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Title: Problem solving medicinal chemistry
Description: It is a problem solving medicinal chemistry in English language in very easy way that can understand to every students and teachers.

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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
...
All rights reserved
...
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retrieval system, without the prior written permission of the publisher
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of the trademark owner, with no intention of infringement of the trademark
...

McGraw-Hill eBooks are available at special quantity discounts to use as premiums and sales promotions, or for use in corporate training programs
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com or (212) 904-4069
...
(“McGraw-Hill”) and its licensors
reserve all rights in and to the work
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AND EXPRESSLY DISCLAIM ANY WARRANTY, EXPRESS OR IMPLIED, INCLUDING BUT
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Neither McGraw-Hill nor its licensors shall be liable to you or anyone else for any inaccuracy, error or omission, regardless of cause, in the work or for any damages resulting therefrom
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Under no circumstances shall McGraw-Hill and/or its licensors be liable for any indirect, incidental,
special, punitive, consequential or similar damages that result from the use of or inability to use the
work, even if any of them has been advised of the possibility of such damages
...

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 problem in Section 3
...
)
EXAMPLE 12 Calculate the empirical formula of a compound composed of 52
...
1% oxygen
...
9 g C

47
...
0 g C

1 mol O
16
...
41 mol C

= 2
...
41 mol C
= 1
...
94

2
...
00 mol O
2
...
50 is equal to 1 12 or 32 , so we multiply every number of moles
by 2 (the denominator of 32 )
...
50 mol C × 2
=
1
...

4
...
The molecular formula gives all the information
52

that the empirical formula gives, and in addition it gives the ratio
of the number of moles of every element to the number of moles
of the compound as a whole
...
Use the integral result to multiply each subscript in
the empirical formula (including the understood values equal to 1)
...
0 amu
...
0 amu + 2(1
...
0 amu
...
0 amu)/(14
...
The molecular formula is thus C7 H14
...
Then we have a two-step solution; ﬁrst ﬁnd the empirical formula as in Section 4
...

Leading Questions
1
...
33 times that of a carbon atom
...
(a) Calculate the number of dozens of oranges in 60 oranges
...
(c) Calculate the number of moles of Al atoms in 3
...
(d ) Calculate the number of moles of H2 molecules in
3
...

3
...
Which section of Chapter 4 is limited to only one type of formula
unit?

Answers to Leading Questions
1
...
33 to 1
...
01 × 1024 Al atoms
= 5
...
02 × 1023 Al atoms

1 mol H2
(d) 3
...
02 × 1023 H2 molecules
= 5
...
(a) 16
...
0 g/mol (of O2 molecules)
4
...
5
...

2
...
In a 5
...
01 amu?
2
...
25 times that of carbon
...
42 times that of the
element in part (a)
...
Calculate the atomic mass of an element if 60
...
9257 amu and the rest have a mass of 70
...

4
...
9183 amu and the percentage that have a mass of 80
...

The atomic mass of bromine is 79
...

5
...

6
...

7
...

(b) Calculate the number of moles of Al atoms in
5
...
(c) Calculate the number of moles of H2
molecules in 5
...
(d ) Calculate the number of
moles of H atoms in 5
...

8
...
0 g
of aluminum
...
Calculate the number of moles of ethylene glycol, C2 H6 O2 , used as
antifreeze in cars, that are in 47
...

10
...
00 × 1020 H2 O molecules
...
Calculate the number of moles of hydrogen atoms in 17
...

12
...

13
...
1% Na, 40
...
4% O
...
Calculate the percent composition of rubbing alcohol, C3 H8 O
...
Calculate the molecular formula of a compound with molar mass
104 g/mol composed of 92
...
7% hydrogen
...
Consider the formula of hydrazinium nitrate, N2 H6 (NO3 )2
...
(b) Calculate the number of moles of
the compound in 17
...
(c) Calculate the number of moles
of nitrogen atoms in that quantity of compound
...

17
...

Solutions to Supplementary Problems
1
...
The 12
...
(The same reasoning tells us that no
American family has 2
...
)
2
...
25(12
...
0 amu
(b) 2
...
0 amu) = 65
...
4%)(68
...
6%)(70
...
7 amu
3
...
Let
x = the percentage of the 78
...

x (78
...
9163 amu)
= 79
...
9183x + 8091
...
9163x
−1
...
9
= −100
...
40%
= 49
...
(a) 2(14
...
0 amu) + 31
...
0 amu)
= 132
...
0 amu
(c) 4(31
...
(a) 132
...
0 g/mol
(c) 124 g/mol
(The numbers are the same as those in the prior problem, but the
units of molar mass are grams per mole
...
(a) 84 oranges
12 oranges

1 mol Al
= 9
...
75 × 1024 Al atoms
6
...
75 × 1024 H2 molecules
6
...
55 mol H2

2
H
atoms
×
(d) 5
...
1 mol H atoms
6
...

6
...
25
...
0 g Al
1 mol Al
= 5
...
47
...
769 mol C2 H6 O2
62
...
5
...
02 × 1023 H2 O molecules
Avogadro s number





18
...
0150 g H2 O
1 mol H2 O
molar mass







1 mol (NH4 )2 SO4
8 mol H


11
...
4 g (NH4 )2 SO4 
132 g (NH4 )2 SO4
1 mol (NH4 )2 SO4
molar mass

12
...
05 mol H atoms

2 Na 2 mol × 22
...
98 g
2 B 2 mol × 10
...
62 g
7 O 7 mol × 16
...
0 g
Total = 179
...
98 g Na
× 100% = 25
...
6 g total

%B =

21
...
04% B
179
...
0 g O
× 100% = 62
...
6 g total
Total = 100
...
Assume 100
...
27 mol Na
23
...
5 g S
= 1
...
1 g S

1 mol O
30
...
90 mol O
16
...
1 g Na

Dividing each of these numbers of moles by the smallest magnitude
yields
1
...
01 mol Na
1
...
90 mol O
= 1
...
26

1
...
00 mol S
1
...
5 is equal to 2 : 2 : 3, and the empirical formula is
Na2 S2 O3
...

3 C 3 mol × 12
...
03 g C
8 H 8 mol × 1
...
064 g H
1 O 1 mol × 16
...
00 g O
Total = 60
...
03 g C
× 100% = 59
...
09 g total

%H =

8
...
42% H
60
...
00 g O
× 100% = 26
...
09 g total
Total = 100
...
Assume 100
...
69 mol C
12
...
7 g H
= 7
...
0 g H

92
...
The empirical formula mass is therefore
13 amu, which divides into 104 amu exactly 8 times
...

16
...
0 g/mol) + 6 mol H(1
...
0 g/mol) = 158 g

1 mol N2 H6 (NO3 )2
(b) 17
...
110 mol N2 H6 (NO3 )2

4 mol N
(c) 0
...
440 mol N
1 mol N2 H6 (NO3 )2

6
...
440 mol N
= 2
...
This problem is similar to the prior problem, but is not stated in
steps
...
0 g/mol) + 4 mol H(1
...
0 g

1 mol NH4 N3
4 mol N
151 g NH4 N3
×
60
...
02 × 1023 N atoms
= 6
...
1 Mole Relationships in Chemical Reactions
Stoichiometry is the subject that tells the quantity of one substance that reacts with some quantity of anything else in a chemical
reaction
...
Therefore it is imperative to write a balanced chemical equation for every problem involving a chemical reaction
...
The ratio of coefﬁcients of any two substances in a
chemical equation can be used as a factor to solve a problem
...
50 mol of Mg with excess N2 ? (b) How many moles of
Mg are required to react with 3
...
50 mol Mg

1 mol Mg3 N2
3 mol Mg

= 0
...

2 Mg(s) + O2 (g) → 2 MgO(s)

2 mol Mg
3
...
00 mol Mg
1 mol O2

5
...

However, stoichiometry problems often give students more trouble
than they should because the problems are often asked in terms of
masses or other quantities that can be related to moles of reactant
or product
...

What mass of Li3 N will be produced by the reaction of
2
...
We must ﬁrst change the mass to moles,
as we did in Section 4
...
Note well: The coefﬁcient in the balanced
chemical equation has nothing to do with the conversion of mass
to moles or vice versa
...
75 g Li

1 mol Li
6
...
396 mol Li

(Note: We use 1 mol of Li in the factor, not the number in the balanced equation
...
396 mol Li
= 0
...
2:

34
...
132 mol Li3 N
= 4
...

Mass
of Li

Molar
mass of Li

Moles
of Li

Balanced
chemical
equation

Moles of
Li3N

Molar
mass of Li3N

Mass
of Li3N

5
...
Examples are number of formula units of reactant or product, or number of moles of
an element in one of the reactants or products, as well as data on
solutions or gases that will be presented later (in Chapters 6 and 7)
...
1 to determine
the number of moles of reactant or product that was asked about,
and ﬁnish the problem as required
...
97 × 1021 formula units of KClO3
...

Solution
heat

2 KClO3 (s) −→ 2 KCl(s) + 3 O2 (g)
9
...
0166 mol KClO3
6
...
0166 mol KClO3

3 mol O2
2 mol KClO3

= 0
...
0249 mol O2

32
...
797 g O2
61

As we gain experience, we may want to combine all three steps into
one, which may help precision by minimizing rounding errors:

9
...
02 × 10 units KClO3
Avogadro’s number

3 mol O2
2 mol KClO3
equation
stoichiometry

32
...
795 g O2

molar mass

EXAMPLE 4 Calculate the number of moles of nitrogen atoms in
the NH4 NO3 produced by the reaction of 2
...

