. c) A hydrocarbon (CxHy) was found to have an elemental mass composition of carbon equal to 89 .49% and a molar ma 268 .4 g/ moi · d) A compound that contain C, Hand F has a molar mass of 1048 g/moi· 110.00 grams of this compound was found to contain 35 .28 g of carbon and 2.96 g hydrogen.

A Variety of Problems: 17. How many grams of BaCO3 (MM = 197) are in 0 .500 mole of BaCO3? 18. How many moles of Ba(NO3)z (MM = 261) are in 10 .0 grams of Ba(NO3)i? 19. How many mole of sodi um are in 6.90 gram of odium (MM= 23 .0)? 20. What is the molar mass of Al2(C2O4)3? A. J. Pappas and M.E. Goicoechea-Pappas

9- 17

21. How many moles of chlorine atoms are in a sample which contains 3.01 x 1Q23 chlorine atom ? 22. How many potassium atom are in 0.249 moles of potassium? 23. What is the average ma s (in grams) of a silver atom? 24. What i the average mass (in f!tomic mass gnits) of a copper atom ? 25. One atom has a ma of 2.33 x 10-22 gram . How many grams would 1 mol of this atom weigh (i.e. , what i the molar mass of this atom)? 26. Calculate the number of H2 molecule in 3.50 mole of H227. How many carbon atoms are in 11.5 g of C2H5OH? 28. In 3.00 x 1020 molecule of C 12H22 O 11 , how many C atom are present? 29. What i the a erage mass (in gram ) of one SO2 molecul ? 30. What i the average ma s (in amu) of one molecule of I2? 31. How many molecule are in 17.2 g of CuS (MM= 95.6)? 32. A molecule is the ba ic repeating unit for which of the following ? a) SO3

b) VI4

c) Cl2

d) CBr4

e) K2S

33. What is the ma s percent oxygen in K2SO4? 34. What i the elemental mas percent composition of sulfur in Al2(SO3)3?

35. A 15.10 gram sample of silver was heated in the pre ence of oxygen. The chemical change that occurred resulted in a compound that had a ma s of 16.12 grams , given thi information what is the ma s p rcent O in the oxide that was formed? 36. Propane i a hydrocarbon (CxHy) that has a ma s percent composition of hydrogen that is equal of 18.3%. In a 45.2 g sample of propane , what ma of carbon (in g) doe it contain? 37. Given that a metal pho phide has a formula of M3P2 and a mass percent of phosphorus that i equal 24.5%, what mass of M3P2 (in g) contain 66.6 g of M? 38. A compound contains 7 .8 g of K, 16.0 g Br, and 6.64 g 0. Find its mpirical formula.

Mo o 39. The mass composition of a 33.34 g ample of a compound containino b o• and His: 13.90 g Mg and 18.30 g 0. Given thi information , determine it empirical formula. A. J. Pappa and M.E. Goicoechea-Pappas

9- 18

40. If l.631 gram of Ga wa found to react with exactly 5 .609 grams of Br. What i the empirical formula of the compound that forms? 41. A compound' mass percent compo ition i 46.7% nitrogen and 53.3 % oxygen. What is its empirical formula? 42. A compound containing the elements tungsten and oxygen ha a mass percent compo ition of 20.7 % 0. What is its empirical formula? 43. A compound (that is not an acid) contain hydrogen, nitrogen and oxygen. The compound's mas percent of hydrogen and nitrogen are 5.04% and 35.00%, re pectively. Given thi information , determine it empirical formula. 44. Vitamin C (MM = 176) is composed of carbon, hydrogen and oxygen. Given that it ma percent elemental compo ition i 40.9% carbon and 4.55% hydrogen , respectively, determine both it empi ri cal and molecular formula. 45. A compound whose molar ma s is 249 g/mot has an elemental percent composition of: 14.46% carbon and 85.54% chlorine. Given this information what i it molecular formula? 46. A chemical reaction wa conducted by mixing 20.00 gram of odium and 10 .00 gram of oxygen. At the end of the reaction, all of the oxygen had reacted and 5 .62 grams of sodium were left unreacted, in addition to the oxide formed . If the molar mass of the oxide that formed i 78 .0 g/m01, determine its molecular formula.

A . J . Pappas and M.E. Goicoechea-Pappas

9-19

ANSWERS 1.

a) 3 H : l P : 4 0 d) 1 Ba: 2 : 6 0

b) 1 Ag : 1 N : 3 0

c) 1 Ca : 2 0 : 2 H

2.

a) 24 .3 g/mol Mg

b) 254. g/mol l2

c) 124. g/mol P4

3.

a) 171. g/mol Ba(OH)2 c) 392. g/mol Crz(SO4)3

4.

a) 1 mol Ba(OH)2 = 171. g = 6.02 x 1023 Ba(OH)2 FU b) 1 Ba(OH)2 FU = 171 . amu Ba(OH)2 c) 1 Na atom = 23.0 amu Na d) 1 P4 molecule = 124. amu P4 = 4 P atom e) 1 mol SiH4= 32.1 g SiH4 = 4 mol H atoms= 4(6.02x 10 23) H atoms f) 1 SiH4 molecule= 32.1 amu SiH4 = 1 Si atom= 4 H atoms

5.

a) 17.4gCl2

b) 34.7gSg

c) 431.gMg(OH)2

6.

a) 0.313 mol Bri

b) 1.70 mol C3Hg

c) 0.119 mol

7.

a) 1.28 x 104 mg NaOH

b) 167. g/mol C7H12Clz

aCIO3

b) 0.409 kg Cr( 03)3

8. a) 3.91 x 1024 Cu atom

b) 120. mol Cu

c) 3.91 x 1024 N2 molecules

d) 120. mol N2

e) 3 .91 x 1024 NaCl FU

f) 120. mol NaCl b) 2.32 x l0-18 mg Fe

9. a) 1.98 x 104 g C c) 4.07 x 10-4 kg Clz

d) 1.52 x 102s H2O molecule

e) 2 .82 x 10-s g NH3

f) 1.41 x 1024 C2H6 molecule

g) 2.41 x 1022 Ca(NO3)2 FU

h) 4 .25 x 102 2 NH4Cl FU

i) 859. g KOH

j) zero (ionic crnpds are not made up of molecu les)

k) 1.14 x 1023 F atoms

1) 22.8 g F2

m) 4.98 x 10-23 g C2H6

n) 9.79 x I0-23 g Co

10. a) 4.40 x 1023 0 atoms

b) 1.45 x 1023 0 atoms

c) 2.49 x 1024 H atoms 11. a) ll.4gS

b) 6.49gS

c) 11.lgS

12. a) 58.5% 0

b) 49.0% 0

c) 24 .7 % 0

13. a) 13.l g Fe and 5.7 g O

d) 43.0% 0

e) 28.5 ~ 0

b) 29.l dg S

14. a) K3B03

b) C4H5N2O

15. a) Sn3N4

b) C5Hg

A. J . Pappas and M.E. Goicoechea-Pappas

9-20

16. a) C2H 4Br 2 C20H2s (EF: C5H7) 17. 9 .7 g BaCO3

b) C6H9Cl d) C 2sH2gF36

(EF: C7H7F9)

18 . 3.83 x 10-2 mol Ba(NO3)2 19. 0.300 mole Na 20. 318. g/mol Al2(C2O4)3 21 . 0 .500 mol Cl atom 22. 1.50 x 1023 K atom 23. 1.79 X l0-22 g 24. 63.5 amu 25. 1 .40 X 10 2 g (MM= 140 8/mol) 26. 2.11 x l024H 2 mol cule 27 . 3.01 x 1023 C atom

28. 3.60 x 1021 C atom 29. 1.06 x 10-22 g SO2 30. 254 amu l2 1. zero· it's an ionic compound (which i not made of up molecules) 32. a c d (A molecule i the ba ic repeating unit of covalent compound .)

