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How to Solve
Word Problems in Chemistry

Other books in the “How to Solve Word Problems” series:
How to Solve Word Problems in Mathematics
How to Solve Word Problems in Arithmetic
How to Solve Word Problems in Calculus
How to Solve Word Problems in Geometry
How to Solve Word Problems in Algebra, Second Edition

How to Solve
Word Problems in Chemistry
David E
...
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DOI: 10
...


For more information about this book, click here
...
C lic k H e re fo r T e rm s o f U s e
...
Look for the material in your text to make sure that
you are responsible for each subject
...

Cover the solutions to the Examples and try to solve them
yourself
...
Do
not merely read the solutions; you must do the problems to really
understand the principles
...
A
given problem can be asked in many different ways, and you must
understand what you are doing in order to succeed
...
These terms are defined in the Glossary
...
For example, a problem may be
presented in parts, then the same problem (perhaps with different
numbers) is presented as a single problem such as might be asked
on an examination
...


vi

Chapter 1

Introduction

1
...
The results are presented with a number and a
unit or combination of units
...

For example, it is very important to the mail carrier to know whether
a new customer has a dog that is 5 inches tall or 5 feet tall! Always
use units
...
2, the units actually
help us figure out how to solve many problems
...
We try to have a different symbol for each
one of these, but there are more things to represent than different
letters
...
For example, the symbol Co represents cobalt, but CO represents carbon monoxide
...
As another example, 1 mg (milligram) is 1-billionth the
mass of 1 Mg (megagram), as introduced in Section 2
...
Do not get
confused; we must take the tiny amount of extra time to do things
correctly from the beginning of our study of chemistry
...
When we learn
a new subject, it might seem hard at first, but remember that it is
presented to enable us to do more things or to do the things we
already know more easily
...
Keep up with the
work if at all possible
...
Missing the background material makes it more difficult to understand the present
material, especially to learn without an instructor
...
Use this book and other study aids to learn
missing material without a teacher
...
There is no use knowing something and applying it to
the wrong thing
...
It is
a mistake to think that hydrogen must be written H2 in all of its
compounds
...

How to Approach a Word Problem
Working word problems requires understanding the principles involved and being able to apply them to the case at hand
...

To do a word problem, follow these steps:
1
...

2
...
Some problems
have values to be determined elsewhere, as from tables of data or
the periodic table (which is always supplied when needed)
...

3
...
For example, if a binary compound of A and B is 25% by mass element A, there is (25 g A)/(100
g total) by definition
...

4
...

5
...
) that we know
which might connect the values given and desired
...

6
...
(If one equation
won’t work, try a different equation
...
Check the answer to see that it is reasonable
...
3
...
For others, we can use the answer to calculate one of the original values, as in empirical formula problems (Section 4
...
Still others require that we know the range
of possibilities for our answer
...
1) we know there is a mistake, because
10,000 moles of anything cannot fit into a liter
...
For most problems, just consider
if the answer is about the right size
...
Perhaps the partial answer will lead to further
steps that will end in a complete solution
...
The bus stopped
at the parking lot, and the troop marched up the “mountain” past
the rock that looked like a lion, down the other side, waded across
the shallow stream, and walked up the next hill past the brokenoff tree
...
They spent the morning playing, had lunch, took a swim
in the pond, and undertook numerous other activities
...
What to do? He did not panic, especially where the boys could
see him
...
He marched his troop up the hill, from where he saw the small
stream and the “lion” rock
...
No one knew that
he had not known all along how to get back
...
The answer to the
first part might suggest what to do next
...
If we know what we need, that
might give us a clue as to what to calculate next
...
)
Here is a problem from the world outside chemistry: “A hunter
aims his rifle due south directly at a bear
...
The hunter fires his rifle due south and kills the bear
...
Let’s do what we can do
...
The hunter
may be standing directly on the north pole, so every horizontal direction is due south
...

(The hunter may also be standing very near the south pole, so that
the bear’s path took it in a complete circle, and the hunter fired without moving his rifle
...
)
We must try to understand the material as we progress
...
There are enough details in chemistry that we
3

must memorize
...

Sometimes it helps to assume a value to work with, especially
with intensive properties such as concentrations
...
4)
...

[For example, to get the value for the ideal gas law constant (Section
7
...
00 mol sample of gas at STP with a volume
of 22
...
] We can then use that
constant in the problem we are trying to solve
...
In
science we could also use such variables, but we find it much easier
to use letters that remind us what the letter stands for
...

We then can write an equation for density, d, in terms of mass and
volume as d = m/V
...
We solve these
equations in the same way that we solve algebraic equations (and
we don’t often use more than simple algebra)
...
We
attempt to expand our list of symbols in the following ways:
Method
1
...

2
...

3
...

4
...

5
...

4

Example
T for absolute temperature
and t for Celsius
temperature
m for mass and m for meter

V1 for one volume and
V2 for a second
MM for molar mass
µ (Greek mu) for micro-

Each such symbol may be treated like an ordinary algebraic
variable
...
2 Dimensional Analysis
An extremely useful tool for scientific calculations (for everyday calculations too) is dimensional analysis, also called the factor label
method
...
For example, if
we have $2
...
We
can change from one of these to the other with a factor—a ratio—of
100 cents divided by 1
...

EXAMPLE 1

Convert 2
...

Solution




100 cents
✥
✥
✥
= 225 cents
2
...

We multiply all the numbers in the numerator and divide by each of
the numbers in the denominator
...
25 dollars, and the ratio had dollars in the denominator
...
(In fact, we use the same abbreviations for singular and plural, and often do not know whether
our answer will be greater than one or not
...
This method tells us to multiply dollars by
100 to convert to cents
...



We can use the reciprocal of that factor to convert cents to
dollars
...


Solution Again we put down the quantity given, and this
time multiply it by a ratio with cents in the denominator:



1 dollar
1535 cents
✥✥✥
= 15
...
In each case, we used the one we needed
to convert from the unit that we had to the one that we wanted
...

EXAMPLE 3

Change 1
...


Solution We know that there are exactly 60 minutes in an
hour, and exactly 60 seconds in each minute:



60 minutes
1
...
60 minutes
1 hour



60 seconds
99
...
660 hours
✥✥✥
✥
1 hour
✥✥
1 minute
✭✭✭✭
(If we know that there are 3600 seconds in an hour, we do not
need two factors, but there will be many problems in chemistry later
in this book in which more than one factor is needed, so it is well
that we learned how to handle more than one factor here
...
We can’t make the mistake of not learning the
method here because we don’t need it yet
...
For example if an elementary school class is 40% girls and
60% boys, we can tell how many children are in a class with 48 boys:



100 children
48 boys
= 80 children
60 boys
In chemistry, if a compound of elements A and B is 25% by mass
element A, there is (25 g A)/(100 g total) by definition
...

6

Calculate the number of people in an audience if 45
people vote
...


EXAMPLE 4

Solution

45 voters




100 people total
75 voters


= 60 people




It is also possible to use the factor label method to convert from
one ratio to an equivalent ratio, using one factor at a time
...
0 miles per hour
into its speed in feet per second
...
Thus 45
...
0 miles divided by 1 hour
...
0 miles
✥✥✥ 5280 feet
1✥
✭✭✭✭
✭
✭
✥
✥
1 hour
✥✥
1 mile
✥✥
60 minutes
✭✭✭
60 seconds

=

66
...
0 feet/second
1 second

Here we needed three factors to convert our ratio to an equivalent ratio with different units
...
We don’t use English system measurements much at all in science, although they are used some in
engineering
...


1
...
However,
it is critical that we know how to use the calculator without thinking about it too much while we are thinking about the chemistry
problems! Read the instruction booklet about how the calculator
works
...
Chemistry requires
principally the arithmetic operations keys ( + , − , × , ÷ ), EE
√
√
or EXP , FLO , SCI , the reciprocal key 1/x , x2 , x , x3 , 3 x ,
LOG , the natural logarithm key LN , the antilogarithm key 10x ,
7

x
the natural antilogarithm key e x , and perhaps y
...
We must practice with each operation using
simple numbers until we are sure that we know how the calculator
works
...
If it displays 4, read
the subsection on precedence rules below
...
There is a special key, called variously 2nd , 2nd F , SHIFT ,
or ALT depending on the model calculator
...
Somewhat like the SHIFT key on a typewriter, pressing this key first makes
the next key pressed perform a different operation than it normally
would
...
)
The following keys are among those that operate immediately
√
x , x3 ,
on whatever value is displayed: FLO , SCI , 1/x , x2 ,
√
x
3 x
x
, LOG , LN , 10 , and e
...
For example, if 2 is in the display and
we push the x2 key twice, we get 16 as an answer
...

Precedence Rules
In calculations that involve more than one operation to be performed, we must know which one to do first
...
For example, in algebra, in the absence of
any other indication, we always multiply or divide before we add
or subtract
...
Thus
2+3
1+9
has a value of 0
...
(Note the difference from 2 + 3/1 + 9, which follows the
normal precedence rules
...
The orders of precedence are presented in Table 1-1
...
(Unary minus is a minus sign that denotes a negative number
rather than a subtraction
...
(The multiplication is done first
...
If the answer
(b) The answer displayed is 2
...