Solution

NH3 (aq) + HNO3 (aq) → NH4 NO3 (aq)

1 mol NH4 NO3
2 mol N
2
...
20 mol N atoms

5
...
We have assumed or have been
told that a sufﬁcient or more than sufﬁcient quantity has been
present of any other reactant(s)
...
For example, if we are making
baloney sandwiches with a slab of baloney and two slices of bread
for each, how many sandwiches can we make (a) with 10 slabs
of baloney and 16 slices of bread? (b) with 10 slabs of baloney
and 24 slices of bread? In case (a), we run out of bread before
we run out of baloney, and we can make only eight sandwiches
(despite the fact that we have more slices of bread than slabs of
baloney)
...
In case (b), we can
make 10 sandwiches before we run out of baloney
...
These are examples of limiting quantities
problems, and the same principles apply to chemical reactions
...
That reactant is called the limiting
quantity, and the reactant that is left over is said to have been
present in excess
...

Calculate the quantity of sodium chloride that can be prepared
by the reaction of (a) 0 mol of sodium and 1 mol chlorine
...
(c) 2 mol sodium
and 2 mol chlorine
...

(b) 2 mol, exactly as predicted by the balanced equation
...

That is, after the 2 mol Na reacts with 1 mol Cl2 , as in part (b),
there is no sodium left to react with the second mole of Cl2
...

To solve limiting quantities problems, the ﬁrst step is to recognize that it is such a problem
...
Make sure that all the quantities
are in moles, or convert them to moles
...
1
...
If we calculated that we have more moles of
the second reactant than is needed, the ﬁrst reactant is in limiting
quantity
...

Note that the balanced chemical equation gives the mole ratios
that react, not necessarily the ratios present at the start of the reaction
...
50 mol of NaOH with 3
...

EXAMPLE 6

Solution

H2 SO4 (aq) + 2 NaOH(aq) → Na2 SO4 (aq) + 2 H2 O(l)
We can start with the quantity of either reactant
...
50 mol H2 SO4

2 mol NaOH
1 mol H2 SO4

= 7
...
00 mol of NaOH is needed to react with all the acid, and 7
...
We base further calculations on
63

the limiting quantity:

3
...
50 mol Na2 SO4

If we had started our calculation with the 7
...
Try it
...
What can be calculated for a chemical reaction from the knowledge
of the number of moles of the ﬁrst reactant that reacts?
2
...
What is the importance of the balanced chemical equation in solving
stoichiometry problems?

Answers to Leading Questions
1
...

2
...

3
...

Supplementary Problems
1
...
18 mol of
oxygen gas with excess titanium
...
Calculate the number of moles of sodium required to react with
17
...

3
...
32 mol of NO according to the following equation, one step in
the industrial production of nitric acid:
4 NH3 (g ) + 5 O2 (g) → 4 NO(g) + 6 H2 O(l)

4
...
00 mol of H3 PO4
reacts with exactly 2
...

5
...

6
...
788 kg of
liquid bromine
...
Calculate the mass of each reagent required to yield 99
...
Calculate the number of moles of aluminum chloride that can be
produced by the reaction of 7
...

9
...
62 × 1020
molecules of liquid bromine
...
Calculate the number of molecules of each reagent required to
yield 1
...
Calculate the number of moles of Na2 SO4 that will be produced by
the reaction of 1
...
76 mol of SO3
...
Calculate the mass of Na2 SO4 that will be produced by the reaction
of 4
...
76 mol of aqueous H2 SO4
...
Calculate the mass of Na2 SO4 that will be produced by the reaction
of 89
...
8 g of aqueous H2 SO4
...
Calculate the mass of H3 PO4 that can be produced by reaction of
water with the quantity of P4 O10 that contains 5
...

15
...
50 g of Ba(OH)2 is treated with a sample of HCl
containing 4
...
(a) What can we tell from the
mass of Ba(OH)2 and its molar mass? (b) What can we tell from the
number of molecules of HCl and Avogadro’s number? (c) What
equation can we predict for a reaction of these compounds from
their formulas? (d ) What can we tell from the results of (a) to (c)?
(e) What ﬁnal conclusions can we draw?
16
...
50 g of Ba(OH)2 is treated with a sample of HCl
containing 4
...
(a) Calculate the number of
moles of Ba(OH)2
...

(c) Write a balanced equation for the reaction that takes place
...
A sample of 4
...
44 × 1021 molecules
...
When 15
...
The water was trapped in one cylinder (by reaction
with a certain compound) and the carbon dioxide was trapped in
another
...
3 mg of mass, and the carbon
dioxide cylinder gained 30
...
(a) What mass of carbon dioxide
was produced by the reaction? (b) What mass of water was
produced by the reaction? (c) How many millimoles of each was
produced? (d ) How many millimoles of carbon was in the original
sample? (e) How many millimoles of hydrogen was in the original
sample? ( f ) How many milligrams of each element was in the
original sample? (g ) How many milligrams of oxygen was there in
the original sample
...
When 25
...
The water was trapped in one cylinder (by reaction
with a certain compound) and the carbon dioxide was trapped in
another
...
7 mg of mass, and the carbon
dioxide cylinder gained 43
...
What is the empirical formula of
the sample?
20
...
0 g
of HClO4 according to the equation
CaCO3 (s) + 2 HClO4 (aq) → Ca(ClO4 )2 (aq) + H2 O(l) + CO2 (g)

(b) Explain why the mass ratio in this reaction is equal to the ratio of
moles of reactants
...
Explain why many of the problems in this chapter seem similar to
others
...
Calculate the number of grams of carbon that react with 2
...
50 × 106 g) of Al2 O3 in the industrial production of
aluminum at high temperature according to the following equation:
Al2 O3 (special solution) + 3 C(s) → 2 Al(l) + 3 CO(g)
...
Calculate the mass of sulfuric acid (the chemical produced in the
largest tonnage in the world) produced by the reaction of
5
...
00 × 106 g) of sulfur in the following sequence
of reactions:
S(s) + O2 (g) → SO2 (g)
SO2 (g) +

1
2

O2 (g) → SO3 (g)

SO3 (g) + H2 O(l) → H2 SO4 (l)
66

24
...
7% Fe2 O3 that
requires 2
...
Ti(s) + O2 (g) → TiO2 (s)

1 mol TiO2
4
...
18 mol TiO2
1 mol O2
2
...
56 mol Br2
= 35
...
3
...
32 mol NH3
4 mol NO

5 mol O2
3
...
15 mol O2
4 mol NO
4
...
The equation represents the reacting
ratio, therefore the coefﬁcient of H3 PO4 is 1 and that of KOH is 2:
H3 PO4 (aq) + 2 KOH(aq) → K2 HPO4 (aq) + 2 H2 O(l)

5
...
13 mol Cl2
293 g Cl2
70
...
13 mol Cl2
3 mol Cl2
1 mol AlCl3
Alternatively, a complete solution in one step may be obtained:

1 mol Cl2
2 mol AlCl3
133 g AlCl3
293 g Cl2
70
...
2 Na(s) + Br2 (l) → 2 NaBr(s)

1 mol Br2
= 17
...
8 g Br2
This problem is now related to Supplementary Problem 2
...
99 g Na
17
...
4 g Na
1 mol Br2
1 mol Na

1 mol NO
7
...
6 g NO
= 3
...
0 g NO
This problem is now related to Supplementary Problem 3
...
32 mol NO
= 3
...
32 mol NO
= 4
...
0 g NH3
= 56
...
32 mol NH3
1 mol NH3

32
...
15 mol O2
= 133 g O2
1 mol O2

8
...
11 × 1024 molecules Cl2
6
...
8 mol Cl2

2 mol AlCl3
3 mol Cl2

= 11
...
87 mol AlCl3

9
...
62 × 1020 molecules Br2

2
...
02 × 1023 molecules Br2

2 mol Na
1 mol Br2

= 2
...
38 × 10−4 mol Na

5
...

23
...
0124 g Na

4 mol NH3
×
4 mol NO

6
...
47 × 1022 molecules NH3
1 mol NH3

1 mol NO
5 mol O2
1
...
0 g NO
4 mol NO

6
...
09 × 1022 molecules O2
1 mol O2
1
...
0 g NO

11
...
It is therefore used for
the rest of the problem:

1
...

1 mol Na2 SO4
1 mol Na2 O

= 1
...
40 mol NaOH
= 2
...
76 mol of H2 SO4 present, but 2
...

1
...
50 × 102 g Na2 SO4

13
...
7 g NaOH
40
...
8 g H2 SO4

1 mol H2 SO4
98
...
24 mol NaOH

= 0
...
12 mol H2 SO4 needed
2 mol NaOH
The H2 SO4 is limiting, so:

1 mol Na2 SO4
142 g Na2 SO4
0
...
24 mol NaOH

= 93
...
P4 O10 (s) + 6 H2 O(l) → 4 H3 PO4 (aq)

98
...
00 mol P
4 mol P
1 mol P4 O10
1 mol H3 PO4
= 4
...
2
...
00 mol of the acid
...
(a) The number of moles of Ba(OH)2
...
(c) Ba(OH)2 + 2 HCl → BaCl2 + 2 H2 O
...
(d ) From the numbers of moles of
each and the balanced chemical equation, we can tell the reactant
that is in limiting quantity
...

1 mol Ba(OH)2
= 0
...
(a) 4
...
44 × 10 molecules HCl
6
...
00738 mol HCl
(c) Ba(OH)2 (aq) + 2 HCl(aq) → BaCl2 (aq) + 2 H2 O(l)

1 mol Ba(OH)2
(d) 0
...
00369 mol Ba(OH)2 needed
The HCl is limiting
...
0263 mol of Ba(OH)2 present − 0
...
0226 mol Ba(OH)2 excess
70

17
...

18
...
0 mg
(b) 12
...
0 mg CO2
= 0
...
0 mg CO2

1 mmol H2 O
12
...
683 mmol H2 O
18
...