33. 34. 35. 36.

36 .7% 0 32.7% S 6.33% 0 36 .9 g C

37. 88.2 g M3P2 38. KBr02 39. MgO2H2 ⇒ Mg(OH)i 40. GaBr3 41. 0 (NOTE: The element that is lea t EN is writt n 1st .) 42. WO3 (NOTE: The element that is 1 a t EN , in thi ca the metal tungsten, i written 1 1.) 43. 2H4O3 ( OTE: Since thi is not an acid, don't list H P 1 • T he ionic compound ammonium nitrate Nl4 03, ha the indicated empirical formul a.) 44. EF = C3H4O3

MF= C6HsO6

45. C3Cl6 46. Na2O2 (NOTE: Thi i

odium peroxide.)

A . J. Pappas and M.E. Goicoechea-Pappa

9-21

BALANCING CHEMICAL EQUATIONS AND REACTION STOICHIOMETRY I.

Balancing Chemical Equations In chemical reaction atom are never de troyed or created; they are only hifted from one ubstance to another. This mean that all atom pre ent at the tart (on th left ide of the chemical equation) must be there at the end (on the right ide of the chemical equation). Coefficients (other than one) placed in front of each of the fonnula are u ed to balance the equation. If no coefficient is present in front of one of the fonnulas of a balanced quation , then it i assumed that the coeffici nt i one. The coefficients of a balanced chemical equation, must be the simplest whole numbers possible. In ord r to balance a chemical equation o that the ame number of atoms of each element are on the left and right hand side of a chemical equation , u e the following a a guide: a) if polyatomic ions are present (on both sides of the equation), balance tho ea a group, b) a e for la t any uncombined elements or atoms that appear in more than one of the formulas (on the same side of the equation). c) if fraction need to be u ed, clear them up by multiplying all co ffici nt by the lowe t common denominator. A fraction i u ed if there are an odd number (other than I) of atoms on one ide of the equation and a diatomic molecule (of the ame atom) on the other ide of the equation (see Example B). If there i an odd number of atom (other than 1) on one side of the equation and an even number on the other (and if those atom appear only once in their respective side of the equation), then the coefficient v ill b the cros multiplication product of the subscripts ( ee Example C). In balancing chemical equations, never change a correct chemical formula to balance an equation. The following chemical equation will be balanced u ing the above procedure: Hg3(P04)2 + Al ➔ AlP04 + Hg (unbalanced) a) Balance polyatomic ion "P04 11 fu t. Th rear two P04 on the left ide of the equation and only one P04 on the right hand ide of th equation; therefore , place a coefficient of 2 in front of AlP04.

The following shorthand notation will mean the same thing: P04 (2 ➔ 1) :. ~ 2 AIP04 A. J. Pappas and M .E. Goicoechea-Pappas

10-1

b) We can either balance Hg or Al next, since both are uncombined. Let' pick Hg next.

Hg3(PO4)2 + Al ➔ 2 AIPO4 + 3 Hg

(3 ➔ I) :. ➔ 3 Hg

Hg

c) Finally balance Al. Al

(1 ➔ 2)

Hg3(PO4)2 + 2 Al ➔ 2 AlPO4 + 3 Hg

➔

:. 2Al

Exam le A: Balance the following chemical equation:

C6H12O2 + 02 ➔ CO2

+ HzO

Solution: Save O for la t because it is uncombined, as well as being pre ent in more than one fommla on the same side of the equation. C

(6 ➔ I) :. ➔ 6 CO2

H

(I 2

0

((2)+02) ➔

➔

C6H12O2 + 02 ➔ 6 CO2 + H2O C6H12O2 + 02 ➔ 6 CO2 + 6 H2O

2) :. ➔ 6 H2O ➔

18):. 802

C6H12O2 + 8 02 ➔ 6 CO2 + 6 H2O

Exam le B: Balance the following chemical equation:

C4H10 + 02 ➔ CO2 + H2O Solution.: Save O for last because it is uncombined, a well a , being present in more than one formula on the ame ide of the equation. ➔

C

(4

H

(10

0

(2

1) :.

➔

➔

2) :.

➔

4CO2

C4H 10 + 02 ➔ 4 CO2 + H2O

5 H2O

C4H10 + 02 ➔ 4 CO2 + 5 H2O

➔ 13

13) :. -

2

02

13

➔

C4H10 + -2 02

Multiply all coefficient by 2 to clear up fractions.

➔

4 CO2 + 5 H2O

2 C4H10 + 1302 ➔ 8 CO2+ 10 H2O

Exam le C: Balance the following chemical equation: Pb

+ AuBr3 ➔ PbBr2 + Au

Solution: Br (3

➔ 2)

:. 2 AuBr3

➔

3 PbBr2

Pb (1 ➔ 3) .-. 3 Pb ➔

Au (2

➔

l) :.

➔

2 Au

A. J. Pappas and M.E. Goicoechea-Pappas

Pb

+ 2 AuBq ➔ 3 PbBr2 + Au

3 Pb+ 2 AuBr3

➔3

PbBr2 + Au

3 Pb+ 2 AuBr3 ➔ 3 PbBr2 + 2 Au 10-2

Problem 1: Balance each of the following equations: a)

C2H6

b)

+ 02

a2B407

c)

CaC03

d)

C3Hg

HCl

+ 02

H20

+

+ H20

HCI

+ +

CO2

➔

CaC]z

➔

CO2

➔

+

aCl

CO2 +

H20

H3B03

➔

+

+

H20

II. Reaction Stoichiometry The quantitative relation hip that exists among the ub tances involved in a chemical reaction (which i established by the balanced chemical equation) is called reaction stoichiometry. The following balanc d chemical equation ➔

5 C + 2 S02 MM:

12.0

64.l

(eq. 1)

CS2 + 4 CO 28.0

76.l

tell u that when 5 mol of C is reacted with 2 mol of S02, l mol of CS2 and 4 mol of CO are produced. Thus, the following equalities can be used in relation to eq. 1: 5 mol C = 2 mol S02 = 1 mol CS2 = 4 mol CO The following concept map can be u ed to solve reaction stoichiometry problem. g MM=> mol

Other Mass d~ -

B

mL

Other Volume

grams X

Use MM of X

Use density of X ml pure X

mol X

Use Coefficients

mol

Use MM

z

grams

of Balanced Eq.

of Z

z

Use Avogadro's

Use Avogadro's

No.

No.

#of @of X

# of @of Z

where,@ = atoms, molecules, FU or ions

Other Mass

Use density of Z

lmL7 . Other ~Volume

1 mol ® = 6.02 x 10 23 ®

NOTE: X and Z are substances undergoing a chemical reaction - they can either be both reactants or products or a combination of reactants and products.

Depending on what is given and/or a ked to be solv d, each arrow repre ents a step (conver ion factor) in the solution to the problem .

A. J. Pappa and M.E. Goicoechea-Pappas

J0-3

IO Balancing Equations and Reaccion_Stoi~me.u;x

Introdu~QC'. Chemistry

a) Mole-to-Mole Conversion Process:

mol

Use Coefficients

X

of Balanced Eq.

mol

z

Example D: Consider eq. J. How many mol of C are needed to react with 6 .00 mol of SO2? Coefficients of Balanced Eq.

Solution:

? mol C = 6.00 mol SO2 (

5

mol C ) = 15.0 mo! C

2 mol S02

Example E: Consider eq. J. How many mo! of CO are produced when 10.0 mol of C are reacted with the appropriate amount of SO2? Coefficients of Balanced Eq.

Solution:

? mol CO = 10.0 mol C (

Problem 2:

4

mol

co) = 8.00 mol CO

SmolC

For the following balanced equation, answer the following questions. 2 Al + 3 NiC}i ➔ 3 Ni MM:

129.6

27.0

+

2 AlCl3

58.7

133.3

a) If 3 .00 mol of Al are used in the reaction , then how many moles of Ni will be produced? b) If 3.00 mol of Al are used in the reaction, then how many moles

of AlCI) will be produced?

b) Mole-to-Gram Conversion Process:

Example F:

Solution:

mol

Use Coefficients

X

of Balanced Eq.