Because the calculator has a different precedence rule for division
and multiplication than we follow in the algebraic expression
ab/cd, where both multiplications are done first, we must divide
the 18 by 2 and then divide that answer by 4
...

6 × 3 ÷ 2 ÷ 4 =
or

6 × 3 ÷ ( 2 × 4 ) =

(c) 512
...

(d) −9
...

9

(e) +9
...

The EE or EXP Key
To enter an exponential number (see Section 2
...
The EE or EXP key
represents “times 10 to the power
...
(Some calculators have the
exponent raised and in smaller numbers
...
66 × 105 into the calculator?
(b) What does the display show?

EXAMPLE 7

Solution

(a) Enter 1


...
If we mistakenly enter
1
...
66 × 10 × 105
...
66 05



The Change Sign Key
The change sign key +/− is used to convert a positive number to a
negative number or vice versa
...
To change the sign of the exponent, press the
change sign key after the EE or EXP key
...
It is not
absolutely essential, but it can save storing a value in memory
...
[The reciprocal of (b + c) is 1/(b + c), which then is multiplied
by a
...
Each key operates immediately
on the value on display
...
For example, the
logarithm of 2 is 0
...
30103 is equal to 2
...
69317
...
)
The antilogarithms reverse the process; they give the value of 10 or
e raised to that power
...
Natural logarithms are as easy to use on the calculator
as common logarithms, and are often more intimately connected
to a chemistry problem
...
There were 20
...
5% girls in a certain class
...
One day, 6 boys and 2 adults were
absent, and only 2 boys attended
...
Calculate the number of hours in 7992 seconds
...
If we spent $1000 per day, how many years would it take us to spend
(a) 1
...
00 billion dollars?
4
...
Use the calculator to compute the value of each of the following
expressions:
6×7
(c) 3x 2 where x = 5
(a) 5 × 7 − 7 × 6
(b)
4 × 14
6
...
11
(d ) 5
...
20 − 7
...
00
(e)
4
...
1
( f ) 3
...
15
6
...
00 × 1014 ÷ 4
...
00 × 1014 − 4
...
00 × 10 − 4
...
02 × 1023 )
11

7
...
0000
(b) x = log(2
...
0000
(d ) x = ln(2
...
699
8
...
65
(b) (7
...
09
9
...
59 × 10−4
(a) 2
...
59 × 10
(d ) 2
...
In a complicated problem, be sure to label the work to know exactly
what each term means
...
5% of the class were adults and 8 class members
were boys
...
5 adults enrolled
8 boys
= 21 adults enrolled
20
...

Alternatively, we could have determined:



52
...
0 boys enrolled (per hundred)
= 21 adults enrolled

Probably, few of us knew how to do this entire problem before
starting any calculations at all
...
220 hours
2
...
74 years
3
...
74 × 103 years
4
...
(We must be sure to label the work so that
we know exactly what each term means
...
(a) −7
(b) 0
...
0 (The subtraction left only two significant digits
...
668 [Not too different from answer (b) because the values
were not too different
...
1 [Not too different from answer (c) because the values were
not too different
...
(a) −3
...
)
(b) 0
...
57 × 10−14 (Note that −14 is larger than −15
...
60 × 10
7
...
0
(b) 4
...
389
(d ) 10
...
Each model calculator is different, so read the instruction booklet if
the instructions here are not applicable
...
377 (Use the 1/x key
...
2 (Use the x 2 key or another method on a more powerful
calculator
...
00 (Use the 2nd F and LOG keys
...
2 × 103 (Use the 2nd F and LOG keys; use only two
significant digits, since the 3 shows the magnitude of this number
...
Each model calculator is different, so the results may be slightly
different from these
...
586700236
(b) −3
...
586700236
(d ) −13
...

The only difference in the logarithms are the integer portions
[except for a round-off in part (d )]
...
We should report the values
(a) −0
...
587
(c) −9
...
587

13

Chapter 2

Measurement

2
...
Once
we have learned it, it is much easier to use than the English
system, as we will see later
...

The units will be introduced in the three following subsections
...
Please note carefully the
abbreviations, and use the proper one for each term
...

It is easy to tell the difference because milli- is a prefix, so an m
before another letter means milli-
...
Please note that of the abbreviations in
Table 2-1, only the L for liter is capitalized
...

For example, capital M stands for another quantity (molarity) or
another prefix (mega-)
...
The meter was originally defined as 1 ten-millionth of the
distance from the north pole to the equator through Paris, France
...
There is an even later definition,
but we will be satisfied that it is the distance between those two
14

Table 2-1

Most Important Metric Terms and Abbreviations

Unit

Abbreviation

Prefix

Abbreviation

meter
gram
liter

m
g
L

kilodecicentimilli-

k
d
c
m

scratches
...
The meter is about 10%
longer than a yard, but that statement is merely to give us some idea
of its length
...

The meter can be divided into subunits (Fig
...
The metric system uses the same prefixes
to define the subunits and multiples for the meter as it does for all its
other units, which is a great advantage
...

The only real use that we will make of the prefix deci- is with
volume measurements, where a cubic decimeter is a useful sized
volume
...

Mass
The unit of mass is the gram
...
2-1 The meter is divided into 10 dm, each of which is divided
into 10 cm, each of which is divided into 10 mm
...
1
0
...
001
0
...
000000001
1 × 10−12

1 Mg = 1 × 106 g
1 km = 1000 m
1 dm = 0
...
01 m
1 mm = 0
...
000001 m
1 ng = 1 × 10−9 g
1 pm = 1 × 10−12 m

∗

The prefixes in boldfaced type are the most important for us to learn first
...
The gram is the unit—the name that the
prefixes are added to—and the kilogram is the mass against which
all other masses are compared
...
)
The same prefixes are used with mass as with distance, and
they have the same meanings
...
In the English system, the subdivisions of a yard
are a foot—one-third of a yard—and an inch—one-thirty-sixth of a
yard
...
The subdivision of a Troy pound is an ounce,
one-twelfth of that pound
...
) Each type of measurement has a different subdivision, and
none is a multiple of 10
...
The symbols for the units and prefixes
are easier to learn than those for the English system units
...
It is easier
to convert metric measurements because the prefixes mean some
multiple of 10 times the fundamental unit
...
275 miles to feet
...
275 km

to meters
...
275 miles
= 6732 feet
1 mile



1000 m
= 1275 m
(b) 1
...



Volume
The metric unit of volume is the liter, abbreviated L, originally defined as the volume of a cube 1 dm on each edge
...
That volume is too large for ordinary
laboratory work, so smaller related units are used—the cubic decimeter (equal to a liter), or the cubic centimeter (equal to a milliliter)
...

Some textbooks use the classical metric unit, the liter, and its
related volumes; others use the SI unit, cubic meters, and its related
volumes
...
For simplicity, after this chapter, we
will use liters (L) in this book rather than cubic decimeters, because
almost everyone is familiar with liters and its subdivisions from everyday use
...
Please note that to convert from cubic meters to cubic
centimeters does not involve a factor of 100, but (100)3
...
We recognize that by definition
1 dollar = 100 cents, and also that 1 cent = 0
...
We can use
a factor corresponding to either of those equalities
...
01
...
01 for the c of cm, 0
...

EXAMPLE 2

Convert 1
...
(b) cm
...


Solution




(a) 1
...
00149 km

(substitute 1000 for the k)

Table 2-3 Comparison of Classical Metric and SI Units
of Volume
SI

Metric

Equivalent

1 m3
1 dm3
1 cm3
1 mm3

1 kL
1L
1 mL
1 µL

1000 L
1L
0
...
000001 L
17

Cubic meter

Cubic decimeter
Liter

Cubic centimeter
Milliliter

100 cm
1m

10 cm

100 cm
1m

1 cm

10 cm
100 cm
1m

10 cm
1 dm

Volume: (100 cm)3
1,000,000 cm3

(10 cm)3
1,000 cm3

1 cm
1 cm
(1 cm)3
1 cm3

Fig
...
(Not
drawn to scale
...
49 m
= 149 cm
(substitute 0
...
01 m



1 mm
= 1490 mm
(substitute 0
...
49 m
0
...
50 m3 to cubic centimeters
...
50 m3

1,000,000 cm3
1 m3


= 2,500,000 cm3 = 2
...
2-2
...
50 m3 to liters
...
2-2) and that it is also 1000 L
...
50 m3
18

1000 L
1 m3


= 2500 L




Units in Scientific Calculations
When arithmetic operations are done with measurements, sometimes the units must be adjusted
...
For
example, to add 2
...
0 cm, we must change one of the
values to the units of the other: 200 cm + 10
...

(2) In multiplication or division of lengths, the square of lengths,
and/or the cube of lengths, the length units must be the same
...
To multiply 2
...
0 cm, again we should change
one to the units of the other: 200 cm × 10
...