1 mmol C
= 0
...
682 mmol CO2
1 mmol CO2
(e) The same number of millimoles of hydrogen atoms are in the
reactant as in the H2 O produced
...
683 mmol H2 O
= 1
...
0 mg C
( f ) 0
...
18 mg C
1 mmol C

1
...
37 mmol H
= 1
...
0 mg −8
...
38
mg = 5
...
4 mg O
= 0
...
0 mg O
There are 0
...
37 mmol H, and 0
...

19
...

1 mmol CO2
1 mmol C
43
...
986 mmol C
44
...
63 mmol H
23
...
0 mg H2 O
1 mmol H2 O

12
...
986 mmol C
= 11
...
63 mmol H

1
...
65 mg H

25
...
8 mg − 2
...
6 mg oxygen

1 mmol O
10
...
663 mmol O
16
...
986 mmol C, 2
...
663 mmol O in the
sample, which gives a ratio of 1
...

1 mol HClO4
1 mol CaCO3
×
20
...
0 g HClO4
100 g HClO4
2 mol HClO4

100 g CaCO3
= 12
...

21
...
The set is intended to demonstrate to us that once we
understand the material, there is not as much to learn as we might
have thought
...
0 g C
6
2
...

2 3
102 g Al2 O3
1 mol Al2 O3
1 mol C
= 8
...
5
...
1 g S

1 mol H2 SO4
1 mol S

98
...
53 × 107 g H2 SO4

6
24
...
00 × 10 g C

1 mol C
12
...
89 × 106 g Fe2 O3

72

1 mol Fe2 O3
3 mol C

100 g ore
13
...
89 × 106 g Fe2 O3
= 6
...
1 Molarity
The most common unit of concentration in chemistry is molarity,
deﬁned as the number of moles of solute dissolved per liter (or cubic
decimeter) of solution
...
The symbol for molarity is
an italic capital M; its unit is molar, symbolized M
...
) Do not use lowercase letters for either! We use mol as an
abbreviation for mole; we do not use either capital M or lowercase m
...

A cup labeled A has two lumps of sugar in it and is
ﬁlled with tea
...
(a) Which cup, if either, has more sugar in it? (b) In which
cup, if either, is the tea sweeter?
EXAMPLE 1

Solution (a) Cup A has more sugar
...
) (b) The tea is equally sweet in each, because the concentration of sugar is the same in each
...
00 L of solution contains 4
...
50 mol
moles of solute
=
= 2
...
00 L

73

EXAMPLE 3 Show that a solution containing 4
...
00 mL of solution is also 2
...

Solution

4
...
00 mL

1 mol
1000 mmol

1000 mL
1L

=

2
...
00 L

= 2
...
Wherever the symbol M appears, it can be replaced by
mol/L or mmol/mL, and for 1/M, their reciprocals can be substituted
...
00 L of 4
...

Solution As with most factor label method solutions, put
down the quantity ﬁrst, then multiply it by the appropriate ratio:

4
...
00 L
= 12
...
6-1 that this answer is correct
...
80 M solution that contains 4
...

Solution

The reciprocal of the ratio corresponding to molar-

ity is used:

4
...
80 mol

= 1
...
00
mol
4
...
00
mol

Fig
...
The number of
moles is the molarity times the volume
...
0 mol in 3
...

74

If a solution is diluted with solvent, its number of moles of
solute does not change, but its molarity gets lower
...
50 L of a 2
...
50 L
...

2
...
50 L
= 3
...
The 3
...
50 L of solution:
3
...
833 M
4
...

6
...

For example, a solution of HCl, whose concentration is known, and
a solution of NaOH, whose concentration is to be determined, are
titrated
...
An indicator, a chemical that changes color at the
point where the proper quantity of one chemical has been added to
the other, signals the end of the titration
...
The end point is a point in the
titration at the point where the ratio of moles of the reactants added
is the same as that ratio in the balanced chemical equation
...
70 mL of NaOH has been added to 25
...
000 M HCl when the end point is reached, what is the
concentration of the base?

EXAMPLE 7

Solution

The balanced equation for the reaction is

HCl(aq) + NaOH(aq) → NaCl(aq) + H2 O(l)
75

The number of millimoles of acid is

3
...
00 mL HCl
= 75
...
00 mmol NaOH

= 1
...
70 mL
If 45
...
00 mL of a
solution of 3
...
000 mmol
25
...
00 mmol H2 SO4
1 mL
According to the balanced chemical equation, the titration is
stopped when the number of millimoles of NaOH is twice the number of millimoles of H2 SO4 , so the concentration of the base is
150
...
282 M NaOH
45
...
200 M solution of NaOH is treated with a 0
...
At the equivalence point (where the reaction is just
completed), (a) what would be the concentration of NaOH if no
reaction had occurred? (b) What is the concentration of the NaCl
produced?

EXAMPLE 9

Solution (a) Because the concentrations of acid and base are
equal, the volume has been doubled by the addition of the HCl
solution
...
100 M
...
100 M
...
200V mol of each reactant
...
200V mol of NaCl is
76

produced, in the 1 : 1 : 1 : 1 ratio of reactants and products:
HCl(aq) + NaOH(aq) → NaCl(aq) + H2 O(l)
The concentration is
0
...
100 M
2
...
If we try using
10
...
0 mL, 2
...
)

6
...
It is deﬁned as the number of moles
of solute per kilogram of solvent
...
(Some texts use regular m for
both
...
Great care must be taken to
avoid confusing molarity and molality because their names as well
as their units and symbols are so similar
...
As usual, be very careful with the units
...
500 mol of solute in 250 g of solvent
...
500 mol
= 2
...
250 kg

EXAMPLE 11 Calculate the mass of water required to prepare a
4
...
0 g of NH3
...
0 g NH3

1 mol NH3
17
...
00 mol NH3

= 0
...
4 Mole Fraction
Another way to measure concentration is with mole fraction, deﬁned as the number of moles of a given component of a solution
77

divided by the total number of moles in the solution
...
The mole fraction of component A is symbolized XA ; it has no units because the
unit moles in the numerator and the unit moles in the denominator
cancel each other
...
00 mol of alcohol and 8
...

EXAMPLE 12

Solution

Xalcohol =
Xwater =

2
...
200
10
...
00 mol water
= 0
...
00 mol total

As is evident from Example 12, the total of all mole fractions
in any given solution is 1
...

Both molality and mole fraction are intensive properties,
which is useful for an easy method to convert from one to the other
...

Calculate the mole fraction of ammonia in a 2
...

EXAMPLE 13

Solution

Assume 1
...
Then there are 2
...
00 kg H2 O

1000 g H2 O
1 kg H2 O

XNH3 =

1 mol H2 O
18
...
6 mol H2 O

2
...
0347
57
...
00 m and 2
...
00 kg
was selected) and in the next
...
100
...

EXAMPLE 14

78

Solution Assume a total of 1
...
Then there
are 0
...
900 mol of water
...
0 g H2 O
1 kg H2 O
0
...
0162 kg H2 O
1 mol H2 O
1000 g H2 O

malcohol =

0
...
17 m
0
...
List the differences between molarity and molality
...
Explain why molality is not used with titrations
...
The one-letter difference in the spelling, the capital letters for
molarity and molar in contrast to the lowercase designations for
molality and molal, the volume for molarity as opposed to mass for
molality, and the fact that solution is designated for molarity and
solvent for molality are the major differences
...
Titrations are done by volume, and molality does not deal easily with
volumes
...
Calculate the molarity of a 250-mL solution containing 80
...

2
...
71 mol of solute
...
Calculate the volume of a 2
...
22 mol of
solute
...
Calculate the number of moles of solute required to make
50
...
500 M solution
...
Calculate the molarity of a solution after 1
...
06 M solution
is diluted to 2
...

6
...
00 mL of the solution takes
41
...
000 M NaOH to neutralize it
...
Calculate the concentration of H3 PO4 if 25
...
98 mL of 4
...

79

8
...
40 mL of 3
...
(b) Calculate the molar mass of the acid if 4
...

9
...
0150 mol of NaCl in 50
...
02 g/mL
...
Calculate the mole fraction of each component in a solution of
50
...
0 g of H2 O
...
A 0
...
100 M solution of
HNO3
...
Calculate the molar mass of an acid, H2 A, if 6
...
22 mL of 3
...

13
...
40 m and contained 245 g of solvent
...

14
...
150
...
Calculate the molar mass of an unknown base, B, with molecules
that each react with two hydrogen ions, if 7
...
44 mL of 3
...

B(s) + 2 HCl(aq) → BH2 Cl2 (aq)

16
...
24 M in sucrose,
C12 H22 O11
...
05 g/mL
...

2
...

4
...

80

80
...
320 M
250 mL
1
...
42 M (Watch the units!)
0
...
22 mol
= 2
...
00 mol

1
...
001 mol
50
...
07500 mol
1 mL
1 mmol
The number of moles of solute in the initial solution is calculated
ﬁrst:

2
...
70 L
= 3
...
50 mol
= 1
...
50 L

6
...
000 mmol NaOH
41
...
9 mmol NaOH
1 mL
HCl(aq) + NaOH(aq) → NaCl(aq) + H2 O(l)

According to the balanced chemical equation, the same number of
millimoles of acid is needed, so the concentration of the acid is:
166
...
676 M HCl
25
...
000 mmol NaOH
1 mmol HCl
41
...
00 mL HCl
= 6
...
The number of millimoles of base is

4
...
98 mL
= 155
...
97 mmol H3 PO4
= 2
...
00 mL

3
...
2 mmol NaOH
1 mL
Because the acid has only one ionizable hydrogen atom, as
denoted by the formula HA,

8
...
40 mL

HA(s) + NaOH(aq) → NaA(aq) + H2 O(l)

the quantity of HA is also 52
...