~

,

mol

z

Use MM of Z

~

,

grams

z

Consider eq. 1. How many grams of SO2 are needed to react with 5 .00 mol of C? Coefficients of Balanced Eq.

MM (S0 2)

? g SO2 = 5.00 mol C (2 mol 502) (64.1 g S02) 128 SO 5 mol C 1 mol S0 2 = ·g 2 A. J. Pappas and M.E. Goicoechea-Pappas

10-4

10 - Balancing Equations and Reaction Stoichiometcy

Introductory Chemistry

Usi ng the balanced equation in Problem 2, answer the following questions.

Problem 3:

a) If 4.00 mol of Al are used in the reaction, then how many grams of the NiCh are needed in the reaction? b) If 5.00 mol of Ni are produced, then how many grams of Al must have been used in the reaction?

c) Gram-to-Mole Conversion Process:

grams

X

Use MM

mol

Use Coefficients

X

of Balanced Eq.

r

of X

-

mol

z

r

Example G : Consider eq. 1. How many mol of C (MM= 12.0) are needed to produce 100. g of CO (MM= 28.0)? Coefficients

Solution:

MM (CO)

? mol C = 100. g CO

Problem 4:

of Balanced Eq.

co) (45mol mol C) _ 4 46 C CO · g

1 mol ( 28.0 g CO

Using the balanced equation in Problem 2, answer the following questions.

a) If 5 .00 g of NiCl2 are used in the reaction, then how many moles of Ni are produced?

b) lf 5.00 g of NiCh are used in the reaction , then how many mo]es of A1Cl3 are produced?

d) Gram-to-Gram Conversion Process:

grams

Use MM

mol

Use Coefficients

X

of X

X

o f Balanced Eq.

.

mol

z

Use MM of Z

.

grams

z

Example H: Consider eq. 1. How many grams of C are needed to react with 100 . g of S02?

Solution:

Coefficients

MM (S0 2)

1 m ol S0 2 )

? C=l00 . 0 S02 ( - - g t:> 64.1 g S0 2 A. J. Pappas and M .E. Goicoechea-Pappas

of Balanced (

Eq.

5 mol C ) -2 mol S0 2

MM (C)

(12. 0g-C) -1 mol C

= 46 8 · gC 10-5

IO - Balancing Equations and Reaction StoicJ,iom®

lntr..Qductory Chemistry

Using the balanced equation in Problem 2, answer the following questions.

Problem 5:

a) If 3.00 g of Al are used in the reaction , then how many grams of Ni are produced?

b) If 3 .00 g of Al are used in the reaction, then how many grams of AlCl3 are produced? e) Using Density When Dealing With Liquid Reactants or Products

When dealing with a substance that is a liquid, it is more convenient to measure its volume than its mass . A substance's density (typical units: g/mL) can be used as a conversion factor to either obtain the mass of a substance from its volume or vice-versa. Example I: How many moles of 0 2 are needed to react with 15.0 mL of C6H6 according to the follo wing balanced equation?

2 C6H6 (I) + 15 0 2 (g) MM:

78.0

d (g/mL):

0.874

Solution:

ml pure X

12 CO2 (g) + 6 H20

➔

Use denisty of X

Use MM of X

grams X

Use CoefficientS of Balanced Eq.

mol X

density (C6H6)

? mol 0 2 = 15.0 mL C6H6 (

18.0 1.00

44.0

32.0

mol

z

Coefficients of Balanced Eq.

MM (C6H6)

0.874 g C6 H 6 ) (l mol C6 H6 )

78.0 g C6 H6

1 ml C6H6

(I')

15 mol 0 2

(

)

2 mol C6H6

1.26 mol 02 Example J: How many mL of water should be produced when 2.16 mol of C6H6 is reacted with a stoichiometric (i.e., appropriate amount) of 02 according to the balanced equation in Example I? Solution:

mol

X

Use Coefficients

mol

Use MM

of Balanced Eq.

z

of Z

Coefficients of Balanced Eq.

? mL H20 = 2.1 6 mol C6H6(

grams

z

MM (H2 0 )

6 mol HzO ) (1 8.0 g HzO) 2 mol C6 H6 1 mol H 2 0

A. J. Pappas and M.E. Goicoechea-Pappas

Use denisty

ml

of Z

pu re Z

density ( H 20 )

( 1 mL H20 ) 1.00 g H 2 0

10-6

uatioos and Reaction Stoichiome _ Problem 6: The following balanced equation illustrates the dehydration of ethanol to produce diethyl ether and water. 2 CH3CH2OH ~ (CH3CH2hO

+

H2O

MM:

46.1

74.l

18.0

d (glee):

0.789

0.713

1.00

a) 5 .0 moles CH3CH 2OH should produce _ _ _ _ mol of H2O.

mol of (CH3CH2hO and

b) _ _ mol ofCH3CH2OH should produce0.217 mol of both (CH3CH2hO and H2O. c) _ _ mL of (CH3CH2hO should form when 4.00 mol of CH3CH2OH is dehydrated. d) When _ __ mL of ethanol are dehydrated , 217 mL of diethyl ether should be produced. Problem 7: When butane , C 4H 10 (MM = 58.l) burns in the flame of a cigarette lighter, water and carbon dioxide are the principal products. 2C4H10 + 1302 ➔

8CO2 + I0H2O

How many grams of oxygen (MM = 32.0) would be required to react with 232.0 mg of butane? Problem 8: The preparation of aspirin (MM = 180.) involves reacting salicylic acid with acetic anhydride. C7H6O3 salicylic acid

+ acetic anhydride

aspmn

acetic acid

How many mL of acetic anhydride (d = 1.082 g/mL, MM = 102.1) would be required to form 5.00 mg of a pirin?

III. Limiting or Excess Reagent Problems A Limiting or Excess Reagent Problem is one in which quant1tat1ve infonnation (e.g., moles, g, mL, etc.) is being provided about more than one of the reactants (reagents or starting materials) in a chemical reaction. As with any type of reaction stoichiometry problem , a balanced equation i required. The excess reagent (xsR) incorrectly predict the formation of too much product, while the limiting reagent (LR) predicts the mailer (and correct) amount of product that should be produced. The theoretical yield i the maximum amount of product that is expected (as preclicted by the LR). A. J. Pappas and M.E. Goicoechea-Pappas

10-7

Introductory Chemistry

Io - Balancing Equations and Reaction Stoichiometry

At the end of the reaction , there should be no limiting reagent (LR) remaining unreacted (i.e., it has all reacted) and some excess reagent (xsR) will remain unreacted (in excess, unconsumed, leftover, unused). Once the limiting reagent (LR) is identified, use the amount of it that was given to predict how much of: •

any other product should be produced

•

the excess reagent (xsR) was only

NOTE Amounts (amt.) can be in any unit (i.e.. mol, g, etc.): however, all

needed (required or necessary).

The equation to the right illustrates how to calculate how much xsR remains unreacted.

must have the same unit. amt.. xsR = initial amt. - needed amt. unreacted (as predicted by LR)

The following "balanced equation" will be used to illustrate the concept of a limiting or excess reagent. Starting Materials, "Reactants" o r "Reagems"

Chassis +

Product

----+>

Car C

(!'