(3) Otherwise, in multiplication or division, the units do not have to
be the same
...
0 g by 23
...


2
...
Every measuring instrument has a limit as to how precisely it can be read
...
Scientists attempt to read
every instrument to one-tenth the smallest scale division
...
1 mm
...
The scientist might report 0
...
We have to recognize which of these digits record
the precision of the measurement, which are present only to specify
the magnitude of the answer, and which do both
...
The word significant in this sense does not mean important; it
means having to do with precision! Every digit serves to report either
the magnitude or the precision of the measurement, or both
...

Significant Digits in Reported Values
First we must learn to recognize which digits in a properly reported
number are significant
...
Zeros are
determined to be significant or not according to the following rules:
1
...
For example, in 1
...

19

2
...
For example,
in 1
...

3
...
For
example, in 0
...

4
...
For example, in 1200 cm, the zeros are undetermined without further
information
...
)
EXAMPLE 5 Underline the significant digits in each of the following measurements, and place a question mark below each digit that
is undetermined
...
0220 m
(b) 10
...
0 L
(d) 100 cm

Solution

(a) 0
...
4 kg

(c) 12
...
In (b), the zero between the 1 and 4 is significant (rule 2)
...
In (d),
the zeros to the right of the other digits in an integer cannot be
determined to be significant or not
...



Significant Digits in Calculations
Warning: Electronic calculators do not consider the rules of significant digits
...

There are two different rules for significant digits in an answer
determined by calculation
...
In addition and/or
subtraction, the number of significant digits in the measurements
is not the deciding factor but their positions are critical
...

20

Determine the answer in each of the following to
the proper number of significant digits: (a) 1
...
52 cm (b)
5
...
921 cm (c) (12
...
42 g)/(1
...
224 cm2 , but because there are
only three significant digits in each factor, we must limit the
answer to three significant digits: 4
...

(b)
5
...
921 cm
19
...
84 cm
The 2 in 5
...
The digit after that represents an estimated
1 in the second measurement added to a completely unknown
value in the first, and thus is completely unknown
...
95 g − 11
...
866 mL) = 0
...
53 g, a value with three significant digits
...
820 g/mL
...



The number of significant digits in the answer is determined
by the numbers in our measurements, not in defined values like the
number of millimeters in a meter
...
2 m, 1
...
200 m
...
Can we tell the number
of significant digits in each answer just by looking at the result or
from the value from which it was calculated?
Solution

(a) Two, three,
andfour, respectively
...
2 m
= 1200 mm
1
...
001 m

0
...
200 m
= 1200 mm
0
...
The values still
have two, three, and four significant digits, respectively, but they all
look the same
...
Just by looking at these values we cannot tell if the
zeros are significant or not; they are undetermined
...
3)
...
We
do that by rounding off
...
If the first
digit that we drop is 5 or greater, we increase the last digit retained
by 1
...

A more elegant method of rounding involves only the case in
which the digit 5 only or a 5 with only zeros is dropped
...
For example, if we are
rounding to one decimal place:
14
...
050, 14
...
would all round to 14
...

14
...
150, 14
...
would all round to 14
...

14
...
250, 14
...
would all round to 14
...

Note that 14
...
1, because it is not covered by this rule
...
) Most courses do not use this rule, and if the first digit to
be dropped is 5, they merely round the last retained digit to the next
higher digit whether it is even or odd
...

Round off the following numbers to three significant
digits each: (a) 12
...
39 g
(d) 0
...
24648 g
EXAMPLE 8

Solution (a) 12
...
4 g
(d) 0
...
246 g
(a) The 4 is merely dropped
...
The incorrect answer 123 g, resulting from merely
dropping the 4, would be very far from the measured value
...

22

(d) We merely drop the last 3, leaving us three significant digits
...
(e) We drop the 48, since the 4 is less than 5
...
We do not round the 4 to 5 by dropping
the last digit and then change the 6 to 7 by dropping that 5
...


Sometimes it is necessary to add digits to obtain the proper
number of significant digits in our answer
...
86 cm2 by 3
...


Solution The answer on our calculator is 2 (cm), but the answer must contain three significant digits, so we add two zeros to
the calculator’s result to get 2
...




The question is often asked “How many significant digits
should we use?” The answer is that we determine how many by
using the measurements given in the problem
...
If a quantitative problem has no
numeric data in its statement, as in a percent composition problem
(Section 4
...


2
...
In order to do so conveniently, we use scientific notation, also known as standard exponential notation
...
The following number is in scientific notation, with its
parts identified:
base

exponent
\ /
1
...

EXAMPLE 10

Which one(s) of the following numbers are in scien-

tific notation?
23

(a) 1
...
246 × 103
...
00 × 10−3

(b) 0
...
246 × 100

(c) 10
...
0 × 10−3

Solution The numbers in (a), (e), and (g) are in scientific notation; (b) is not because its coefficient is not as great as 1; (c) and
( f ) are not because their coefficients have two integral digits each;
(d) is not because it has a fractional exponent
...
The electronic calculator will do the
arithmetic with numbers in scientific notation, but we still have to
know how the process works because the calculator does not consider significant digits
...
3 for a discussion of calculator
processing of numbers in exponential form
...
67 × 102 cm + 2
...
67 × 10−2 cm + 2
...
67 × 102 cm)(2
...
67 × 102 cm2 )/(2
...
90 × 102 cm
...
29 × 101 to 0
...
If we are not convinced, we can change each number to
a decimal number and add:
267

cm

22
...
256 cm = 2
...
Again watch the significant digits
...
11 × 103 cm2
...
Caution: We must watch
out for the units even while considering a completely different
part of the problem
...
7 cm
...


24

2
...
To get density, we merely divide the mass by the volume
...

The subject is used here to review all the material covered in Sections
2
...
3
...
3 g/cm3 )
...
Was the sample gold?
Solution The density of the sample was (256 g)/(51 cm3 ) =
5
...
The sample was not gold
...
”)



Because density is a ratio, it can be used as a factor in dimensional analysis problems
...
3 g/mL)
...
6 g/mL)
...
2 mL
19
...
6 g
(b) 153 mL
= 2080 g = 2
...
5 Time, Temperature, and Energy
Time
The basic unit of time is the second
...
), but
shorter periods use the regular metric prefixes
...
001 second
...
However, watch out for times stated in
two units, such as “an hour and 15 minutes
...
15 hours
...




1 hour

60 minutes
1 hour




15 minutes




60 seconds
1 minute

60 seconds
1 minute


= 3600 seconds

= 900 seconds




The total time is exactly 4500 seconds
...
We can prove this to
ourselves by heating a pan with 1 inch of water in it on a burner at
home for 2
...
With a thermometer, we measure the rise in
temperature
...
00 minutes
...
The pan with less water
was warmed to a higher temperature by the same quantity of heat
...
The Fahrenheit scale is in common use in the United
States
...
The metric system scale is the Celsius
scale, on which the freezing point of water is 0◦ C and its normal
boiling point is 100◦ C
...
15 K and 373
...
These
latter temperatures are often rounded to three significant digits for
ease of use
...
On the
Kelvin scale, the “degree” sign is not used, and the units are called
kelvins
...

Table 2-4

Temperature Scales

Fahrenheit (F )
Celsius (t)
Kelvin (T)
26

Freezing Point
of Water

Normal Boiling
Point of Water

32◦ F
0◦ C
273 K

212◦ F
100◦ C
373 K

To convert from Fahrenheit to Celsius or back, use the following equation, where F stands for the Fahrenheit temperature:
t = (F − 32◦ )/(1
...
Ask the instructor
if the conversions between Fahrenheit and Celsius are necessary to
learn
...
A joule is the energy required
to move a force of 1 Newton through a distance of 1 meter, and a
Newton is the force required to accelerate a 1 kg mass 1 meter per
second every second
...
184 J to heat 1
...
000◦ C
...
The kinetic energy
of a body is
KE = 12 mv 2
These energy and temperature relationships will be developed more
fully in later chapters, where they are used
...
What is the SI equivalent of (a) 1 L? (b) 1 mL? (c) 1000 L?
2
...
0987 g
(b) 1
...
1 g
(d ) 0
...
Calculate the sum of exactly 1 m + 2 dm + 3 cm + 4 mm
...
What is the difference between 2 mg and 2 Mg?
27

5
...
Which ones of the following sets of units are the dimensions of
density?
g/cm3

g/mL

mg/mL

kg/m3

kg/L

kg/dm3

g/cm

Answers to Leading Questions
1
...
(a) 3 significant digits, 4 decimal place digits
(b) 4 significant digits, 3 decimal place digits
(c) 2 significant digits, 1 decimal place digit
(d) 3 each
...

3
...
234 m = 12
...
4 cm = 1234 mm
4
...
002 g; 2 Mg = 2,000,000 g = 2 metric tons
5
...
All but g/cm (which might be the basis for pricing a gold chain for a
necklace)
...
Convert (a) 2
...
(b) 2
...

2
...
852 km to meters
...
66 mm to meters
(c) 10
...