81

4170 mg HA
= 79
...
9 g/mol
52
...
The molality may be calculated immediately:
(b)

m=

0
...
300 m
0
...
0 g
= 2
...
0 g
XNaCl =

0
...
00536
(2
...
0150 mol)
(Note the signiﬁcant digits
in the denominator
...
The
mass of the NaCl is

58
...
0150 mol
= 0
...
9 g, and the volume is
1 mL
50
...
9 mL
1
...
0150 mol)/(0
...
301 M

10
...
0 g CH3 OH
= 1
...
0 g CH3 OH

1 mol H2 O
75
...
17 mol H2 O
18
...
17 mol H2 O + 1
...
73 mol total
X H2 O =

4
...
728
5
...
000 − 0
...
272
82

11
...
Therefore the volume of the solution has doubled, and
the concentrations of the sodium ion and the nitrate ion have each
been halved to 0
...
The hydrogen ion and hydroxide ion have
been completely used up, so the sodium nitrate is 0
...

12
...
000 mmol NaOH
22
...
66 mmol NaOH
1 mL

The quantity of H2 A is

66
...
33 mmol H2 A

6
...
00 × 102 g/mol H2 A
0
...
The number of moles of solute in the initial solution was

2
...
245 kg solvent
= 0
...
588 mol of solute, but
now it is in 0
...
The new molality is
0
...
59 m
0
...
Assume that we have a total of 1
...

Then there are 0
...
0 g H2 O
0
...
3 g H2 O = 0
...
150 mol alcohol
= 9
...
0153 kg H2 O
83

15
...

3
...
44 mL
= 124
...
3 mmol HCl
2 mmol HCl

= 62
...
99 g B
= 129 g/mol B
0
...
Assuming that we have 1
...
05 kg = 2050 g
of solution and 2
...

344 g C12 H22 O11
2
...
28 kg
...
24 mol)/(1
...
75 m
...

84

Chapter 7

Gas Laws

7
...
(We could have used equations earlier, for example, as early as density calculations in Section 2
...
)
It would be good to review the section Scientiﬁc Calculations
(Section 1
...
Pay careful attention
to the units
...
It makes a great deal of
difference if the problem states that the volume increased 5
...
0 L
...
00-L sample of gas at 150 kPa and 25◦ C is compressed to 0
...

A 1
...
500 L
at constant temperature
...
000 atm = 101
...
0 torr = 760
...
We are to determine
that variable
...
Know the conditions applicable for each of these types
of problems:
Boyle’s law

given sample of gas,
temperature constant
Charles’ law
given sample of gas,
pressure constant,
temperature in kelvins
combined gas law given sample of gas,
temperature in kelvins

P1 V1 = P2 V2
V1 /T1 = V2 /T2
P1 V1 /T1 = P2 V2 /T2

Every gas law that includes temperature must have the temperature
in kelvins—absolute temperature!
In Boyle’s law, pressure and volume are inversely proportional
...

(a) Calculate the ﬁnal volume if a 1
...
22 atm at constant temperature
...
50-L sample of gas at 788 torr is
changed to 1
...

EXAMPLE 1

Solution

(a) Using Boyle’s law:
86

P1 V1 = P2 V2

Dividing both sides by P2 yields:
P
1
2

105 kPa
1
...
50 L
V2

V2 = P1 V1 /P2 = (105 kPa)(1
...
27 L
This volume is reasonable, since the volume will decrease because of the increased pressure
...
22 atm = 927 torr

P1 V1
P2 V2
=
= V2
P2
P2

V
1
...
50 L)/(927 torr) = 1
...

From now on, we will follow our usual practice of using liters
rather than cubic decimeters, but we will use some pressures in kilopascals and some in torr, as well as some in atmospheres
...
50 L at 25◦ C at constant pressure
...
50 L

V1 = T1 V2 /T2 = (348 K)(1
...
75 L
This volume is reasonable, since the volume will decrease because of
the decreased temperature
...

87

Calculate the ﬁnal volume if a 1
...
000 atm at 100◦ C
...
000 atm = 760
...
50 L
V2

T
10◦ C = 283 K
100◦ C = 373 K

P1 V1 T2
(905 torr)(1
...
35 L
=
P2 T1
(760
...

An 8
...
0 kPa and 25◦ C is compressed
...
00 L? (b) is compressed to 2
...
00 L and it started at 8
...
00 L − 2
...
00 L
...
0 kPa)(8
...
00 L)

(b) V2 is 2
...
Again P2 is given by Boyle’s law:
P2 =

P1 V1
(99
...
00 L)
= 396 kPa
=
V2
(2
...
2 Moles of Gas
There are at least two distinct methods of calculating the number of
moles in a sample of gas
...

Molar Volume Calculations
A gas at 0◦ C and 1
...
3 kPa is said to be at standard temperature and pressure, abbreviated STP
...
00 mol of an ideal gas occupies 22
...
All real gases occupy approximately 22
...
This volume is called the molar volume
of a gas, but the word molar here means per mole, and has nothing to
88

do with molarity (Section 6
...
Thus measurement at STP of the volume of a sample of gas enables calculation of the number of moles
of gas in the sample, and vice versa
...
750 mol of O2
at STP
...
00 L
at STP
...
4 L (at STP)
(a) 0
...
8 L
1
...
313 mol
(b) 7
...
4 L (at STP)

If the sample of gas is not at STP, we can use the combined gas
law to calculate what its volume would be at STP, then convert that
volume to number of moles
...
(See Fig
...
)
Calculate the number of moles of O2 that occupies
17
...

EXAMPLE 6

Moles of
gas

Molar
volume

Volume
at STP
Combined
gas law
Volume
at any
pressure and
temperature

Fig
...
To convert moles of
a gas to a volume at any temperature and pressure, ﬁrst convert
the moles to volume at STP with the molar volume (ﬁrst arrow),
then convert the volume to the required volume at the given conditions with the combined gas law
...
Where do we
start? Where the complete data are given
...

89

Solution

First, calculate the volume that the gas would oc-

cupy at STP:
P1 V1 T2
(751 torr)(17
...
6 L
P2 T1
(760 torr)(298 K)

V2 =

Then calculate the number of moles:

1
...
6 L
= 0
...
4 L (at STP)

Calculate the volume occupied by 0
...
7 kPa
...
151 mol

22
...
00 mol

= 3
...
3 kPa)(3
...
35 L
P2 T1
(73
...
(Real gases obey this law best under low-pressure and hightemperature conditions
...
If R is given in L·kPa/mol·K,
we will express the volume in liters and the pressure in kilopascals
...
3 kPa/atm
...
0821 L·atm/mol·K
or

(note well the zero after the
decimal point)

R = 8
...
(See Fig
...
)
90

Molar
volume

Moles of
gas

Volume
at STP
Combined
gas law

Ideal
gas
law

Volume
at any
pressure and
temperature

Fig
...
In addition to the method in Fig
...

EXAMPLE 8 Calculate the number of moles of gas in a sample occupying 245 mL at 409 torr at 38◦ C
...
538 atm

V = 245 mL = 0
...
538 atm)(0
...
00516 mol
RT
(0
...
555 atm at −14◦ C
...
555 atm
◦

V = 532 mL = 0
...
555 atm)(0
...
0139 mol
n=
RT
(0
...
7-3 Additional Conversions
...
Conversely, these values may
be determined from gas data by working in the opposite direction
on the ﬁgure
...
(Remember the boy scouts
...
For example, we can
use it in a stoichiometry problem
...
7-3
...
0 L of oxygen at STP
...
0 L

or

n=

1 mol O2
22
...
01 mol O2

(1
...
0 L)
PV
=
= 2
...
0821 L·atm/mol·K)(273 K)

To ﬁnd the number of moles of KClO3 from the number of moles of
92

oxygen requires the balanced chemical equation (Section 5
...
01 mol O2
= 1
...

On a space station, a 10
...
3 kPa pressure springs a leak, and 8
...
What is the ﬁnal pressure of the
oxygen in the drum at 25◦ C?

EXAMPLE 11

Solution Even if we cannot see how to complete this problem, we do know that from the initial P -V-T data, we can solve for
the initial number of moles of oxygen:

n=

PV
(101
...
0 L)
=
= 0
...
31 L·kPa/mol·K)(298 K)

We also can tell what fraction of a mole of oxygen escaped:

1 mol O2
8
...
250 mol O2
32
...
409 mol − 0
...
159 mol
Because the drum is steel, it does not change volume; it is still 10
...
159 mol)(8
...
4 kPa
V
(10
...
3 Dalton’s Law of Partial Pressures
When gases are mixed, the volumes of the individual gases assume
the total volume of the mixture, and therefore they are equal to each
other and to the volume of the mixture
...

However, the pressures of the individual gases add up to the total
pressure of the mixture, and the numbers of moles of the individual
gases add up to the total number of moles of the mixture
...
For example, for
a mixture of two gases, using the subscripts 1 and 2 to denote the
individual gases:
V1 = V2 = Vtotal
P1 + P2 = Ptotal
T1 = T2 = Ttotal

n1 + n2 = ntotal

The ideal gas law can be used with any individual gas in the
mixture or with the mixture as a whole
...
)
In a mixture of argon and helium, the volume of the
argon is 2
...
(a) What is the volume of the mixture? (b) What
is the temperature of the mixture? (c) What is the number of moles
of gas in the mixture?
EXAMPLE 12

Solution (a) Since gases in a mixture each have the volume
of the mixture, in this case the volume of the mixture is equal to
the volume of the argon, 2
...
(b) The temperature of the mixture
(and of the argon) is the same as that of the helium, 27◦ C
...
954 atm
760 torr

n=

(0
...
00 L)
PV
=
= 0
...
0821 L·atm/mol·K)(300 K)

EXAMPLE 13 Show that the ratio of the pressures of two gases in a
mixture is equal to the ratio of their numbers of moles
...

and
T1 = T2 = T:
Since V1 = V2 = V

n1 RT
P1
P1 V
n1
=
=
=
P2 V
n2 RT
P2
n2
And similarly, starting with Ptotal Vtotal = ntotal RTtotal instead of
94

P2 V2 = n2 RT2 :

P1
n1
=
Ptotal
ntotal

EXAMPLE 14 Calculate the oxygen pressure in a mixture of
0
...
750 mol of nitrogen with a total pressure
of 40
...