G If 9 Chassis and 28 tires are available: a)

What is the maximum number of cars that should be produced? Set up two stoichiometry problems each starting with the give n initial amount of stai1ing material (i.e., reagent or reactant) and then calculate how much product each predicts will be produced. Excess IncorrectI redicts the Rea~rmation of too much product

. (- lcar-) = 9 cars ? cars = 9 chassis 1chassis

Limiting

Reag,m

~:lie,

Predicts the formation of the

? cars = 28 tires

,(m:u:~:f )product

- .4 llres

= 7 cars

As predicted by the limiting reagent , the maximum number of cars expected to be produced is 7. The theoretical yield of cars is: 7

b)

Which is the limiting reagent (LR) and the excess reagent (xsR)? LR: Tires (reagent that predicts the smaller amount of product) xsR: Chassis (reagent that pred icts the formation of too much product)

c)

How much of the limiting reagent remains unu sed (i.e., unreacte d) ?

All of the limiting reagent (i .e ., the tires) is expected to be used up; thu s, at the end of the "reaction" , no tires are expected to remain uncons umed . A. J. Pappas and M.E. Goicoechea-Pappas

10-8

IO - Balancing Equations and_Reaction Stoichiometry

Introductory Chemistry

d) How much of the excess reage nt was only necessary (consumed , needed or used)? Starting with the a mount of limiting reagent (28 tires), solve fo r how much of the excess reagent (chassis) was on Iy necessary. . ( l chassis ) = 7 chassis. ? chassis = 28 tires . /" 4 tire s '\_ "'-Only 7 chassis were needed

Start w ith amo unt of / Lim iting Reagent given

e)

How much of the excess reagent is expected to remain unused (i .e., unreacted, leftover, uncon sumed, in excess)? amt. xsR unreacted

=

initial amt. - needed amt. (as predicted by LR)

9- 7 = 2 c hassis re main unused

Example K : If 5.00 g of benzene are reacted with 6.00 g of oxygen accord ing to the fol1owing balanced equation , then: 2 C6H6 + 15 0 2 ➔ 12 CO2 MM:

78. l

32.0

+ 6 H20

44.0

18.0

a) Ide ntify the limiting and excess reagent.

Solution: To identify the limiting and excess reagent, you need to set up two stoichiometry problems to solve for how much product (either CO2 or H 2O) each given initial amount of reagent predicts will be produced. We'll detennine how many mol of CO2 (simply because it was the first product listed) each given amount of reactant predicts should be produced.

? mol CO2

= 6.00 g/ 0 2

Limitino e

Rea8'

1 mol 0 2 ) ( ---32.0 g 0 2

( 1 2 m ol CO 2 ) ----15 m ot 0 2

=0 .1 50 mol

Predicts the formation of the smaller (and correct) amount of product

Oxygen (0 2) is the liiniting reagent and benzene (C6H 6) is the excess reagent. The theoretical yield (i.e., the maximum amoW1t) of CO 2 that is

expected to be produced is 0.150 mol as predicted by the limiting reagent. A. J. Pappas and M .E. Goicoechea-Pappas

10-9

Introductory Chemistry

IO - BaJaociog Equations and ReacLion Stoicbiometcy

b) Calculate the theoretical yield of water (in g). Solution: Once the identity of the limiting reagent (i.e., 0 2) has been established, we can use the initial amount of it that was given to solve for how much water (i.e., a product) should be produced.

(1

o2 ) . g H2O =6.00gO2 -mol - -/' 32.0 g 0 2

?

Start with amount of Limiting Reagen t given

MM

(6 H-0) (18.0 ---~ - -g- -mol 2 15 mol Oz coefficients

HzO) = 1 ·35 g

1 mol H 2 0 MM

'\

Theoretical Yield of H 20

c) How many grams of excess reagent were only required?

Solution: Once the identity of the limiting reagent (i.e., 02) has been established, we can use the initial amount of it that was given to solve for how much of the excess reagent (i.e., C5H6) was only required.

(1

mol 0 2 ) ? g C5H6 = 6.00 g 02 - - - 32.0 g Oz /

(2-mol - -C6H6) - - (78.1 - --g C6H6) - - = 1.95 g 15 mol Oz

1 mol C6 H 6

coefficien1s

MM

MM

Start with amount of Limiting Reagent g iven

/

Mass of excess reagent (i.e. , C6H 6) that was only needed

d) How many grams of excess reagent are expected to be left unconsumed?

Solution: To find out how much of the excess reagent (i.e., C6H6) remains unused (unreacted, in excess, leftover, unconsumed), use the following equation: amt. xsR = initial amt. - needed amt. unreacted

(as predicted by LR)

= 5.00 g - 1.95 g = 3.05 g C6H6 remain unconsumed Example L:

If 15.6 g of phosphoric acid are reacted with 18.8 g of copper(II) hydroxide according to the following balanced equation, then:

2 H3PO4 + 3 Cu(OHh MM:

98.0

97.6

➔

6 H2O + Cu3(PO4h 18.0

381

a) How many grams of each product are expected to be produced? Solution: Because quantitative infonnation is being given about more than one of the reactants, this is a Limiting Reagent Problem. This being the case, two stoichiometry problems need to be set-up each starting with the given initial amount of reagent and then in each case solving for how much of one of the products [e.g. H20] is expected to be produced. A. J. Pappas and M.E. Goicoechea-Pappas

10-10

lntroductory Chemistry

10 Balancing Equations and R~acti.Qn--5Joichiometrx

Excess Reagent ~

? g HzO

= 15.6 g H3PO4

In~orrectly predicts the formation of too much product

1 mol H3P04) ( - - -~~

~

( 6 mol HzO) (18.0 g HzO )

- - - -2 mol H-:i P0 4

98.0 g H-:iP04 MM

- -- 1 mol H 2 0

coefficients

=

8.60 g

MM

l mol Cu(0H)i ) ( 6 mol H2 0 ) (18 .0 g H2 0) ? g H?O = 18.8 g Cu(OHh ( - - - - ---- - -- - ----/ 97.6 g Cu(0~) 7 3 m~l Cu(OH)J 1 mol HzD L .um.t.mo R eaoenr ,.., '"'

Predicts the formation of the smaller- - ---....,.-....,.----,-----,,-----:--

= 6.93 g

t

Theoretical Yield of H 2 0

(and correct) amount of product

Now that we have established which is the limiting reagent [i.e., Cu(OHh], we can use the initial amount of it that was given to solve for how much of the other product [i.e., Cu3(PO4) 2] in the reaction is expected to be produced. 1 mol Cu(OH) 2 ) ?g= 18.8gCu(OH)i ( - - - - cu (P0 )i ' 97.6 g Cu(OH) 7 4

3

(1

mol Cu3 (P0 4

)z) (381 g

3 mol Cu(0H) 7

Cu 3 (P0 4

)z) =

24.Sg / Theoretical Yie ld of Cu/P04 )i

1 mol Cu1 (P0 4 )J

Start with amount of Limiting Reagent given

Thus, 6.93 g of H2O and 24.5 g of Cu3 (PO4) 2 are expected to be produced. b) How many grams of the excess reagent were only required? Solution: Once the identity of the limiting reagent [i.e., Cu(OHh] has been established, we can use the initial amount of it that was given to solve for how much of the excess reagent [i.e., H 3PO4] was only required. ? (1

H ;

= 18.8 g Cu(OH)z ( '

3 04

1 mol Cu(OH)z ) ( 97.6 g Cu(OH) 7

2 mol H3P04 ) 3 mol Cu(OH) 7

(98.0 g

H3P04)

1 mol H-:iP0 4

= 12.6 g J'

Mass of excess reagent (i.e., H3 P0 4) that was only needed

Start with amount of Limiting Reagent given

The Law of Conservation of Mass could also be used to solve this problem.

2 H3PO4 + 3 Cu(OHh ~

6 H2O +

18.8 g

?

31.2g - 18.8g = l2.6g

➔

Cu3(PO 4) 2

6.93 g

24.5 g 31.2g

31.2 g

c) How many grams of the excess reagent are expected to be left unreacted?