3
...
852 kilowatts to watts (W)
...
2 megahertz to
hertz (Hz)
...
Convert 3
...
(b) to cubic decimeters
...

5
...

6
...
68 km to (a) centimeters
...

7
...
55 cm2 times 2
...

8
...
0 g by 23
...

9
...
200 cm by 8
...
5 mm
...
)
10
...
200 cm by 8
...
5 mm
...
Underline the significant digits in each of the following
...
(a) 2
...
721 cm
(c) 22
...
0◦ C
28

12
...
What is the sum, to the proper number of significant digits,
of 1
...
74 × 109 m?
14
...
83 × 1011 cm and 6
...
Calculate the density of a rectangular solid that is 40
...
0 cm by 5
...
50 kg
...
Which has a greater density, a sample of oxygen gas at 2
...
00 g/cm3 ? Explain
...
A rectangular drinking trough for animals is 2
...
1 cm
wide, and 21
...
A 2
...
73 kg/dm3 is placed in it
...
If light travels 3
...
00 year?
19
...
00 × 108 m/s and it takes light about 500 s to get
from the sun to the earth, how far away is the sun?
20
...
00 × 108 m/s
...
A rectangular drinking trough for animals is 2
...
1 cm
wide, and 21
...
A 2
...
73 kg/dm3 is placed in it
...

(b) Calculate the volume of the trough
...
(d ) Are the volumes of the trough and the liquid the
same? (e) Calculate the height of the liquid
...
Calculate the volume of 2
...
6 g/mL)
...
Calculate the mass of 1
...
86 g/mL)
...
A sample of a pure substance has a mass of 329 g and a volume of
41
...
Use a table of densities to determine the identity of the
substance
...
Convert 35◦ C to the Kelvin scale
...
Convert 422 K to Celsius
...
Convert the density 5
...
(b) to kg/dm3
...
Under certain conditions, air has a density of about 1
...

29

Calculate the mass of air in a lecture room 10
...
0 m
by 3
...

29
...
90 m long, 53
...
7 cm deep
...
95 × 105 g sample of liquid with
density 1
...
Calculate the height of the liquid
in the trough
...
A certain ore is made up of 17
...
9% iron
...


Solutions to Supplementary Problems



1
...


3
...


5
...
44 m
= 244 cm

0
...
44 m
= 2,440,000 cm3 = 2
...
852 km
= 4852 m
1 km



0
...
66 mm
= 0
...
01 m
(c) 10
...
103 m
1 cm
The metric prefixes mean the same thing no matter what unit they
are attached to
...
]
(a) 4
...
2 MHz
= 4
...
50 L
= 0
...
50 L
= 3
...
50 L
= 3500 cm3 = 3
...
331 L
1000 cm3







1000 m
1 cm
= 468,000 cm = 4
...
01 m





1000 m
1 mm
(b) 4
...
001 m
= 4
...
55 cm )(2
...
88 cm (Watch out for units and
significant digits
...
0 g)/(23
...
98 g/cm3 (Watch out for units and
significant digits
...
200 cm)(8
...
15 cm) = 1
...

(b) 0
...
402 m
(d ) 30? 0? L
(a) 2
...
0◦ C
Three (0◦ C + 273
...
83 × 1011 m + 0
...
90 × 1011 m
1
...
74 × 109 m = 1
...
74 × 109 m
= 8
...
0 cm)(10
...
00 cm) = 2000 cm3 = 2
...
(a) 4
...

8
...

10
...

12
...

14
...


(4
...
00 dm3 ) = 2
...
)
16
...
00 g
= 0
...

17
...
Because the lengths are given in
different units, we must convert them to comparable units
...
(c) The volume of the liquid
...
(e) Solve the equation V = lwh for h
...
00 × 108 m
365 days
18
...
00 year
1 year
1 day
1 hour
1s
= 9
...
00 × 108 m
19
...
)







365 days
24 hours
3600 s  3
...
4 years
1 year
1 day
1 hour
1s
16
= 4 × 10 m (about 25 thousand billion miles)



1 cm
21
...
10 m
= 2
...
01 m





1 dm3
1
...
73 g/cm3
1 dm3
1 kg
1000 cm3
(b) V trough = (2
...
1 cm)(21
...
96 × 105 cm3



1 cm3
5
(c) V liquid = 2
...
50 × 105 cm3
1
...

(e) h = V/lw = (1
...
10 × 102 cm)(43
...
6 cm





1 mL
1000 g
= 184 mL
22
...
50 kg
1 kg
13
...
86 g
23
...
75 L
= 13,800 g = 13
...
d = (329 g)/(41
...
85 g/mL (The substance is iron
...
35◦ C + 273◦ = 308 K
26
...
94 kg 1000 g
27
...
94 × 10−3 g/cm3
1 m3
1 kg
1 × 106 cm3



1 m3
5
...
94 × 10−3 kg/dm3
1 m3
1 × 103 dm3
3
V = (10
...
0
 m)(3
...
3 kg
450 m3
= 590 kg (over half a metric ton)
1 m3



1 cm
29
...
90 m
= 1
...
01 m





1 dm3
1
...
55 g/cm3
1 kg
1000 cm3
1 dm3

28
...
26 × 105 cm3
1
...
26 × 105 cm3 )/(1
...
1 cm) = 12
...




69
...
2 g iron
17
...
2% iron
=
30
...
95 × 105 g

33

Chapter 3

Classical Laws of
Chemical Combination

3
...
That means that the total mass
of the reactants is equal to the total mass of the products
...
54 g
of zinc with 3
...

Solution The compound produced has a mass equal to the
total mass of the reactants:

6
...
21 g = 9
...
Also, this law can be
used to solve for the masses of reactants as well as those of products,
just as the algebraic equation x = a + b can be solved for x if a and
b are given as well as it can be solved for b if a and x are given
...
24 g
of methane (natural gas) to form 3
...
79 g
of water
...
See Supplementary Problem 2
...
41 g + 2
...
20 g
...
20 g, so the
oxygen has a mass of 6
...
24 g = 4
...



34

What mass of aluminum oxide must be electrolyzed
with carbon electrodes to yield 1
...
48 × 106 g of carbon monoxide in the Hall process for the industrial production of aluminum
...
06 × 106 g of mass
...
59 × 106 g) + (2
...
06 × 106 g)
= 3
...
2 The Law of Definite Proportions
The law of definite proportions states that the elements in any
given compound are in definite proportions by mass
...

EXAMPLE 4 If 6
...
0906 g of chlorine to form the only compound of chlorine and zinc, how much
zinc will react with (a) 14
...
36 g of chlorine? (c) with 100
...
07 g
...
15 g
...
0)/(7
...
0)/
(7
...
537 g

100
...
0906


= 92
...
18 g
28
...
0 g
7
...
085
6
...
07 g
26
...
19 g




The reaction of 6
...
20 g of oxygen
produces 9
...

(a) How much zinc oxide would be produced if 6
...
00 g of oxygen were mixed and allowed to react? (b) What law
enables us to answer this question?
Solution (a) 9
...
(Zinc and oxygen react in a ratio of 6
...
20 g, no matter how much extra oxygen is present
...



EXAMPLE 6 (a) In the experiment of Example 5, how much oxygen
did not react? (b) What law enables us to answer this question?
Solution (a) Since 3
...
00 g reacted, 1
...
(b) The law of conservation of mass
...
00 g of oxygen
...
00 g oxygen




6
...
20 g oxygen


= 2
...
3 The Law of Multiple Proportions
The law of multiple proportions states that when two or more compounds consist of the same elements, for a given mass of one of the
elements, the masses of the other elements are in small, whole number ratios
...
In a certain sample of carbon
monoxide, 1
...
33 g of oxygen
...
00 g of carbon, there is 2
...
Thus, for a given mass of carbon (1
...
33 g) : (2
...
That is a small, whole
number ratio
...

EXAMPLE 8 A sample of a compound of sodium, chlorine, and oxygen contains 2
...
08 g of chlorine, and 1
...
A second compound made with these same elements contains
1
...
54 g of chlorine, and 2
...
Show that
these data support the law of multiple proportions
...

(Any one will do
...
00 g of sodium in each
...
00 g sodium, 1
...
695 g of oxygen
36

For the fixed mass (1
...
54 g
=
1
...
695 g
=
2
...



Show that the following data are in accord with the
law of multiple proportions:

EXAMPLE 9

Compound 1
Compound 2

Element 1
29
...
4%

Element 2
40
...
6%

Element 3
30
...
0%

Solution Assume that we have 100
...
To get a fixed mass of one of the elements, it
is easiest to divide each mass in each compound by the magnitude
of the mass of one of the elements in that compound
...

Element 1
Element 2
Element 3
Compound 1 29
...
1 = 1
...
5 g/29
...
39 g
30
...
1 = 1
...
4 g/32
...
00 g 22
...
4 = 0
...
0 g/32
...
39 g

The ratio of masses of element 2 in the two compounds is
1
...
698 g = 2 : 1
...
)
The ratio of masses of element 3 in the two compounds is
1
...
39 g = 0
...
0
...
75
3
=
1
...