Solution

Poxygen
noxygen
=
Ptotal
ntotal
Poxygen
0
...
0 kPa
1
...
0 kPa

When a gas is collected over water, some of the water evaporates and its vapor forms a mixture with the other gas
...

A mixture of a normal gas and water vapor behaves just like any
other gas mixture as long as no more water can evaporate and no water
vapor can condense
...
Any attempt at increasing the water vapor pressure (by reducing the volume, for example)
will result in water vapor condensing
...
Tables of water vapor pressure at various temperatures
are given in texts and typically the water vapor pressure at a speciﬁed
temperature is given on exams
...
2 kPa
...
(b) Oxygen is collected over water at 25◦ C
under a barometric pressure of 107 kPa
...
2 kPa) Calculate
the pressure of the oxygen
...
2 kPa = 104 kPa
95

(b) This problem is merely another statement of the problem of
part (a)
...
4 The Law of Combining Volumes
The law of combining volumes states that if all the gases involved
in a chemical reaction are measured (separately) at the same temperature and pressure, their volumes will be in the same ratio as their
numbers of moles in the balanced chemical equation
...
However, it can help do certain problems quickly
...

Because every term on the right side of the last equation is the same
for the two gases, the number of moles is the same also
...
Under these conditions, if the numbers
of moles of two gases in a chemical reaction are equal, their volumes
would be the same
...
Because this is true, then the gases in a chemical reaction
involving separate gases at equal temperature and equal pressure
have their volumes proportional to the number of moles of gas—
the coefﬁcients in the balanced equation
...

(a) If 6
...
(b) Can the volume of the water be determined?

EXAMPLE 16

96

Solution

2 H2 (g) + O2 (g) → 2 H2 O(l)

(a) Two types of solution are possible
...
00 L
...
00 L
VH2 (P/RT)
2 mol H2
=
=
=
nO2
VO2 (P/RT)
VO2
1 mol O2
VO2 = 3
...
)
(b) The volume of the water cannot be determined because, in this
problem, it is not a gas
...
5 Graham’s Law
Effusion is the escape of a gas through small pores in its container
...
(For example, ammonia gas diffuses through air, and can be
smelled on the far side of a room from where it is allowed to escape
...
The most useful
mathematical form of this law involves the rates of two gases:

r1
MM2
=
r2
MM1
The usual problem involving Graham’s law asks for the ratio of rates
of effusion or diffusion of two gases, or it gives the rate for one gas
and asks the rate for another
...
0 g/mol √
= 15
...
98
2
...
98(r oxygen )
Hydrogen effuses about four times as fast as oxygen under the same
conditions of temperature and pressure
...
00 mmol/minute
...
9 g/mol
= 0
...
9 g/mol

r chlorine = (0
...
00 mmol/minute) = 3
...
Once again, it is important to remember that hydrogen, nitrogen,
oxygen, ﬂuorine, and chlorine, when uncombined with other
elements, are gases that exist as diatomic molecules
...
For ease of solution, let the unknown rate appear in the numerator
...
The proportion is inverse; if the rate of a given gas appears in
the numerator, its molar mass is in the denominator, and vice
versa
...
(The faster something moves,
the less time it takes to reach its destination
...

If escaping ammonia gas can be smelled across a room
in 5
...
9 g/mol
r ammonia
= 2
...
0 g/mol

Solution

98

Because the ammonia diffuses 2
...
04 times as long to get there:
Time = (2
...
00 minutes) = 10
...

7
...

KE = 3RT/2N
where R is expressed in joules per mole per kelvin (1 J = 1 L·kPa)
...
A line (called
a bar) over a variable designates the quantity as an average
...

If two samples of gas are at the same temperature, the average
kinetic energies of their molecules are equal
...
If
the gases are not the same, their molecules have different molecular
masses
...

KE = 12 mv 2
Since KE for the two gases is the same, but m is different, v 2 must
also be different
...
(This is the basis for Graham’s law
...
(a) Compare the average kinetic energies of their
molecules
...

Solution

(a) Since the temperatures are the same, the average kinetic energies
are the same
...

99

Leading Questions
1
...

2
...
What calculation differences are there between two gases in a
mixture and two gases in a chemical reaction measured separately?
4
...
00-L sample of gas being
expanded (a) 5
...
00 L?
5
...

6
...

Answers to Leading Questions
1
...

2
...
(b) Yes
...

3
...
Two separate samples of gas may have equal pressures
and equal temperatures (as in problems involving the law of
combining volumes)
...
(a) The ﬁnal volume is 7
...
(b) The ﬁnal volume is 5
...

5
...

P 1 + P 2 + P 3 = P total
6
...
(a) Calculate the ﬁnal pressure of a gas if a 22
...
0 kPa is expanded to 0
...

(b) Calculate the ﬁnal pressure of a gas if a 49
...
123 L at constant
temperature
...
Calculate the initial temperature of a gas if a 0
...

100

3
...
25 atm and 50◦ C is changed to 1
...

4
...
790-L sample of gas at 255 kPa and 25◦ C is compressed
...
200 L? (b) was lowered 0
...
200 L?
5
...
444 mol of gas in a sample occupying
666 mL at −8◦ C
...
To show that the law of combining volumes works, do each of the
following parts by calculating the number of moles of each reactant
...
00 L of hydrogen gas reacts with oxygen gas to form liquid
water, what volume of oxygen will react, assuming that the oxygen
and hydrogen are both measured at 25◦ C and 785 torr
...
00 L
of hydrogen gas reacts with oxygen gas to form liquid water, what
volume of oxygen will react, assuming that the oxygen and hydrogen
are both measured at 53◦ C and 805 torr
...

7
...
250 mol of a gas in 10
...
Explain why in Dalton’s law problems
=
n2
P2
n1
V1
but in combining volume problems
=
n2
V2
9
...
88 g of a gas occupies 2
...
(b) Calculate the molar mass
...
Calculate the volume at STP of 26
...

11
...
00 g of nitrogen gas at 25◦ C
and 1
...

12
...
9 kPa
containing 0
...
200 mol of nitrogen
...
(a) Calculate the number of moles in 4
...
995 atm
...
Calculate the rate of effusion of helium from a porous cup under
the same conditions that CH4 effuses at a rate of
20
...

15
...
The molar masses of 235 U and 238 U are
235
...
05 g/mol, respectively
...
Calculate the time it would take for SO2 to diffuse across a room
under the same conditions that it takes an equal number of moles
of NH3 5
...

17
...
(b) Calculate their “average” velocities (u), using
KE =

1
2

mu 2

and

1 J = 1 kg·m2/s2

18
...
00 L of CO2 , all measured at the same temperature and pressure
...
As far as possible, determine the relative volumes of the substances
involved in the following reaction in the open atmosphere
at 500◦ C:
2 C(s) + O2 (g) → 2 CO(g)

20
...
00 g of KClO3
(P H2 O = 24 torr)
...
Calculate the molar mass of 4
...
75 L at
1
...

22
...
00 atm pressure is 1
...

23
...

24
...
00-g sample of gas is contained in a 2
...
10 atm pressure
...
8% carbon and the rest
hydrogen
...
A 5
...
51-L vessel at 25◦ C and
1
...
The gas contains 81
...
(a) Calculate the number of moles of gas in the sample
...
(c) Calculate the
empirical formula of the sample
...

26
...
70-g sample of gas contained
in a 2
...
09 atm pressure
...
7% carbon and the rest hydrogen
...
(a)
1
2

P
78
...
4 mL
0
...
0 kPa)(22
...
85 kPa

This pressure is reasonable, since the volume increased because
of the decreased pressure
...
1 mL = 0
...
123 L
2 P2
P 2 = P 1 V 1/V 2 = (755 torr)(0
...
123 L) = 301 torr

2
...

T
V
0
...
555 L
T1 = T2 V 1/V 2 = (248 K)(0
...
555 L) = 1
...

This temperature is reasonable, since the volume will increase
because of the increasing temperature
...
25 atm
953 mL 50◦ C = 323 K
1550 mL 75◦ C = 348 K
2 P2
P2 =

P 1 V 1 T2
(1
...
828 atm
=
V 2 T1
(1550 mL)(323 K)

This pressure is reasonable, since the volume will increase
because of the decreased pressure and also because of the
increased temperature
...
Note the wording of the three parts: “(a) was lowered by 0
...
200 L? (c) was lowered to 0
...
200 L” and “lowered 0
...
“Lowered to
0
...