Solution:

amt.xsR

=

initial amt. -

unreacted

needed amt. (as predicted by LR)

15.6 g - 12.6 g 3.0 g H 3P04 remain unreacted A. J. Pappas and M.E. Goicoechea-Pappas

10-11

Problem 9:

If 25.2 g of aluminum are mixed with 125 g of ·ulfuric acid, then how many grams of aluminum sulfate hould be produced?

2 Al

+ 3 H2SO4

27.0

MM:

➔

98.l

3 H2 + Al2(SO4)3 342

2.02

Problem 10: If 36.0 g ample of calcium hydroxide was allowed to react with 54.0 g of phosphoric acid, then what i the theoretical yield (in g) of calcium pho phate?

3 Ca(OH)i ( ) + 2 H3PO4 (aq) ➔ Ca3(PO4h (s) + 6 H2O (-t'.) MM:

74.1

98.0

310.

18.0

Problem 11: Consider the following balanced equation: Mg (s) + H2SO4 (aq) ➔ MM: 24.3 98.1

H2 (g) + MgSO4 (aq) 2.02 120

a) If 1.50 g of Mg are treated with 8.30 g of H2SO4, how many grams of H2 hould be produced (i.e., what is the theoretical yield of H2 (in g)? b) Calculate both the number of gram and mol of the exce reagent which remains unconsumed at the end of the reaction. Problem 12: Consider the following balanced equation: 3 CuS

(s)

+ 8 HN03 (aq)

➔

3 Cu( 03)i (aq) + 3 S (s) + 2

O (g)

+4

H 20

(-f'J

a) If 0.600 mol of copper(II) ulfide is treated with 1.40 mol of nitric acid, what is the theoretical yield (in mol) of copper(II) nitrate? b) Calculate both the number of mole and gram of the excess reagent remaining unreacted at the end of the reaction. IV. Percent Yield The product yield can be less than expected for the following reasons: a) the reaction did not go to completion (i.e., the reactants did not completely get converted into products); b) a reaction other than the one de ired (side reaction ) also occurred; c) the separation of the desired product was so difficult that all of the product formed couldn't be uccessfully i olated and/or d) the person prefonning the experiment had poor lab technique. A. J. Pappas and M.E. Goicoechea-Pappa

10- 12

Introductory Chemistry

IO - Balancing Equations and Reaction Stoichiometrv

The percent yield (of a particular product) gives an indication of how successful product isolation proceeded and can be obtained by using the following equation. Units for both the actual yield and theoretical yield must be the same .

Actual Yield Theoretical Yield

% Yield= - - - - - - • 100

Actual Yield - the amount of specified product actually obtained (isolated, collected, procured) after performing the reaction. Theoretical Yield - the maximum amount of specified product that should be obtained as predicted by the limiting reagent. Example M: Consider the following balanced chemical equation: 2 C 6H 6 (t) + 15 0 2 (g) ➔ 12 CO 2 (g) + 6 H2O (/) MM: 78.1

32 .0

44.0

18.0

When 2.6 g of C6H 6 were reacted with 0.88 mol of 0 2, 1.0 g H 2O of were isolated. Calculate the percent yield of H 2O? First find the theoretical yield of water (in g) in this Limiting Reagent Problem.

Solution:

1 mol C6 H 6 ) (

? g H2O = 2.6 g C6H6 ( - - - ;t 78.1 g C6 H 6

6mol H20) (18 .0 g H20) - - - - = 1.8 g

- - --

2 mol C6 H 6

1 mol H 2 0

LRimiting Predicts the formation of the smaller eagent - - - - - - - - - - - - - -- - - ' (and correct) amount of product

\

Theoretical Yield ofH20

6 mol H 2 O ) (18.0 g H 2 0) 15 mol 0 2 1 mol H 2 0

? g H2O = 0.88 mol 0 2 ( - - - /

=6.4 g

Excess _ __Predicts the formation Reaoent _ __ _ _ _ ____,/ of too much product

0

Actual Yield (amount of product , H2O, that was

isolated) ⇒

1.0 g

%Yiehl = _ A_c_tu_al_Y_ie_ld_ • l00 ::: l.OgH2 0 • 100=S 6 _0 % Theoretical Yield 1.8 g H 20 Problem 13: Consider the following balanced chemical equation: CsHs MM:

a)

68.l

+ 2 H2 2 .02

➔

CsH12 72.l

What is the% yield of CsH 12 if 3.00 g of it were obtained in the reaction of 5.00 g of CsHs with excess hydrogen?

A. J. Pappas and M.E. Goicoechea-Pappas

10-13

Introductory Chemistry

10 - Balancing Equations and Reaction Sroichiometo:

b) If 4.80 rnL of C5H 12 (d = 0.626 &/mL) were obtained in the reaction of 5.00 g of C 5H8 with 0.445 g ofH2, then what is the % yield of C5H 12? Problem 14: Consider the fo llowing balanced chemical eq uation:

CH4 + 4 Ch MM: 16.0

a)

➔

CC4 + 4 HCl 154.

71.0

36.5

If 10 .0 g of HCl were obtained in the reaction of 5 .00 g of C~ with excess Cb, then what is the percent yield of HCl?

b) What is the % yield of CCl4 if 50 .2 g of it were isolated in

the reaction of l 0.0 g of CH4 with 3.0 mol of Ch? A Medley of Problems: 15. Balance the following chemical equations: a)

Na (s)

+

b)

Ga (s)

+

c)

Al (s)

+

F2 (g) ➔

AlF3 (s)

d)

Mg (s)

+

N2 (g) ➔

Mg3N2 (s)

e)

Li20 (s)

+

H 20 (I) ➔

f)

CO2 (g)

+

g)

C5H1 2(f)

h)

Cl2O3 (g)

+

H20 ( I) ➔

i)

Fe20 3 (s)

+

CO2 (g)

j)

P 4O 10 (s)

Cl2 (g)

➔

NaCl (s)

0 2 (g) ➔

Ga203 (s)

LiOH (aq)

H20 ( I') ➔ +

0 2 (g) ➔

➔

H2O (I) ➔

+

H2C03 (aq) C0 2 (g) +

H20 (t'J

HCIO2 (aq)

Fe2(C0 3)3 (s) H3PO4 (s)

k) C5H10Os (s) 16. How many moles of oxygen are needed to complete ly burn 100 . g of CgHL8 (MM= 114.) CgH 18

+

02 ➔

A. J. Pappas and M.E. Goicoechea-Pappas

CO2 +

H 20

(unbalanced) 10 - 14

10 - Balancing Equations and Reaction Stoichiometry

Introductory Chemjstry

17. What mass (in g) of CO 2 (MM = 44 .0) should be produced from the complete combustion of 100. g of C 3Hs (MM= 44.1)? C3H8

+ 502

➔

3 CO 2 +

4 H2O

18. What mass (in grams) of hydrogen gas can be produced by reacting 6.0 moles of aluminum with excess hydrochloric acid? 2 Al + 6 HCI

➔

2 AICl3 + 3 H2

19. Calculate the number of moles of 0 2 that should be produced as a result of decomposing 1.226 g of KCIO 3 (MM = 123). KCIO3

➔

KCI +

02

(unbalanced)

20. Using the following balanced equation , how many moles of calcium chloride would be necessary to prepare 94.0 g of calcium phosphate (MM= 310.)? 3 C aCl2 (aq)

+ 2 Na3P04 (aq)

➔

Ca3(P04)i (s) + 6 NaCl (aq)

2 1. Consider the following balanced equation: K2Cr2O7 + 6 KI + 7 H2SO4 ➔ Cr2(SO4)3 + 3 12 + 7 H 2O + 4 K2SO4 Calculate: a) the number of moles of potassiu m dichromate that will react with 2.0 moles of potassium iodide, assuming H2SO4 is present in excess. b) the number of moles of potassium sulfate which can be produced from 100 grams of KI (MM= 166 .) reacting with excess sulfuric acid and potassium diclu·omate. 22. What m ass (in g) of nitric acid (MM = 63.0) will be needed to convert 0.400 mole of copper into copper(II) nitrate?