To convert ratios containing decimal fractions to whole number ratios, convert them to common fractions and multiply the
numerator and denominator by the denominator of the common
fraction
...
75 is 34 , so we can multiply the 0
...
75
by 4 (the denominator of the common fraction) to
1
37

Table 3-1 Some Common Fraction Equivalents
to Decimal Fractions
0
...
2

1
5

0
...
4

2
5

0
...
6

3
5

0
...
8

4
5

0
...
(We also in simple cases merely use the common
fraction, 34
...

Convert each of the following ratios to an integral
ratio: (a) (0
...
50 g)/(1 g); (c) (2
...


EXAMPLE 10

Solution

(a) (0
...
50 g)/(1 g) =

1
2
3
2

(c) (2
...
)
2 23

=

8
3

(The fractional part is 23 or 83 ,
so multiply by 3
...
What mass of an element is present in a 100
...
1% of the compound?
2
...
4, what happens to our 100
...
100
...
1 g
100 g


= 29
...


(The number of grams is equal in
magnitude to the percentage
...
We have reduced the 100
...
0 g)/(32
...
09 g
...
09 g
...
How much oxygen is required to convert 11
...
84 g of cadmium oxide?
2
...
547 g of methane is burned in
excess oxygen, carbon dioxide and water vapor are formed
...
The tube containing the phosphorus
pentoxide increases in mass by 1
...
50 g
...
A 10
...
3% sodium and
the rest chlorine
...
00-g
sample? (b) What is the mass of sodium in a 4
...
Show that the following data support the law of multiple
proportions:
Element 1
Element 2
Element 3

Compound 1
1
...
00 g
4
...
88 g
7
...
76 g

5
...
8%
28
...
1%

Compound 2
18
...
7%
48
...
8%
49
...
9%

6
...
40 g of aluminum reacts with 9
...
02 g of the only compound of just these two elements
...
Convert each of the following ratios to integral ratios:
(a) (1
...
00 g B); (b) (2
...
00 g B);
(c) (3
...
00 g B)
...
Convert each of the following ratios to integral ratios:
(a) (2
...
00 g B); (b) (4
...
00 g B);
(c) (1
...
00 g B)
...
(a) Would a set of mixtures of carbon monoxide (42
...
1% oxygen) and carbon dioxide (27
...
7% oxygen)
be expected to have a definite composition? (b) What are the
extreme limits on the percentage of carbon in such a set of
mixtures?
39

10
...

11
...
53 g
50
...
56%

Element 3
1
...
4%

(a) What should we do about the units in the data of compound 1
to simplify any calculations to be done? (b) How can we get mass
ratios from percentages? (c) How can we get a fixed mass of one
element in the two compounds? (d ) Do we consider the mass ratio
of element 1 to element 2 in each compound to establish the law of
multiple proportions? (e) What ratios do we consider?
12
...
53 g
50
...
56%

Element 3
1
...
4%

(a) Calculate the mass in grams of element 2 in compound 1
...

(c) Calculate the mass of elements 1 and 3 in each compound per
gram of element 2
...

13
...
1279 kg
62
...
68 g

Element 2
10
...
34%
1
...
43 g
27
...
117 g

Show that these data support the law of multiple proportions
...
The law of conservation of mass requires that
12
...
24 g = 1
...

2
...
23 g + 1
...
547 g = 2
...

40

3
...
3%
...
)



39
...
00 g NaCl
= 1
...
Dividing each element in compound 2 by 4
...
00 g)
mass of element 1 in the two compounds yields

Element 1
Element 2
Element 3

Compound 1
1
...
00 g
4
...
00 g
1
...
00 g

The ratio of element 2 in the two compounds is 2
...
50 g =
4 g : 3 g, an integral ratio
...
00 g : 2
...

5
...
Then we divide
each element in each compound by the magnitude of the mass of
element 1 in the compound to get a fixed (1
...
00 g
1
...
55 g

Compound 2
1
...
78 g
2
...
00 g
3
...
67 g

For the fixed mass (1
...
33 g : 1
...
00 g of element 3 in
the compounds
...

6
...
40 g of aluminum to
9
...

7
...
5 is equal to 12 , multiply each ratio by 2, to get
(a) (3 g A)/ (2 g B); (b) (5 g A)/(2 g B);
(c) (7 g A)/(2 g B)
...
(a) (2
...
00 g B)
0
...
01 g A)/(3
...
75 g A)/(1
...
75 is equivalent to
by 4:

3
4

so multiply numerator and denominator

(19
...
00 g B)

(c) (1
...
00 g B)
0
...
00 g A)/(5
...
(a) No, mixtures do not obey the law of definite proportions
...
99% carbon
monoxide, with 42
...
99% carbon dioxide, with 27
...

10
...
9 g
27
...
1 g
72
...
00 g
1
...
00 g
2
...
66 g) : (1
...
(a) We should convert the mass of element 2 in compound
1 to grams
...
0 g of compound, we
merely change the percent signs to grams
...

(d ) No
...




0
...
627 g
12
...
These conversions yield:
42

Compound 1
Compound 2

Element 1
7
...
0 g

Element 2
0
...
56 g

Element 3
1
...
4 g

Compound 1
Compound 2

Element 1
12
...
99 g

Element 2
1
...
00 g

Element 3
2
...
99 g

(c)

(d) For a fixed (1
...
0 g : 8
...
66 g : 7
...

13
...




1000 g
= 127
...
1279 kg
1 kg
Compound 1
Compound 2
Compound 3

Element 1
127
...
07 g
10
...
66 g
10
...
334 g

Element 3
28
...
59 g
7
...
00 g
6
...
006 g

Element 2
1
...
000 g
1
...
667 g
2
...
335 g

For a fixed (1
...
00 g : 6
...
006 g = 2 g : 1 g : 1
...
That of element 3 in the three compounds is
2
...
668 g : 5
...


43

Chapter 4

Formula Calculations

4
...
The unit of atomic mass is called, fittingly
enough, the atomic mass unit
...
(A few chemists use the dalton as the unit of atomic mass, in
honor of John Dalton
...
For example, 12 C atoms have a mass of exactly 12 amu
each, whereas 13 C atoms have a mass of 13
...
It turns
out that the ratio of isotopes of each of the elements in all naturally
occurring samples is very constant (to three or more significant
digits), so the weighted average of the masses of the atoms of an
element is constant, which is why Dalton’s hypotheses worked
...
The mass number refers to
a specific isotope, and is an integer—the number of protons plus
neutrons in each atom
...
Atomic masses
for almost all the elements are presented in the periodic table; mass
numbers are presented there only for elements that do not occur
naturally
...
If we ever solve a problem
and get an atomic mass outside this range, we know we have likely
made a mistake
...
That might be the mass of a mole of atoms
(Section 4
...

44

The weighted average of several sets of items is the average
with regard to the number in each set
...

(a) The atoms of a certain element have a mass 2
...
What is the atomic
mass of the element? (b) Which element is it? (c) What is the best way
to make sure that we get equal numbers of atoms of two elements
to compare total masses?
EXAMPLE 1

Solution

(a) The mass of the average atom is 2
...
026(12
...
31 amu
(b) Magnesium (see the periodic table)
...



Naturally occurring magnesium consists 78
...
98504 amu, 10
...
98584 amu, and 11
...
98259 amu
...


EXAMPLE 2
24

45

Table 4-1

Types of Formula Masses

Formula Unit

Name

Example

Atom
Molecule
Molecule

Atomic mass
Molecular mass
Molecular mass

Hg
NH3
H2

Formula unit of an
ionic compound

Formula mass

MgCl2

200
...
0 amu
2
...
2 amu

Solution
(78
...
98504 amu) + (10
...
98584 amu) + (11
...
98259 amu)
100
...
31 amu


Atomic masses are used to describe combined as well as uncombined atoms
...
The collection of atoms written to represent the compound is defined as one formula unit
...
The
term formula mass (sometimes called formula weight) refers to the
sum of the atomic masses of every atom (not merely every element)
in a formula unit
...
For uncombined atoms, the
formula mass is the atomic mass
...
For ionic compounds, there is no special name for formula
mass
...

It turns out that determining formula masses does not depend
on the nature of the formula unit; merely add the atomic masses of
each atom present no matter what the nature of the formula unit
...
0 amu
(e) 238
...
98 amu + 3(35
...
34 amu
(c) 18
...
8 amu

Calculations for (c) and (d) are done the same way that the calculation for (a) is done
...




4
...
Their formula masses are measured in atomic mass
units, which are useful for comparison purposes only
...
The mole is defined as the number of 12 C atoms in
exactly 12 grams of 12 C
...
001 mol, and is useful
for calculations with small quantities of substances
...

That is, one 12 C atom has a mass of 12 amu;
one mole of 12 C atoms has a mass of 12 grams;
one millimole of 12 C atoms has a mass of 12 mg
...
0 g is 6
...
It turns out
that this number is the number of atomic mass units in 1
...
02 × 1023 12 C atoms 1 mol 12 C
12
...
0 g
1 12 C atom
= 6
...