103

(a) V 2 is 0
...
200 L = 0
...
P 2 is given by the combined
gas law:
P2 =

P 1 V 1 T2
(255 kPa)(0
...
590 L)(298 K)

(b) V 2 is again 0
...
200 L = 0
...
P 2 is again 399 kPa
...
200 L
...
790 L)(348 K)
= 1180 kPa
=
V 2 T1
(0
...
n = 0
...
666 L
T = −8◦ C + 273◦ = 265 K
P =

nRT
(0
...
31 L·kPa/mol·K)(265 K)
=
= 1470 kPa
V
(0
...
(a) The number of moles of hydrogen is given by
n=

[(785/760) atm](6
...
253 mol
RT
(0
...
127 mol
...
127 mol)(0
...
01 L
P
[(785/760) atm]

(b) The number of moles of hydrogen is given by
n=

[(805/760) atm](6
...
237 mol
RT
(0
...
119 mol
...
119 mol)(0
...
01 L
P
(805/760) atm

(c) No matter what values are chosen, the same answer is found
each time (within rounding error)
...
P = nRT/V = (0
...
0821 L·atm/mol·K)(296 K)/(10
...
608 atm
Alternatively, at STP, the gas would occupy

22
...
250 mol
= 5
...
00 atm)(5
...
607 atm
=
V2T1
(10
...
Dalton’s law involves a mixture of gases, in which the volumes and
temperatures must be the same; the law of combining volumes
involves separate gases, with pressures and temperatures speciﬁed as
equal
...
75 L)
PV
=
= 0
...
(a) n =
RT
(8
...
88 g
(b) MM =
= 42
...
116 mol
10
...
4 L at STP
...
The number of
moles is calculated from the mass with the molar mass:

26
...
0 g O2

0
...
834 mol O2

22
...
7 L

11
...
00 g N2
(Remember that
= 0
...
0 g N2
nitrogen gas is N2 !)
The volume is given by the ideal gas law:
V =

(0
...
0821 L·atm/mol·K)(298 K)
n RT
=
= 5
...
03 atm
105

or by the molar volume and the combined gas law:

0
...

22
...
60 L

P1V1T2
(1
...
60 L)(298 K)
= 5
...
03 atm)(273 K)

n O2
P O2
=
P total
n total
P O2 =

(105
...
400 mol O2 )
= 70
...
600 mol total

13
...
995 atm)(4
...
65 L
=
P STP T2
(1
...
65 L

1 mol
22
...
163 mol

(b) It does
not makeany difference
...

r He
=
r CH4

MMCH4
=
MMHe

16
...
00 = 2
...
00 g/mol

r He = (2
...
0 mmol/minute) = 40
...
MM235 = 235
...
998 g/mol) = 349
...
05 g/mol + 6(18
...
04 g/mol for 238 UF6

r 235
=
r 238

352
...
0086 = 1
...
03 g/mol

This very small difference is sufﬁcient to separate the isotopes with
repeated effusions
...
1 g/mol √
=
=
16
...
77 = 1
...
0 g/mol
Since the NH3 travels 1
...
94
times as long:
1
...
00 minutes) = 9
...
31 J/mol·K)(295 K)
3RT
=
= 6
...
02 × 1023 /mol)
(b) The mass of a molecule in kilograms is

17
...
016 amu

1g
1 kg
= 3
...
02 × 1023 amu
1000 g

2KE
2(6
...
35 × 10−27 kg
(about 4300 miles per hour)

18
...
00 L of O2 to react to
produce the 4
...
00 L O2
4
...
The ratio of volumes of O2 to CO is 1 : 2, according to the law of
combining volumes
...

20
...
00 g KClO3

1 mol KClO3
122 g KClO3

3 mol O2
2 mol KClO3

= 0
...
0123 mol O2

The volume of oxygen gas at STP is determined from its number of
moles:

22
...
0123 mol O2
= 0
...
276 L)((298 K)
= 0
...

n=

MM =

(0
...
0821 L·atm/mol·K)(298 K)
n RT
=
= 0
...
971 atm

(1
...
75 L)
PV
=
= 0
...
0821 L·atm/mol·K)(295 K)
4
...
1 g/mol
0
...

22
...
00-L volume
...
72 g
...
00 atm)(1
...
0409 mol
RT
(0
...
72 g
= 42
...
0409 mol

23
...
00 mol)(8
...
4 L)
= 101 kPa

24
...

(b) The molar mass of the gas
...

(d ) The molecular formula
...
)
(1
...
51 L)
PV
=
= 0
...
(a) n =
RT
(0
...
00 g)/(0
...
2 g/mol

1 mol C
(c) 81
...
82 mol C
12
...
2 g H

1 mol H
1
...
1 mol H

The mole ratio of H to C is
18
...
65 mol H
2
...
82 mol C
1 mol C
1 mol C × 3
3 mol C

The empirical formula is C3 H8
...
0 g/mol, equal within two
signiﬁcant digits to the molar mass, so the molecular formula
is C3 H8
...
This problem is similar to Problem 25, but it is not stated in parts
...
09 atm)(2
...
112 mol
RT
(0
...
70 g)/(0
...
0 g/mol

The empirical formula and molecular formula are calculated as in
Chapter 4
...
7 g C
= 7
...
0 g C

1 mol H
14
...
2 mol H
1
...

The empirical formula mass is 14
...
0 g/mol, so there are (42
...
0) = 3 empirical formula units per
molecule
...

109

Chapter 8

Thermochemistry

8
...
The heat required to change the temperature of a sample by a certain number of degrees is given by
q = mct
where q is the heat added to the sample
...
The m is the mass of the sample,
c is its speciﬁc heat capacity, called speciﬁc heat for short, and t
is the change in temperature
...
) The change is always deﬁned as the final value minus the
initial value:
t = tﬁnal − tinitial

or

t = t2 − t1

The system is deﬁned as the portion of the universe under investigation
...
When
heat is added to a system and the temperature rises, t is positive,
and therefore q is positive because m and c are always positive
...
Heat added to a system is positive; heat removed from a
system is negative
...

The specific heat of a substance is deﬁned as the heat required
to raise the temperature of 1 gram of the substance 1◦ C
...
) It takes 4
...
000 g of water from
14
...
5◦ C, so the speciﬁc heat of water (at that temperature) is
4
...
The speciﬁc heat of liquid water at other temperatures
110

Table 8-1

Speciﬁc Heat Capacities

Metals

c ( J/g·◦ C)

Aluminum
Chromium
Copper
Gold
Iron
Monel metal
Platinum
Silver
Zinc

0
...
45
0
...
129
0
...
427
0
...
24
0
...
5% Cu)

Water
Solid (ice)
Liquid
Vapor

2
...
184
2
...
Remember the value 4
...

(Perhaps a more familiar unit of heat is the calorie; 1 calorie is
deﬁned as 4
...
000 cal/g·◦ C
...
Be sure to distinguish between
the change in temperature and the individual temperatures, as well
as between heat and speciﬁc heat!
Differences in temperature or changes in temperature are the
same on the Celsius scale and the Kelvin scale
...

Solution

t = 100◦ C − 0◦ C = 100◦ C
T = 373 K − 273 K = 100 K

Calculate the quantity of heat required to raise the
temperature of 2
...
00◦ C
...
00 g)(4
...
00◦ C) = 33
...

EXAMPLE 3 Calculate the quantity of heat required to raise the
temperature of 2
...
00◦ C to 25
...

The change in temperature is 4
...

Solution

Calculate the quantity of heat required to change the
temperature of 2
...
00◦ C to 21
...

EXAMPLE 4

Solution This time, the change in temperature is 21
...
00◦ C = −4
...
00 g)(4
...
00◦ C) = −33
...

EXAMPLE 5 (a) Calculate the temperature change when 55
...
0 g of water at 22
...
(b) What is the ﬁnal
temperature?
Solution

(a) q = mct = (12
...
184 J/g·◦ C)(t) = 55
...
11◦ C
(b) The ﬁnal temperature is 1
...
0◦ C + 1
...
1◦ C
Note the difference between the change in temperature and the
ﬁnal temperature!

To what temperature would the water in Example 5 be warmed
if 55
...
7 J of energy from a chemical
reaction had been added? It doesn’t matter to the water what form
of energy has been used; just the quantity of energy is important
...
)
What mass of water is heated 2
...
4 J of
heat is added to it?

EXAMPLE 6

112

Solution

q = mct = (m)(4
...
30◦ C) = 87
...
08 g

EXAMPLE 7 Calculate the heat capacity of a metal if 157 J raises the
temperature of a 53
...
7◦ C to 21
...

Solution

q = mct = (53
...
8◦ C − 15
...
48 J/g·◦ C
(The temperature change has only two signiﬁcant digits
...
They will both wind up at the same temperature
...

EXAMPLE 8 Calculate its speciﬁc heat if a 35
...
0◦ C is immersed in 52
...
3◦ C, warming the
water to 20
...

Solution

The ﬁnal temperature of the metal is also 20
...

q = 0 = mwater cwater twater + mmetal cmetal tmetal
0 = (52
...
184 J/g·◦ C)(20
...
3◦ C)
+ (35
...
7◦ C − 58
...
1)(4
...
4) + (35
...
3◦ C)
c = 0
...

EXAMPLE 9 Calculate the ﬁnal temperature after 30
...
950 J/g·◦ C) at 71
...
3◦ C
...
(Since the two terms add up to zero, one must
113

be positive and the other negative
...
184 J/g·◦ C)(t − 21
...
0 g)(0
...
3◦ C)
Canceling the units grams in each term and dividing the entire equation by joules yields an equation that looks easier to work with:
0 = (155)(4
...
3◦ C) + (30
...
950/◦ C)(t − 71
...
5/◦ C)t − 2030
Then
(678/◦ C)t = 15,800
t = 23
...
)
Check:
0 = (155 g)(4
...
3◦ C − 21
...
0 g)(0
...
3◦ C − 71
...
3 − 21
...
)

8
...
That happens most of the time, but not
when a pure substance is heated at a temperature at which it will
change phase
...

(It takes some energy to rearrange the particles that constitute the
substance—it takes some potential energy rather than the kinetic
energy of heat
...
For example,
114

gas

sublimation

condensation

vaporization

fusion
solid

liquid
solidification

Fig
...

when sublimation occurs (a solid is changed directly to the gas
phase), the heat involved is called the heat of sublimation
...
8-1
...
The word for changing from gas to either liquid or
solid is condensation
...