3 Cu (s) + 8 HNO3 (aq) ➔ 3 Cu(NO3)z (aq) + 2 NO (g) + 4 H2O (f) 23. Calculate the number of grams of hydrogen (MM = 2.02) that can be produced if 8.40 grams of aluminum (MM = 27.0) is reacted with a stoichiometric (i.e., appropriate) amount of NaOH :

2 Al (s) + 6 NaOH (aq) ➔ 2 Na3AIO3 (aq) + 3 H2 (g) A . J. Pappas and M.E. Goicoechea-Pappas

10-15

24. A convenient laboratory method for ge nerating oxygen invol ve thermally decompo ing pota sium chlorate. MM

2 KCIO 3 ➔ 123

2 KCl 75 .6

+

302 32.0

If 5.00 g of KCIO3 are thermally decomposed, then: a) b) c) d) e) f)

What i the expected maximum yield of oxygen (in g)? How many 0 2 molecules should be produced? How many KCl formula units hould be produced? What i the theoretical yield of KCl (in mol)? If 1.88 g of 0 2 were procured, what is it % yield? What i the % yield of KCl if 4.00 x 10-2 mol of it were isolated?

25. If 50.0 g CaCO3 are treated with 35.0 g of H3PO4, then: 3 CaCO3 + 2 H 3PO4 ➔ Ca3(PO4)i + 3 CO2 + 3 H2O 18.0 MM: 100. 44.0 98.0 310. a) b) c) d)

What is the theoretical yield (in g) of Ca3(PO4)2? Which is the excess reagent? What is the theoretical yield (in mol) of CO2? How much of the excess reagent (in g) was only needed to conduct thi reaction? e) How many gram of the exces reagent will remain unreacted? f) If 49 .9 g of Ca3(PO4)i were i olated, then what is it % yield? g) What is the % yield of CO2 if 0.485 mol of it were obtained? 26. If 10.0 mL ofC6116 (d = 0.874 glmL) is combusted in the presence of 30.0 g of 02 under typical laboratory condawns (i.e., T = 25°C and P = 1 atm), then: MM:

2 C6H6 (f) + 15 0 2 (g) ➔ 78.1 32.0

12 CO2 (g) + 6 HzO (.f') 44.0 18 .0

a) Calculate the theoretical yield of water (in mol). b) Which is the excess reagent? c) How much of the excess reagent (in g) was only required? d) How much of the excess reagent (in g) remain uncon urned? e) If 0.250 mol of water were i olated, then what i its perc nt yield?

f) How many rnL of CO2 (d = 1.98 x 10-3 glee) hould be produced? g) If 1.21 x 104 rnL of CO2 were co11ected, what is it % yield? A. J. Pappas and M.E. Goicoechea-Pappa

10-16

Introductory Chemistry

1O- Balancing Equations and Reaction Stoichiometry

ANSWERS l.

a) 2 C2H6 + 7 02 ➔

4 CO2 + 6 H2O

b) Na2B4O7 + 2 HCl + 5 H2O ➔ 4 H3BO3 + 2

c) CaCO3 + 2 HCJ ➔

aCl

CaC!i + CO2 + H2O

2.

d) C3Hg + 5 02 ➔ a) 4.50 mol i

3.

a) 778. g NiC'2

b) 90.0 g Al

4

a) 0.0386 mol Ni

b) 0.0257 mol AlCl 3

5.

a) 9.78 g

6.

a) 2.5 mol (CH3CH2hO, 2.5 mol H 2O

b) 0 .434 mol CH3CH2OH

c) 208 mL (CH3CH2hO

d) 244 mL CH3CH2OH

3 CO2 + 4 H2O b) 3.00 mol AlCb

i

b) 14.8 g AlCh

7.

0.831 g oxygen

8.

2.62 x 10-3 mL acetic anhydride

9.

145. g aluminum sulfate

10.

50.2 g calcium pho phate

11.

a) 0.125 g H2

b) 2.25 g H2SO 4 = 0.0229 mol H2SO4

12.

a) 0 .525 mol Cu(NO 3)z

b) 0 .075 mol CuS = 7 .17 g CuS (MM = 95 .6)

13.

a) 56.7%

b) 56.7% (C 5 H 8 i the limiting reagent and H2 is in excess)

14.

a) 21.9%

b) 52.2% (C~ is the limiting reagent and C}z is in excess)

15.

a) 2

a (s) + C}z (g)

b) 4 Ga (s)

2 NaCl (s)

3 02 (g) ➔ 2 Ga2O3 (s)

+

c) 2 Al (s) +

➔

3 F2 (g)

➔

d) 3 Mg (s) + N2 (g) ➔

2 AIF3 (s) Mg3N2 (s)

e) LiiO (s) + H2O (I) ➔

2 LiOH (aq)

f) CO2 (g) + H2O (I) ➔ H2CO3 (aq) g) C5H12 (f)

8 02 (g) ➔ 5 CO2 (g) + 6 H2O (t)

+

h) Cl2O3 (g) + H2O ( I) ➔ 2 HClO2 (aq) i) Fe2O3 (s) + 3 CO2 (g)

➔

j) P4O10 (s) + 6 H2O (I) ➔ k) C5H10O5 (s)

16.

11.0 mol 02

17.

299. g CO2

18.

18 .0 g H2

19.

0.0150 mol 02

+

Fe2(CO3h (s) 4 H3PO4 (s)

5 02 (g) ➔

A. J. Pappas and M.E. Goicoechea-Pappas

5 CO2 (g) + 5 H2O (I)

10- 17

arions and Reac.t· 20.

0 .910 mol CaCl2

21.

a) 0.33 mol K2Cr2O1

22.

67 .2 g HNO3

23.

0.943 g H2

24.

a) 1.95 g 0 2

25.

d) 0.0407 mol KCl a) 51.7 g Ca3(PO4 )z e) 2.3 g H3PO4

b) H3PO4

a) 0.336 mol H2O

b) 02

26.

e) 74.4%

b) 0.40 mo! K2SO4

b) 3 .67 x 10 e) 96.4%

22

f) 96.5%

c) 2.45 x 1022 KC! FU 0 2 molecule f) 98.3% d) 32.7 g H3PO4 c) 0.500 mo! CO2 g) 97.0%

c) 26.9 g 02 4

t) 1.49 x 10 mL CO2

A. J. Pappa and M.E. Goicoechea-Pappa

d) 3.1 g 02

g) 81 .2%

10-18

CHEMICAL REACTIONS I.

Reactions There are several reaction types that you will be expected to recognize and in ome ca es be able to predict the final products. The e reaction types are: a) combination/ synthesis (recognize) b) decomposition (recognize) c) combustion (recognize and predict products) d) ionization (recognize and predict products) e) single replacement (recognize and predict products) f) double replacement/ metathe is (recognize and predict products) Examples of each will be shown below. Note that in some of the reactions each reactant and product has its physical state indicated by the use of the following symbols: (g)

II.