EXAMPLE 4 Calculate the number of (a) lemons in 3
...
(b) atoms in 3
...

Solution


(a) 3
...
50 mol

12 lemons
1 dozen


= 42 lemons

6
...
11 × 1024 atoms




Just as with dozens, the mass of a mole of atoms depends on
which atoms are specified
...
(b) The mole of uranium has a greater mass despite
there being equal numbers of atoms, because each uranium atom
has a greater mass than each lithium atom
...
Molar mass has the same numeric value as
the number of atomic mass units in a formula unit, but it is expressed
in units of grams per mole
...
0 g/mol because the formula mass of water is 18
...

Because molar mass is a ratio, it can be used as a factor in problem
solving
...
50 dozen lemons, assuming that the average weight is 3
...
(b) the mass
of 3
...

Solution





3
...
5 pounds
1 dozen
(b) The atomic mass of uranium is found on the periodic table
...
50 mol U
= 833 g U
1 mol U

(a) 3
...
The following figure may help
us remember how to convert moles to numbers of individual items
or to mass, or vice versa
...
00 mol of
NH3
...
00 mol of NH3
...
02 × 1023 molecules NH3
1 mol NH3
= 3
...
0 g NH3


= 85
...
00 mol NH3
1 mol NH3

(a) 5
...

EXAMPLE 8

Calculate the number of molecules in 56
...


Solution




56
...
34 mol NH3

1 mol NH3
17
...
34 mol NH3

6
...
01 × 1024 molecules NH3
Once we get more experience doing these types of problems, we may
solve them in a single step:





1 mol NH3
6
...
7 g NH3
17
...
01 × 1024 molecules NH3




The subscripts in the chemical formula tell us how many moles
of atoms of each element are present in a mole of the compound
...
In doing problems involving the numbers of moles
of atoms in a given number of moles of compound, be sure to identify the substance after writing the unit involved
...
40 mol
of K3 PO4 , a compound used as a fertilizer?
Solution

1
...
20 mol K



49

4
...
For example, H2 SO4 has a
mole ratio of 2 mol of hydrogen atoms to 1 mol of sulfur atoms to
4 mol of oxygen atoms
...
To get the percent
composition, take an arbitrary quantity of the compound (1
...
00 mol)
...


EXAMPLE 10

Solution

The masses are calculated as shown above:




14
...
02 g N
1 mol N



1
...
032 g H
1 mol H



16
...
00 g O
3 mol O
1 mol O
Total = 80
...
00 g/mol of oxygen atoms; this problem has
nothing to do with oxygen molecules, O2
...
032 g H
× 100
...
037% H
80
...
02 g N
× 100
...
00% N
80
...
00 g O
× 100
...
95% O
80
...
99%
50

We should always check our answer to see that it is reasonable! A total
between 99
...
5% is reasonable
...
4 Empirical Formulas
Empirical formula problems should be done with at least three significant digits in each value
...

We have learned how to convert a formula to percent composition; we will now do the opposite—convert a percent composition
to the empirical formula
...
Thus CH2 is an empirical formula, but C2 H4 is not, because
its subscripts can both be divided by 2
...
If we start with a set of masses for the elements in the compound, we can change them to moles as shown in
Section 4
...
We then have to make that set of moles into an integral
set of moles, and use those integers as subscripts in our formula
...
4 g of carbon, 66
...
2 g of oxygen
...
4 g C
= 32
...
01 g C



1 mol H
66
...
55 mol H
1
...
2 g O
= 32
...
00 g O

We now have a mole ratio of these elements, but it is not an integer
ratio
...
76 mol C
= 1
...
76

32
...
000 mol O
32
...
55 mol H
= 2
...
76
51

The ratio is close enough to a 1 : 2 : 1 ratio to deduce the


empirical formula to be CH2 O
...
Instead of masses of the elements, a problem is usually stated in terms of percentages of the
elements
...
0-g sample, the percentages are equal to the masses in grams
...
The
second complication is the division by the magnitude of the smallest number of moles may not give all integers, but some decimal
fractions in addition to integers
...
(We had the
same pro
solution of hydrogen ions, but neither of these can be connected to
a wire, so an inert piece of metal (usually platinum) is used as the
electrode
...
9-2
...
When the
standard hydrogen half-cell is used with another half-cell, the cell
potential is equal to the potential of the other half-cell, because the
standard hydrogen half-cell potential is zero
...
000 M Cu2+ solution suitably connected to a
standard hydrogen electrode just described has a potential of 0
...
We assign the standard reduction potential of the Cu2+ /Cu half-cell to be 0
...
The potential
131

Salt
bridge
H2 gas
Pt

H⫹

Fig
...

of a cell consisting of a zinc electrode immersed in 1
...
76 V, with the zinc metal being oxidized to Zn2+
...
76 V
...
The standard reduction potential of
the zinc half-cell is therefore −0
...

The concentrations of its components affect the potential of a
half-cell
...

2
...

4
...
000 M concentration is at unit activity
...
000 atm (101
...

Any pure solid is at unit activity
...

(Water in dilute aqueous solution is included as at unit
activity
...
See Table 9-1
...
However, not only reductions of cations to metals,
but any reduction half-reaction can be included in the table of standard reduction potentials
...

132

Table 9-1

Standard Reduction Potentials at 25◦ C
o (V)

F2 (g) + 2 e− → 2 F− (aq)
MnO4 − (aq) + 8 H+ (aq) + 5 e− → Mn2+ (aq) + 4 H2 O(l)
Ag+ (aq) + e− → Ag(s)
Fe3+ (aq) + e− → Fe2+ (aq)
Cu2+ (aq) + 2 e− → Cu(s)
2 H+ (aq) + 2 e− → H2 (g)
2 H2 O(l) + 2 e− → H2 (g) + 2 OH− (aq) (pure water)
Fe2+ (aq) + 2 e− → Fe(s)
Zn2+ (aq) + 2 e− → Zn(s)
2 H2 O(l) + 2 e− → H2 (g) + 2 OH− (aq) (1 M OH− )
Na+ (aq) + e− → Na(s)

2
...
51
0
...
771
0
...
0000
−0
...
44
−0
...
828
−2
...
(2) If we multiply the coefficients in the
equation by some number, we do NOT change the potential
...
(3) When we add chemical equations for half-cells, we add
the corresponding potentials
...

To get an equation for an overall reaction with a positive potential from two half-reactions, reverse the half-reaction from the
table with the smaller (or more negative) potential before adding
the half-reactions
...

In a table of reduction potentials listed in order of decreasing
the number
...
)
2
...

A(aq) + 2 B(aq) 
 C(aq) + 2 D(s)
Initial
concentration (M) 1
...
70
0
...
100
0
...
100
Equilibrium
concentration (M) 1
...
50
0
...
100)
=
= 0
...
10)(1
...

[C]2
(3
...
(a) K =
2
[A][B]
(0
...
456)2
(b)

A
+ 2B
 2C
Initial concentrations (M)
0
...
456
3
...
10
+0
...
20
Equilibrium concentrations (M) 0
...
66
3
...
02)2
=
= 65
[A][B]2
(0
...
66)2

(c) Since the rise in temperature caused the equilibrium to shift left
(the K is smaller, so the concentration of product is smaller), it
is an exothermic reaction
...

 CHO2− (aq) + H+ (aq)
HCHO2 (aq) 
Initial
concentrations (M) 0
...
0000
0
...
2 × 10−3
4
...
2 × 10−3
Equilibrium
4
...
096
4
...
2 × 10−3 )2
=
= 1
...
096

6
...
250

0
...
000

x

x

x

0
...
150 + x

x
163

Neglecting x when added to or subtracted from a larger
number, just as we did with regular equilibria, yields
Ka =

(0
...
8 × 10−5
[HC2 H3 O2 ]
0
...
0 × 10−5 M

7
...
It has a constant
(5
...
Reaction (c) proceeds much less than
reaction (b) because water is much less strong an acid than H+ is
...
The equilibrium that we are considering is still the ionization of the
ammonia
...
150

0
...
000

x

x

x

0
...
150 + x

x

Neglecting x when added to or subtracted from a larger
quantity yields
Kb =

x (0
...
8 × 10−5
[NH3 ]
0
...
8 × 10−5 M
[H+ ] = (1
...
8 × 10−5 ) = 5
...
25

9
...
53 × 10−8 M; (b) 1
...

−
+
HBH2 O3 (aq) 
 BH2 O3 (aq) + H (aq)

10
...
100
Change due
to reaction (M)
x
Equilibrium
concentrations (M) 0
...
000

x

x

x

x

x2
[BH2 O3 − ][H+ ]
=
= 7
...
100

x 2 = 7
...
000

x = 8
...
07

11
...

(a) All the HCl will react with half the C2 H3 O2− from the sodium
salt, yielding HC2 H3 O2 and leaving half the C2 H3 O2− , so this is a
buffer solution
...