Some heats of phase change are presented in Table 8-2
...
For example, the
heat of fusion of water at 0◦ C is 6
...
00 kJ/mol
...
00
40
...
3
29
...
5

801

Melting

115

the phase change takes place at constant temperature, there is no
◦
C as part of the unit and there is no t term in the equation to
calculate the associated heat
...

Calculate the heat required to melt 14
...
00 kJ

14
...
90 kJ
18
...
7 g of ice to 0◦ C,
how much heat would it take to warm the ice to the freezing point
and then melt it? (See prior example
...
As usual, watch out for the units and
the signiﬁcant digits!

4
...
90 kJ + 0
...
34 kJ

The last example indicates that to calculate the total heat required to heat a sample and to change its phase requires a separate
calculation for each
...
Then we add the three terms to get the total heat
for the process
...
0 g of liquid
water at 80
...
0◦ C
...

EXAMPLE 13

Solution To calculate the heat used to warm the liquid water
to 100◦ C takes a speciﬁc heat calculation:

q = mct = (45
...
184 J/g·◦ C)(20
...
6 kJ
1 mol
= 102 kJ
45
...
0 g
1 mol
116

To calculate the heat used to warm the water vapor takes a speciﬁc
heat calculation:
q = mct = (45
...
042 J/g·◦ C)(10
...
77 kJ + 102 kJ + 0
...

Note that we cannot even combine the two speciﬁc heat calculations because liquid water and water vapor have different speciﬁc
heat capacities
...
3 Enthalpy of Reaction
We are greatly interested in determining the energy associated with
chemical reactions
...
Even better, they would like to use the energy produced by some reactions to
make other reactions go so that they will not have to pay for extra
fuel nor dump unwanted heat into the ecosphere
...
The price we pay for manufactured goods includes
a good percentage for the energy that is used in the production and
transportation of the goods
...
We loosely use the
terms heat and enthalpy change interchangeably
...
The enthalpy change of a process is named for the
process; for example, the enthalpy of fusion is the enthalpy change
associated with a fusion (melting) process
...
Enthalpy of combustion is the enthalpy of the burning
process
...
For example, carbon burns
in limited oxygen supply to yield carbon monoxide, and in excess
oxygen it produces carbon dioxide
...

117

The enthalpy of the reaction
CO(g) +

1
2

O2 (g) → CO2 (g)

is the enthalpy of combustion of CO
...
Thus the enthalpy of formation of carbon monoxide
is the enthalpy change associated with the reaction
C(s) +

1
2

O2 (g) → CO(g)

and that of carbon dioxide is the enthalpy change associated with
the reaction
C(s) + O2 (g) → CO2 (g)
The term standard state signiﬁes the condition in which a substance
is most stable at the temperature of the problem
...
) Thus oxygen at 25◦ C is in its standard state as a gas in
diatomic molecules
...

There can be more than one name associated with the enthalpy
change of a single reaction
...

Enthalpy of formation data such as are presented in Table 8-3
allow us to calculate the enthalpy change of any reaction having its
substances listed
...
(The enthalpy required to change an element in
its standard state to the element in its standard state is zero
...
5
−83
...

EXAMPLE 15

Solution

CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2 O(l)
H = H f (products) − H f (reactants)
= H f (CO2 ) + 2H f (H2 O) − H f (CH4 ) − 2H f (O2 )
= (−393 kJ) + 2(−286 kJ) − (−74
...
The other enthalpies of formation are obtained from Table 8-3
...
Note speciﬁcally that the enthalpy of
119

formation of water is −286 kJ/mol and that there are two moles
present in the equation
...

EXAMPLE 16 Calculate the enthalpy change for the reaction of
23
...

Solution The number of kilojoules is calculated for 1 mol,
just as was done in Example 15
...
7 g
speciﬁed in the problem:

H = 23
...
0 g CH4

−891 kJ
1 mol CH4

= −1320 kJ

from the
prior problem

Calculate the enthalpy change for the reaction of CH4
with oxygen to yield carbon dioxide and 75
...

EXAMPLE 17

Again the result of Example 15 is used:

1 mol H2 O
−891 kJ
75
...
0 g H2 O
2 mol H2 O

Solution

Note that the −891 kJ enthalpy change of Example 15 is associated

with 2 mol of water in the balanced chemical equation!
Hess’s Law
If enthalpy of formation data are not available but other types of
enthalpy data are, and if we can combine the chemical equations for
the available reactions to give the desired reaction, we can combine
the enthalpy changes in the same way
...
For example, consider the general reactions
A+2B→C+2D
C+2D→E

H = −20 kJ
H = 15 kJ

We can add these equations and add their associated enthalpy
120

changes to get the enthalpy change for the reaction

A + 2B + C + ✏
2✏
D → E+C +✏
2✏
D
A + 2B → E
H = −5 kJ
If we are given equations that add up to the one desired, it is
simple to add them and their associated enthalpy changes
...

EXAMPLE 18

Calculate the enthalpy change for the reaction
3 C2 H2 (g) → C6 H6 (l)

given the enthalpies of combustion of these compounds:
C2 H2 (g)

− 1305 kJ/mol

C6 H6 (l)

− 3273 kJ/mol

Solution Note that the data given are not enthalpies of formation; we cannot merely subtract the enthalpy of the reactants from
the enthalpy of the products
...
What are the two main mathematical equations presented in this
chapter?
2
...

3
...
q = mct
and
H = H f (products) −H f (reactants)
2
...
6 kJ/mol, so the
heat of condensation is −40
...
(It is negative because heat is
removed from the vapor to condense it to liquid
...
(a) The metal loses heat; q is negative
...
(c) q is zero
...
Calculate the value in Kelvin of (a) 35◦ C
...
(c) Calculate the
difference between these temperatures on each scale
...
Calculate the quantity of heat required to raise the temperature of
42
...
00◦ C
...
Calculate the quantity of heat required to raise the temperature of
52
...

4
...
0◦ C
...
Calculate the ﬁnal temperature after 127 g of a metal
(c = 0
...
4◦ C is immersed in 338 g of water at
55
...

6
...

7
...
)
8
...
5 g of the metal at 62
...
4 g of water at 19
...
9◦ C
...
Calculate the heat required to change 35
...
0◦ C to ice at −10
...
Use the data of Tables 8-1 and 8-2
...
Write an equation to represent each of the following: (a) formation
of PCl3
...
(c) combustion of CH4
...

11
...

12
...
0 g of water
from 273
...
15 K
...
A chemical reaction raised the temperature of 200
...
17 J/g·◦ C) by 1
...
(a) Calculate the quantity of heat
added to the solution
...

14
...
3◦ C in temperature when 1
...
Calculate the enthalpy change for the reaction of barium oxide with
carbon dioxide to yield barium carbonate
...
Calculate the enthalpy change for the reaction of 175 g of barium
oxide with carbon dioxide to yield barium carbonate
...
Calculate the enthalpy of combustion of carbon monoxide from the
enthalpies of formation of carbon monoxide and carbon dioxide
...
Calculate the enthalpy of the following reaction given the enthalpies
of combustion in equations 1 to 3 below:
C2 H2 (g) + 2 H2 (g) → C2 H6 (g)
(1) C2 H2 (g) + 2
...
5 O2 (g) → 2 CO2 (g) + 3 H2 O(l)

H = −1305 kJ
H = −286 kJ
H = −1560 kJ

19
...
9 g
of oxygen to yield carbon dioxide and water
...
Calculate the ﬁnal temperature of the water if 100
...
0◦ C and 100
...
0◦ C are mixed
...
The enthalpy of neutralization of a strong acid with a strong base is
−55
...
If 100
...
00 M NaOH and
100
...
00 M HCl, both at 25
...
18 J/g·◦ C and that the
density of that solution is 1
...
(a) Write a balanced chemical
equation for the reaction
...
(c) Determine the quantity of heat that
the reaction will release
...

( f ) Calculate the change in temperature of the solution
...

123

22
...
2 kJ/mol of water formed
...
0 mL of 1
...
0◦ C and 100
...
00 M HCl at 23
...
Assume that the heat capacity
of each initial solution and of the ﬁnal solution is 4
...
02 g/mL
...
Calculate the enthalpy change of the metal in Example 9
...
(a) 308 K
(b) 288 K
(c) 20◦ C = 20 K difference in
temperature
...
q = mct = (42
...
442 J/g·◦ C)(5
...
4 J
3
...
9 g)(2
...
5 kJ
4
...
184 J/g·◦ C)(t) = 935 J
t = 2
...
00◦ C higher than the initial temperature,
or
19
...
00◦ C = 21
...
q = 0 = m water c water twater + m metal c metal tmetal
0 = (338 g)(4
...
5◦ C) +
(127 g)(0
...
4◦ C)
t = 53
...
6 kJ
= 311 kJ
18
...
311 kJ + 533 J = 311 kJ + 0
...
q = 0 = m water c water twater + m metal c metal tmetal
6
...
4 g)(4
...
9◦ C − 19
...
5 g)(c metal )(21
...
1◦ C)
0 = (77
...
184 J)(2
...
5 g)(c metal )(−40
...
1 J/g·◦ C
124

9
...
0 g)(4
...
0◦ C) = −3220 J

To calculate value of q to freeze the water takes a heat of
solidiﬁcation calculation:

35
...
0 g

−6
...
7 kJ

To calculate the value of q for cooling the ice takes a speciﬁc heat
calculation:
q = mct = (35
...
089 J/g·◦ C)(−10
...
22 kJ + (−11
...
731 kJ )
= −15
...
(a) P(s) + 2 Cl2 (g) → PCl3 (l)
(b) H2 O(s) → H2 O(l)
(c) CH4 (g) + 2 O2 (g) → CO2 (g) + 2 H2 O(l)
(d ) CO2 (s) → CO2 (g)
11
...
7 kJ) − 72 (0 kJ) = −1560 kJ

Alternatively,
2 C2 H6 (g) + 7 O2 (g) → 4 CO2 (g) + 6 H2 O(l)
H = H f (products) − H f (reactants)
= 4H f (CO2 ) + 6H f (H2 O) − 2H f (C2 H6 ) − 7H f (O2 )
= 4(−393 kJ) + 6(−286 kJ) − 2(−83
...
Either method gives the same
result
...
0 g)(4
...
15 K − 273
...
)
13
...
0 g)(4
...
43◦ C) = 1190 J
(b) q = −1190 J
(The heat gained by the water, and therefore
positive, was released by the chemical reaction, a loss of heat, with
q having a negative value
...
Again, watch the units!