⇒

gas

(t')

⇒

pure liquid

(s)

⇒

solid

(aq)

⇒

aqueous

Combination or Synthesis Reactions This type of reaction is characterized by the formation of one compound from simpler materials. A general example is: A+B ➔ AB

where A and B may be elements or compound

Specific examples of combination reactions are shown below: 2 K (s) + F2 (g)

➔

6 Ca (s) + 2 N2 (g)

2 KF (s) ➔

2 Ca3N2 (s)

P406 (s) + 6 H20 (t') ➔ Na20 (s)

+ CO2 (g)

➔

4 H3P03 (aq)

Na2C03 (s)

III. Decomposition Reactions This type of reaction is characterized by a compound being broken down into simpler compounds or all the way down to its component elements. A general example is: AB

➔

A + B

where A and B may be elements or compounds

Specific examples of decomposition reaction are shown below: Cs2C03 ( ) ➔ Cs20 (s)

+ CO2 (g)

Ba(Cl03)2 (s) ➔ BaC}i (s)

2 K20(s)

➔

2 KOH( ) ➔

+ 3 02 (g)

4 K (s) + 02 (g) K20 (s)

A. J. Pappas and M.E. Goicoechea-Pappas

+ H20 (g) I 1- 1

IV. Combustion Reactions Compounds containing C and H in their formula, when burned in the presence of 02 , form CO2 (g) and H20 (t'J· A general example is: CxHy + 0 2 (g) ➔ CO2 (g) + H20 (f)

(not balanced)

Combustion reactions are exothermic (i.e., they release heat) . Example A: Complete and balance the following equations a) CH4 (g)

+ 0 2 (g)

➔

b) C4H10 (g)

+ 0 2 (g)

➔

Solution: The products of a combustion reaction are CO2 (g) + H20 (f )• a) CH4 (g)

+ 2 0 2 (g) ➔ CO2 (g) + 2 H20

b) 2 C4H10 (g)

(t)

+ 13 0 2 (g) ➔ 8 CO2 (g) + 10 H20

(t')

Problem 1: Complete and balance the following combustion reactions . a) C3Hg (g) + 02 (g) ➔ b) C3H6 (g) + 0 2 (g) ➔ V.

Ionization Reactions Ionization refers to the process in which ionic compounds, when dissolved in water, separate to form solvated ions in solution. Solvated ions are ions that are ® surrounded by solvent aB O:> EP,_ ~ Na• 1 molecules. If the solvent is + H0 er/£> l"I:) + 0 t> oclJ c1water then these ions are NaCl Na+(aq) 1L ci-(aq) a--o said to be hydrated. solid crystal 1 H10 1 lattice NaCl(aq) Substances that are watersoluble can be classified as either electrolytes (weak or strong) or nonelectrol ytes. Electrolytes are substances that conduct an bright dim electrical current when they are dissolved in water.

t,

I I

e;

101

Electrical current is carried through aqueous solutions by

Strong

WNk

the movement of ions. The Elect.rolyu Electr0fyte EB= cation strength of the electro!yte e =~nion depends upon the number of ions in solution, as well as the charges of these ions. A. J. Pappas and M.E. Goicoechea-Pappas

11-2

- Chemical 8-eac.ti.oo.s. tr NOTE: All ionic compounds have the potential to conduct electricity ivhen in a molten state. In a molten state, there is free movement of ions, which makes it possible for current to flow .

Strong electrolytes are substances that completely ionize in water and conduct electricity strongly. A general example of the di ssociation / ionization process for strong electrolytes is ,

Ax By

H20

Jlo

x Ay+

+

y a x-

Aqueous solutions of the followi ng substances are considered strong electrolytes. a) Strong Acids (and Strong Electrolytes) HCl , HBr, HI, HCI03, HCl04, HN0 3, H2S0 4 b) Strong Bases (and Strong Electrolytes) IA hydroxides,

IIA hydroxides except Be(OHh and Mg(OHh

c) Soluble Salts - Ionic compou nds containing a cation other tha n H+ and an anion other tha n 0 2- or OH- . To predict whether salts [in addition to oxides (rule 7), hydroxides (rule 8) and acids (rule 9)) are soluble in water, use the solubility rules given below. WATER SOLUBILITY RULES l . All salts of alkali metals (IA) are soluble.

2. All N H4+ salts are soluble . 3. All salts containing the anions: N03-, Cl03-, Cl04- , C2H30 2- are soluble. 4. All c1-, Br-, and rare soluble except for Ag+, Pb2 +, and Hg22+ salts. 5. All SO42- are soluble except for Pb2+, Sr2+, and Ba2+ salts.

6. All o2- are insoluble except for IA metals, Ca2 + , Sr2+ , and Ba2+. [Soluble metal oxides, form hydroxides (e.g., Bao+ H2O ➔ Ba2+ + 2 OH-)]. 7. All Otr are insoluble except for those bases* containing IA metals, NH4+. Ba2+ , Sr2 +and Ca2 +.

8. Unless other wise informed , assume that salts contai ning anions not 3 menboned above [e.g., P04 -, As043-, s2- , S032- , Cr042-, etc.] are insoluble except for IA metals and NH4+.

col-,

9. All inorganic acids* are soluble. Only those organic acid s* having less than 6 carbons are soluble. ,.

Even though acids (e.g., HF, CHJCOOH = HC1 Hp z> and bases (e.g., NH,,OH) may be soluble in water, only the ones that are strong completely ionize in water.

A. J . Pappas and M.E. Goicoechea-Pappas

I l-3

11 - Chemic~ctio.ns

Introductory Chemistry

If an ionic compound is a strong electrolyte, that means that it has dissolved (dissociated/ionized) in water (i.e., LiCl (aq) ➔ Lt (aq) + Cl -(aq))- [NOTE:

For dissociated/ionized substances, the physical state (aq) of the ions is ofte n omitted, i.e., Li+ + Cl- .]

Weak electrolytes are substances that partially ionize in water (i .e ., substances that conduct electricity poorly in an aqueous solution). A general example of the dissociation / ionization process for a weak electrolyte is,

Two arrows pointing in opposite directions mean that the reaction is reversible (i.e., the reaction can occur in both directions). With weak electrolytes, the un-ionized species, AxBy is still presen t in sol uti on . The following are considered weak electrolytes: a) weak acids - any acid that is not one of the strong acids. Weak organic acids have the following general formula - CxHyCOOH. Acetic acid is an organic acid whose formula can be written as: HC2H302 or CH3COOH. b) weak bases - any base that is not one of the strong bases c) insoluble salts - any salt that is not water soluble according to the solubility rules d) water (water only undergoes ionization to a very small extent). In some textbooks water is considered a non-electrolyte.

Non-electrolytes do not ionize in water at all (i.e., they do not conduct electricity because no ions are produced). Such being then we1d write NR. (no reaction has occwred). N .R . (no reaction -- no d issociat ion/ ioni7.ation)

The follo wing are considered non-electrolytes: a) organic compounds such as alcohols (general formula: CxHyOH) and sugars (general formula: CxH2xOx)- These substances dissolve in water; however , they do not ionize / dissociate in water. Example B: Complete and balance those reactions in which the substance completely ionizes in water (i.e., are strong electrolytes) or partially ionizes in water (i.e., weak electrolyte). For those that are weak electrolytes use two arrows pointing in opposite directions. a) V(N03)3 c) Al(C2H302)3 A. J . Pappas and M .E. Goicoechea-Pappas

d) HCl 11-4

Introdyctocy Chemistry e) AgBr H,o

I 1 - Chemical Reactions

f) CuC03

Solution: a) V(N03)3

(Solubility Rule 3)

sol· (aq)

b) (NH4)2S04

2 NH4+(aq) +

c) Al(C2H302)3

A13 \aq) + 3 C2H302-(aq)

d) HCl

e) A gB r

(Solubility Rule 2 or 5) (Solubility Rule 3) (Strong Acid)

H ,O

~~➔

f) CuC03

H,O

Ag

+

(aq)

+ Br-

(Solubi lity Rule 4)

(aq)

➔ Cu2+ (aq) + C03 2-

(aq)

(Solu bility Rule 8)

Problem 2: Complete and balance those reactions in which the substance completely ionizes in water (i .e., are strong electrolytes) or partially ionizes in water (i.e., weak electrolyte). For those that are weak electrolytes use two arrows pointing in opposite directions.

c) CsCl

d) FeF3

e) K2C20 4

f) NaOH

g) Mg(C2H30 2)2

h) Hg2Br2

i) CoS

j) C6H1206

k) C2HsCOOH P roblem 3: Categorize the substances in Problem 2 into one of the following categories: a) strong electrolyte, b) weak electrolyte, or c) non-electrolyte. VI.