(c) Two salts do not make a buffer solution
...

12
...
100

0
...
000

6
...
6 × 10−3

6
...
093

6
...
6 × 10−3

(6
...
7 × 10−4
[CH3 NH2 ]
0
...
(a) Solution (iv) is the most basic
...
(b) We convert from each
solution to the next by adding 0
...
(c) That much
base would completely neutralize the HCl and still be 0
...
30
...
We use the pH to calculate the hydrogen ion concentration, and
use that and K w to determine the OH− concentration:
[H+ ] = 5
...
0 × 10−14 )/(5
...
9 × 10−4

H2 O(l) + B(aq)
Initial
concentration (M)
Change due
to reaction (M)
Equilibrium
concentration (M)

+
−

 BH (aq) + OH (aq)

0
...
000

0
...
9 × 10−4

1
...
9 × 10−4

0
...
9 × 10−4

1
...
9 × 10−4 )2
=
= 3
...
100)

15
...
0 × 10−6
+
HA(aq) 
 H (aq)

Initial
concentration (M) 0
...
0 × 10−6
Equilibrium
concentration (M) 0
...
000

0
...
0 × 10−6

6
...
0 × 10−6

0
...
0 × 10−6 )(0
...
0 × 10−6
[HA]
(0
...
The H+ from the ionization of HCl (a strong acid) represses the
ionization of the acetic acid, so the concentration of H+ is 0
...

−
+
HC2 H3 O2 (aq) 
 C2 H3 O2 (aq) + H (aq)

Initial
concentration (M) 0
...
100 − x

0
...
150

x

x

x

0
...
150)[C2 H3 O2 − ]
=
= 1
...
100)

[C2 H3 O2− ] = 1
...
100 M acetic acid alone, the acetate ion concentration (equal to
the hydrogen ion concentration) is 1
...
The hydrogen ion
from the strong acid has lowered it to 1
...
The presence of any stronger acid will
repress the ionization of any weaker acid in the same solution
...
(a) This is not a limiting quantities problem because the two do not
react
...
(c) H+ and OH− , as in every aqueous
+
solution
...
The chloride ion has no effect
...
100
0
...
000
Change due
to reaction (M)
x
x
x
Equilibrium
concentrations (M)
0
...
150 + x
x

Neglecting x when added to or subtracted from a larger
quantity yields
Kb =

[NH4+ ][OH− ]
x (0
...
8 × 10−5
[NH3 ]
0
...
2 × 10−5 M
(g )

pH = 9
...


Beginning (mol)
0
...
050
End of acid-base
reaction (mol)
0
...
150

0
...
050

0
...
100

0
...
150

0
...
000

x

x

x

0
...
100 + x

x

Neglecting x when added to or subtracted from a larger
quantity yields
Kb =

x (0
...
8 × 10−5
[NH3 ]
0
...
7 × 10−5 M
pH = 9
...
08 before the addition of NaOH to 9
...
The small value of the increase is caused by the
buffering action of the equilibrium
...
All but (c)
...
In (b), the
ammonium ion hydrolyzes
...
In (e), the acid and base react completely to form
NH4 Cl, and the solution is the same as in (b)
...
The number of moles of each reactant is



0
...
100 L
= 0
...
200 L
...
0200 mol NaC2 H3 O2 , so the solution contains 0
...
0200 mol
= 0
...
200 L

The hydroxide ion concentration is 7
...

21
...
200 mol
0
...
0200 mol
1L
The total volume is 0
...
The acid and base react to form
0
...
100 M ammonium
ion (plus chloride ion):
0
...
100 M
0
...
12, as shown in Example 21
...
There are four
such properties, and they utilize three different concentration units
(C
(b) t = 0
...
279◦ C = −0
...
The freezing point of the solution is equal in
magnitude to the freezing-point depression, because the freezing
178

point of water is 0
...
No other solvent has a freezing point
of zero, so don’t expect the magnitude of the freezing point to
be equal to the magnitude of the freezing-point depression in
general
...
Using data from Table 11-1, we find that the freezing-point
depression is
t f = 6
...
3◦ C = 3
...
0◦ C/m)(m) = 3
...
16 m

4
...


(0
...
0821 L·atm/mol·K)(298 K)
nRT
=
= 3
...
55 L

π V = nRT
π
n
0
...
00593 M
RT
V
(0
...
12 g solute
7
...
00593 mol of solute per liter and 7
...
13 g
= 1200 g/mol = 1
...
00593 mol

6
...
55◦ C = (0
...
03 m

Then we calculate the t f from the molality:
t f = k f m = (1
...
03 m) = 5
...
00◦ C − 5
...
64◦ C
7
...
0◦ C/m)(m) = 6
...
7◦ C = 3
...
0◦ C/m)
m = 0
...
31 g solute 1000 g solvent
27
...


27
...
5 × 102 g/mol
0
...
655 mol C6 H12
55
...
0 g C6 H12
o
P solvent = Xsolvent Psolvent

0
...
106 atm)
Xsolvent = 0
...
655 mol
0
...
637 + 0
...
655
0
...
018
y = 0
...
44 g)/(0
...
Freezing-point depression is directly proportional to molality, so we
first change the mole fraction of glucose to molality of glucose
...
0000 mol total, there are present
0
...
0 g H2 O
1 kg H2 O
0
...
0170 kg H2 O
1 mol H2 O
1000 g H2 O
The molality of glucose is then
(0
...
0170 kg H2 O) = 3
...

t f = k f m = (1
...
26 m) = 6
...
06◦ C

10
...
00 kg of water, we have 2
...
6 mol H2 O
18
...
00 mol)/(57
...
0347
...
0347)(3
...
111 kPa

11
...
5 kPa
π
=
=
= 0
...
31 L·kPa/mol·K)(298 K)

Assume that we have 1
...
01 kg of solution and
0
...




344 g C12 H22 O11
0
...
71 g C12 H22 O11
1 mol C12 H22 O11
The mass of water then is 1
...
00771 kg = 1
...
0224 m, and its freezing-point
depression is
t f = k f m = (1
...
0224 m) = 0
...
0417◦ C

Notice how a solution with a significant osmotic pressure has an
extremely tiny freezing-point depression
...

12
...
86◦ C/m)(m) = 0
...
4 × 10−5 m
If we assume that we have 1
...
6 mol of
water and 5
...
The mole fraction of
sucrose is
(5
...
6 mol total) = 9
...
7 × 10−7 )(24
...
3 × 10−5 torr

Vapor-pressure lowering is not significantly more precise than
freezing-point depression is
...
πV = nRT
In such a dilute solution, the ions act almost independently
...
00200 mol of sodium ions and 0
...
00400 mol of solute particles:
π=

(0
...
0821 L·atm/mol·K)(298 K)
nRT
=
V
0
...
284 atm

14
...
Therefore, there are 0
...

πV = nRT
π=

(0
...
0821 L·atm/mol·K)(298 K)
nRT
=
V
0
...
356 atm

15
...
The molality of particles is
(0
...
500 kg) = 0
...
Neglecting the interionic
attractions,
t f = k f m = (1
...
400 m) = 0
...
744◦ C (or perhaps a little higher due to
interionic attractions)
...
(a) The molality of the solute
...

(c) The empirical formula
...
(See the next two problems
...
(a) t f = k f m = (20
...
5 C − 2
...
8 C
= m(20
...
19 m

(b) The number of grams of solute per kilogram of solvent is



4
...
9 g solute
=
149 g solvent 1 kg solvent
1 kg solvent
182

MM =

28
...
5 × 102 g/mol
0
...
0 g C
= 3
...
0 g C



1 mol H
6
...
62 mol H
1
...
3 g O
= 3
...
0 g O

The mole ratio is 1 mol C : 2 mol H : 1 mol O
...

(d) The empirical formula mass is 30
...

18
...

The molecular formula is C6 H12 O6
...
The number
ofmoles of solute is


1 mol
2
...
33 × 10−4 mol
8580 g
The molality is (2
...
1000 kg) = 2
...
86◦ C/m)(2
...
33 × 10−3 )◦ C
(b) t f = k f m = (14
...
33 × 10−3 m) = (3
...

20
...
00 L, that is 1
...
00224 mol sucrose
...
71 g C12 H22 O11
0
...
01 kg − 0
...
00 kg
The molality of the sucrose is 0
...


183

Chapter 12

Thermodynamics

Before studying this chapter, review the measurement of enthalpy
change (Chapter 8), potential (Chapter 9), and equilibrium (Chapter
10)
...
For example, Tables 12-1 and 12-2 include
entropy in units including joules and free energy change in units
including kilojoules
...
1 Entropy
Entropy, denoted S, is a quantitative measure of randomness
...
) In order to confirm that our answers to
problems are reasonable, we must know (1) that greater numbers
of moles, especially of gases, have greater randomness than smaller
numbers of moles (other factors being equal), (2) that gases have
much greater randomness than liquids, which have greater randomness than solids, (3) that greater volumes of gases have greater
randomness than smaller volumes (other factors being equal),
and (4) similarly, higher temperatures imply greater randomness
...
Absolute entropy, is a measure of the actual randomness
of a substance or a system
...
Standard means the
substance is at unit activity, where the activity of a pure solid or
liquid is defined as 1, the activity of a solute is equal to its molarity,
and that of a gas is equal to its pressure in atmospheres or its
number of moles per liter
...
Warming the substance to room temperature gives it an
entropy above zero
...
69
197
...
6
222
...
4
130
...
69
205
...
62
240
...
83
69
...
6
51
...
33
191
...
Note especially that the entropies of elements are
not zero at room temperature, as are their enthalpies of formation by
definition
...

Calculate the standard entropy change for the reaction of a mole of carbon monoxide with oxygen to produce carbon
dioxide at 25◦ C
...


Note the units
...
6 J/mol·K) − (1 mol CO)(197
...
0 J/mol·K) = −86
...

Note that the units include kelvins
...




12
...

Its value enables us to predict in which direction an equation will
proceed, or if the system is at equilibrium
...

Calculate the free energy change for a reaction at 25◦ C
in which the enthalpy change is 24
...
1 J/K
...
4 kJ − (298 K)(0
...
9 kJ
Note that it was necessary to convert joules to kilojoules (or vice


versa) and also that the kelvins in the second term canceled
...
Standard free energy of formation, G◦f , of a substance
is the free energy change of the reaction of the elements in their
standard states to produce the substance in its standard state, quite
analogous to the enthalpy of formation
...
Standard free energy changes of selected
substances are presented in Table 12-2
...
Any equation that is true for G in general is also
true for G ◦ (under the special conditions of unit activities for all
reactants and products)
...


186

Table 12-2 Standard Free Energies of Formation of Selected
Substances at 298 K

∆G◦f

∆G◦f

Substance

State

(kJ/mol)

Substance

State

(kJ/mol)

CO
CO2
HCl
NaCl
Na2 CO3
CH4

g
g
g
s
s
g

−137
...
4
−95
...
0
−1048
−50
...
71
51
...
57
−237
...
4

EXAMPLE 3 Calculate the standard enthalpy of formation of NO(g)
at 298 K from data in Tables 12-1 and 12-2
...
62 J/mol·K) − 12 mol N2 (191
...
0 J/mol·K) = 12
...
71 kJ
G ◦ = H ◦ − TS ◦ = H ◦ − (298 K)(12
...
0124 kJ/K)
H ◦ = 86
...
70 kJ = 90
...



We can determine the free energy change of any reaction if
we have the free energies of formation of all the substances in the
reaction, analogous to H ◦ :
G ◦ = Gf◦ (products) − Gf◦ (reactants)
The principles of Hess’s law (Chapter 8) also apply to free energy
change calculations
...


EXAMPLE 4

Solution

NO(g) + 12 O2 (g) → NO2 (g)
For 1 mol of NO(g):
G ◦ = Gf◦ (products)−Gf◦ (reactants)
= Gf◦ (NO2 ) − Gf◦ (NO)
= 51
...
71 kJ) = −34
...
0 g NO




−34
...



The sign of the free energy change enables us to predict if a
reaction (or other process) proceeds as written, goes in the opposite
direction, or neither (because the system is at equilibrium)
...
In a given process, if the energy of a system decreases (H is
negative) and its randomness increases (S is positive), the process
is spontaneous as written at all temperatures (because G is always
negative)
...
If both H and S increase or both decrease,
the term with the larger magnitude determines in which direction
188

the process is spontaneous, and that depends on the temperature
...

H2 O(s) → H2 O(l)
The randomness increases as the solid melts, but the energy increases
also
...
G is 0
...
In contrast, if we decrease the temperature from 0◦ C slightly, the randomness term decreases, making the
energy term more important, G becomes positive, and the equilibrium shifts to the left (toward lower energy)
...

Neither H nor S change much with temperature
...

EXAMPLE 5 Given the following values for a certain reaction at
25◦ C: H = −10
...
0 J/K
...

Solution

(a) G = H − TS = −10
...
0 J/K)
= −10
...
0250 kJ/K)
= −10
...
45 kJ = −2
...
)
(b) G = H − TS = −10
...
0 J/K)
= −10
...
95 kJ = −0
...




12
...
The relationship is
G = G ◦ + RT ln Q
189

where R and T have their usual meanings and Q is the concentration
ratio as described in Chapter 9
...
0 kJ, if the A concentration is 0
...
150 M, and the C concentration is 2
...

Solution

G = G ◦ + RT ln Q
= G ◦ + (8
...
0 kJ + (8
...
00)2
= 3
...
500)(0
...
4 kJ) + 2 mol(−237
...
8 kJ)
−2 mol(0 kJ/mol) = −817
...



We can derive another useful relationship if we solve the equation for G for the special case of a reaction at equilibrium
...

Calculate the value of the equilibrium constant for a
reaction with G ◦ equal to 255 J at 25◦ C
...


EXAMPLE 8

Solution

G ◦ = −RT ln K = 255 J = −(8
...
103
K = 0
...





12
...
We may wonder if there is any
connection between free energy change and potential, and indeed
there is:
G = −nF
where  is the potential, n is the number of moles of electrons, and F
is the Faraday constant, 96,500 C/mol e−
...

EXAMPLE 9 Calculate the free energy change for a two-electron
reaction in which the potential is −5
...

Solution

G = −nF = −(−5
...
A negative
191

potential has produced a positive free energy change, both indicating that the reverse reaction is spontaneous
...
Simplify the following equation for standard conditions:
G = G ◦ + RT ln Q

2
...
3), substitute
−n F and − ◦ n F for G and G ◦ , respectively
...
At standard conditions, the value of Q is 1, its natural log (ln) is 0,
and G = G ◦ , as expected, leading to the identity G ◦ = G ◦
...

G = G ◦ + R T ln Q
−n F = − ◦ n F + R T ln Q
Dividing each side by −n F yields
 = ◦ −

RT
ln Q
nF

which is the Nernst equation (in the form of its natural logarithm
instead of its common logarithm)
...
(a) Calculate the value of S ◦ for the reaction of carbon with
oxygen to produce carbon dioxide
...

2
...

3
...
88 kJ and the entropy change is 7
...

4
...

5
...
65 g of
CH4 with O2 to produce CO2 (g) and H2 O(l)
...
Given the following values for a certain reaction at 25◦ C:
H = 4
...
10 J/K
...

192

7
...

9
...

11
...


for which G = −877 J, if the A concentration is 0
...
75 M, and the C concentration is 0
...

Calculate the potential for a three-electron reaction in which the
free energy change is −12
...

Calculate the value of Q at which a reaction with G ◦ = −1
...

Calculate the value of T at which a reaction with H = −5
...
0 J/K would not be spontaneous
...
130 V
...

A voltaic cell
3 M(s) + 2 AuCl4− (aq) → 8 Cl− (aq) + 3 M2+ (aq) + 2 Au(s)

operating at 25◦ C has a potential of −0
...
110 M, [Cl− ] = 1
...
500 M
...
(c) Calculate the value of G ◦
...

13
...
25 V when
[AuCl4 − ] = 0
...
00 M, and [M2+ ] = 0
...

Calculate the value of K
...
Derive one equation relating the variables of the voltaic cell of the
prior problem to its equilibrium constant
...
Calculate the value of K for the following reaction at 25◦ C:
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2 O(l)

16
...
00100 V

17
...
5 O2 (g) → CO(g) + 2 H2 O(l)
193

18
...


Solutions to Supplementary Problems
1
...
6 J/mol·K) − (1 mol C)(5
...
0 J/mol·K) = 2
...

2
...
57 kJ) − (−237
...
56 kJ

3
...
88 kJ−(353 K)(0
...
25 kJ
4
...
45 J/mol·K) − (1 mol NO)(210
...
0 J/mol·K) = −72
...
84 kJ) − (86
...
87 kJ
G ◦ = H ◦ − TS ◦ = H ◦ − (298 K)(−72
...
87 kJ
H ◦ = (−34
...
7 kJ = −56
...
For 1 mol of CH4 (g):
CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2 O(l)
194

G ◦ = G f◦ (products) − G f◦ (reactants)
= G f◦ (CO2 ) + 2G f◦ (H2 O) − G f◦ (CH4 ) − 2G f◦ (O2 )
= (−394
...
13 kJ/mol) − (−50
...
9 kJ

For 2
...
65 g CH4




1 mol CH4
16
...
9 kJ
1 mol CH4


= −135 kJ

6
...
22 kJ − (398 K)(−6
...
22 kJ + 2
...
65 kJ
◦

7
...
31 J/K)(298 K) ln

[C]
[A]2 [B]

= (−877 J) + (8
...


(0
...
750)2 (1
...
0439 V

Note that 1 J divided by 1 C equals 1 V
...
For the reaction to be nonspontaneous, G must be zero
(or positive)
...
31 J/K)(298 K) ln Q
ln Q = 0
...
50

10
...

G = H − TS
0 = (−5010 J) − T (−10
...


G �

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