Single Replacement Reactions Reactions in which one element displaces another from a compound are called replacement (or displacement) reactions. Active metals displace less active metals or hydrogen from their compounds in solution. Active metals are those with low ionization energies that readily lose electrons to form cations. Also, acti ve halogens can displace less active halogens from their compounds in solution.

A. J . Pappas and M .E. Goicoechea-Pappas

11 -5

The Electr motive or Activity Serie indicate which el ment are more active and which are le active.

Electromotive (Activity) Series* for metals Li>K>Ba>Sr>Ca> a>M

Al>Zn>Cr>Fe >Cd>Co> i> n>Pb>(H)>Sb>Bi>Cu>Hg>Ag>Pd>Pr>Au

for halogens F> Cl> Br> I

* with metal having variable oxidation number , one of its lower oxidation tates is often formed:

cr3+, Fe 2+, Co 2+, * Reaction of Metal

i2+. n 2+. Pb 2+, b3+. Cu+, Hgi2+, Pd 2+, Pt 2+

with (H):

Bold: React with H20 (i), with team (i.e., H20 (g)). and acid ; Bold- nderline: React only with . team and acid [ OT with HzO (t)l; Double Under.Line: React only with acid

[ OT with H20 (f) or team].

There are basically three type of ingle replacement reactions: A. A free and chemically active metal displacing a less active metal from a compound In general, this type of single replacement reaction can be illustrated a follows: A + BC (aq) ➔ AC + B where A (a metal) di places B from an aqueou

elution containing BC

Exam le C: Complete and balance the following reaction a) Mg ( ) + AuBr3 (aq)

➔

b) Ag () + Hg(N03)2 (aq) ➔ c) Pb ( ) + Au(Cl04)3 (aq)

➔

Solution: a) Mg i more active than Au; therefore, it di plac s Au. The compound that form i MgBQ (Mg 2+ - group IIA & Br- - group VIIA); it i water soluble according to olubility rule 4.

3 Mg ( ) + 2 AuBr3 (aq) ➔ 3 MgBri (aq) + 2 Au ( ) b) Ag i I s active than Hg· therefore, it doe not di place Hg. Ag (s) + Hg(N03)2 (aq) ➔ A. J. Pappa and M.E. Goicoechea-Pappas

.R. (No R action) 11 -6

I c) Pb i more active than Au; therefore, it di places Au. Even though Pb has a variable oxidation number; as tated above, the lower +2 oxidation state is often formed . The compound that form i Pb(Cl04)z (Pb 2+ and Cl04-); it is water oluble according to solubility rule 3. 3 Pb

(s) +

2 Au(Cl04)3

(aq) ➔ 3

Pb(Cl04)2

(aq)

+ 2 Au ( )

B . A free and chemically active metal displacing hydrogen from acids or water In general this type of ingle replacement reaction can be illu trated as follows: ➔

M + HX (aq) M + H20(t org)

MX + H2 (g) ➔

MOH + H2(g)

where M (a metal) di places H from an acid (HX) or HzO (liquid water or steam) Recall that hydrogen (Hz) is a diatomic element wh ich is a ga under typical lab condition .

Exam le D: Complete and balance the following reactions. a)

K (s) + H20 (f)

b)

i (s) + HN03

➔

(aq) ➔

Solution: a) K is a very reactive metal ; therefore, it displaces H from water, steam or acids. The compound that forms is KOH (K+-group IA, & OH-); it is water soluble according to solubility rules 1 and 7. 2 K (s) + 2 H20 (f) ➔ 2 KOH b)

(aq)

+ H2 (g)

i is more active than H; therefore, it displaces H. The compound that forms is Ni(N03)2 (Ni 2+ - +2 oxidation tate is often formed and N03-); it is water soluble according to solubility rul 3. i ( ) + 2 HN03

(aq) ➔

Ni(N03)2

(aq)

+ H2 (g)

C. An active halogen (VI/A non.metal) in the uncombined state displa,cing a less active halogen This type of single replacement reaction can be illu trated as follows:

X2 + MY (aq)

➔

MX + Y2

where X2 (a halogen) displaces Y (a less active halogen) the metal does not change oxidation tates A. J. Pappas and M.E . Goicoechea-Pappas

11-7

mical Reac_.ti.ons Since all halogens form an ion that has a -1 charge, the formula of the salt that forms basically remains the same except for the replacement of Y (in MY) with X. Recall that halogens (either X2 or Y2 in the above reaction) are

diatomic and have the following physical state: F2 (g), Cl2 (g), Br2(t), h (sJ· Example E: Complete and balance the following reactions. a)

h () +

FeF3 (aq) ➔

b) Cb (g) + FeBr3 (aq) ➔

Solution: a) I is le

active than F; therefore, I doe not di place F.

Ii ( ) +

FeF3 (aq) ➔ N .R. (No Reaction)

b) Cl i more active than Br; therefore, Cl di places Br. The compound that forms is FeBr3 (formed by the combination of Fe 3+ and Br- ); it water soluble according to solubility rule 4.

3 Ch (g) + 2 FeBr3

(aq) ➔

2 FeCb (aq) + 3 Bri (tJ

Problem 4: If a reaction occurs, complete and balance the following equation , assuming a single replacement reaction. If no reaction occurs, write N .R. a) F2 (g) + AICl3 (aq) ➔

b) K ( ) + H20 (t') ➔ c) Ba (s) + H20 (t') ➔ d) Zn (s) + CuS04 (aq) ➔ e) Al (s) + CuC}i (aq) ➔

f) Cu (s) + H20 (t ) ➔ g) Cd (s) + HCl (aq) ➔

h) Al ( ) + Ba(N03)2 (aq) i) BQ (t')

+ NaCl (aq)

➔

➔

j) Mg (s) + A1I3 (aq) ➔ k) Fe (s) + H2S04 (aq) ➔ A. J. Pappas and M.E. Goicoechea-Pappa

11-8

lntm

N a•

Cl . Na• Cl '

NaCl

d...,,., . dwater= 1 g/mL) 2.

5. 6.

12.8 L 302.8 amu

7.

At. No.= 51 I) 4d

0

#e- "'46

# n = 72

I) O g) Hg h) o) 1 amu, positive, nucleus b) molecule (the element oxygen= 02)

9. a) atom 10. a) 23.2 g XF

1

e) Br

:c

Approx mass"' 123 am u i) ScJ• j) sez- k) 4s, 4p, 4d, 4f; 2• 4 1 = 32 p) electron d) molecule c) FU

b) 15.6 g X

11. [Ar) 4s = [Ar] 3dlO 12. a) 2 b) paramagnetic 13. a)

#p• = 51

b) I c) Sb d) Si m) Pb n) Ga or Ge

a) Al

8.

Mass No.= 123

3d 10

[NOTE: Exceptions occur in Groups VIB and 18]

~

f!

.!.: .!: .!.:

H

ls

2s

2p

3s

.!,:

.!: _!:

n

tJ. H H

3p

4s

3d

b) PbSb>As>Br

n

_!: _!: :!:_

u

b) Yes

4p

b) Pb< Bi a (As - P) > c (F - F) (Note: most polar ⇒ least covalent character ⇒ sharing electrons the least =;, biggest difference in electronegatlvity between the elements ⇒ most Ionic character) A. Br03·

16.

['..0-8,i --ol..

a)

..

D.

C. BaF2

8. OClz

..

..

IBa2+1[ :~= -]2

:c1-o-c1 .. :

:o;

POI

=o=P-·1·: ..

ionic compound

b)

tetrahedral

tetrahedral

trigonal planar= planar triangle

c)

trigonal pyramid

V-shape = bent

V-shape = bent

d)

< 109.S-

< 109.5°

17. a) N, P, As (VA non-metals) 18. Ni= +2 0 = -2

P = +5

19. a) mercury(II)

b) chlorite