12
...
184 J/g·◦ C)(10
...
8 g

15
...
The enthalpy change for 1 mol of BaO was calculated in the prior
problem
...
CO(g) +

1
2

O2 (g) → CO2 (g)

H = H f (products) − H f (reactants)
= (−393 kJ) − (−110 kJ) − (0 kJ) = −283 kJ

18
...
We need 2 mol of hydrogen on
the left, so we double equation (2) and its H value
...
We then add the three resulting H values to get
(−1305 kJ) + 2(−286 kJ) + (+1560 kJ) = −317 kJ

(We could manipulate the equations and cancel the species that
appear on both sides to reassure ourselves that the required
equation is actually obtained
...
C2 H6 (g) + 3
...
7 kJ) − 3
...
9 g of O2 :

46
...
0 g O2

−1560 kJ
3
...
0 = m hot c hot thot + m cold c cold tcold
Since the masses are the same and the speciﬁc heats are the same,
the magnitudes of the changes in temperatures must be the same:
thot = −tcold
...
0◦ C
...
(a) HCl(aq) + NaOH(aq) → NaCl(aq) + H2 O(l)
(b) There are present 0
...
100 mol of HCl,
so 0
...

(c) (0
...
2 kJ/mol) = −5
...
52 kJ
...
0 mL)(1
...
18 J/g·◦ C)t = 5520 J
t = 6
...
0◦ C + 6
...
5◦ C
22
...
0◦ C
...

23
...
0 g)(0
...
3◦ C − 71
...
(2) A chemical reaction can produce an electric
current
...
Two entirely different sorts of calculations are generally used for the two kinds of cells
...
)
To calculate the quantities of electricity and electrical energy interacting with matter, we must learn the electrical units
involved:
1
...
The charge on
1
...
60 × 10−19 C
...
The ampere, A, is the unit of electric current
...

3
...
The volt, V, is the unit of potential, which is
the “driving force” that causes charge to ﬂow: 1 J = (1 V)(1 C) =
(1 V)(1 A)(1 s)
...
1 Electrolysis Cells
In the middle of the nineteenth century, Michael Faraday discovered
that in the electrolytic reduction of metals from their compounds,
the mass of metal produced was directly proportional to the electric
charge that passed, directly proportional to the atomic mass of the
metal, and inversely proportional to the oxidation state of the metal
128

in the compound
...
(2) the greater the atomic mass of M, the greater was the
mass of metal produced
...

These generalities are known as Faraday’s laws
...
All that we need to
be able to do is to write net ionic equations for the reactions and
remember the values for the constants listed above
...

Calculate the mass of copper metal produced from
CuSO4 solution by passage of a 3
...

EXAMPLE 1

Solution

7250 s

3
...
00 A

Cu2+ (aq) + 2 e− → Cu(s)

1 mol e−
1 mol Cu
63
...
16 g Cu
96,500 C
2 mol e−
1 mol Cu
from the
deﬁnition

from the
equation

atomic
mass

EXAMPLE 2 Calculate the time required to deposit 7
...
00 A
...
00 g Ag
108 g Ag

1 mol e−
1 mol Ag

96,500 C
1 mol e−

1s
4
...
1 minutes)

If two quantities are given and the current is to be calculated,
we convert the quantities given to coulombs and seconds and divide
them to get the answer
...
0 g of gold
from AuCl3 solution in 7
...

EXAMPLE 3

129

Solution

Au3+ (aq) + 3 e− → Au(s)

1 mol Au
3 mol e−
96,500 C
= 58,800 C
40
...
00 hours
= 25,200 s
1 hour

The current is (58,800 C)/(25,200 s) = 2
...

EXAMPLE 4 Calculate the time required with a current of 10
...
0 g of PbO2 to PbSO4 in the lead storage cell (found in
most automobile batteries)
...
0 g PbO2

1 mol PbO2
239 g PbO2

2 mol e−
1 mol PbO2

96,500 C
×
−
1 mol e

1s
= 3850 s
10
...
The number of moles
of electrons involved is clearly given in the balanced equation no
matter what the ﬁnal product
...

9
...
These combinations are called either
130

Light
bulb

e⫺
e⫺

e⫺

NO3⫺

K⫹

Zn
SO

2⫺
4

Zn2⫹

Salt
bridge

Cu
2⫺

SO4

Cu2⫹

Fig
...

half-cells or, somewhat ambiguously, electrodes
...
9-1)
...

The quantitative measure of the tendency of a metal to lose
electrons is the potential of that electrode
...
However, there cannot be a loss of electrons by any
species unless there is a simultaneous gain of electrons by another
...

Therefore, by convention, we set the potential of a standard hydrogen/hydrogen ion electrode at 0
...
This standard hydrogen electrode consists of hydrogen gas at 1
...
000 M
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 potentials, the reactants in a higher equation react spontaneously with
the products in any equation below
...
Also, Fe3+ reacts spontaneously with Cu when the
species are at unit activity
...
(b) State in which direction the
spontaneous reaction occurs
...
51 V
0
...
80 V

We must multiply this equation by 5 to get the same number of
electrons as are present in the reduction half-reaction
...
80 V

All that is left to do is to add these equations, and add the corresponding potentials:
5 Ag(s) + MnO4 − (aq) + 8 H+ (aq) →
Mn2+ (aq) + 4 H2 O(l) + 5 Ag+ (aq)

0
...

The Nernst Equation
The actual potential of a cell () depends not only on the standard
potential ( o ) but also on the concentrations of the reactants and
products in solution or their pressures in the gas phase (the things
that can vary)
...
The pH meter is based on this principle
...
The Nernst equation for the general reaction
a A + b B → c C + d D + n e−
at 25◦ C is
= o −

134

[C]c [D]d
0
...
The
value 0
...
The value of n for a complete reaction
is the number of moles of electrons in either half-reaction, since
the same electrons are involved in both
...
The number of moles of electrons can be calculated by
determining the total change in oxidation number for either the
oxidizing agent or the reducing agent
...
Each concentration is
raised to the power equal to its coefﬁcient in the balanced chemical
equation
...

EXAMPLE 7 Calculate the value of for the oxidation of iron(II) to
iron(III) in a solution of 0
...
500 M Fe3+ according
to each of the following equations:

(a) Fe2+ → Fe3+ + e−
(b) 5 Fe2+ → 5 Fe3+ + 5 e−
Solution

(a)

= o −

0
...
771 − 0
...
500)
= −0
...
0414
(0
...
812 V

135

(b)

= o −

[Fe3+ ]5
0
...
771 −

(0
...
0592
log
5
(0
...
0592
log 55
5
= −0
...
0414 = −0
...
771 −

As is apparent, there is no difference in the ﬁnal potential
...
) is an in
tensive property
...
3 Relationship of Potential and Electrolysis
The tendency of an oxidation or reduction half-reaction to proceed
can tell us what products to expect in an electrolysis reaction
...
Electrolysis of a concentrated solution of sodium chloride produces chlorine
along with the hydrogen, whereas electrolysis of a dilute solution
of sodium chloride produces oxygen and hydrogen
...
Do not forget that aqueous solutions
contain water as well as any solutes
...
0 g of sodium chloride dissolved in water by passage of
1
...
0 minutes
...
Hydrogen will be produced instead
...
What is the difference between and o ?
2
...

3
...
is the actual potential, o is the potential if all the reagents are 1 M
(for solutes) or 1 atm (for gases)
...
03 × 1023 electrons/mol
2
...
60 × 10−19 C/electron
(Avogadro’s number, within rounding error)
3
...
Standard solutions are always 1 M in solute no matter what the
coefﬁcients are in the balanced chemical equation
...
Calculate the mass of copper metal produced from CuCl4 2−
solution by passage of a 2
...
55 hours
...
Calculate the current required to deposit 145 g of gold from
AuCl4 − solution in (a) 225
...
(b) 3 hours and 45
...

3
...
Does the spontaneous reaction produce or use
up iron(II)?
4
...
303(RT/F ), where 2
...

5
...
Calculate the standard cell potential of a cell composed of the zinc
and silver half-cells
...
Is hydrogen more easily produced from 1 M hydrogen ion solution,
from 1 M hydroxide ion solution, or from pure water?
8
...
00 V when connected to a standard copper electrode
...
Calculate the concentration of hydrogen ions in a solution in
contact with hydrogen gas at 1
...
040 V
...
Two electrolysis cells are connected in series to the same source of
electricity
...
) If 10
...
Calculate the time required to reduce 17
...
45 A
...
A 1% change in concentration of which of the species in the
following half-reaction will make the greatest change in potential?
Cr2 O7 2− + 14 H+ + 6 e− → 2 Cr3+ + 7 H2 O

Solutions to Supplementary Problems
1
...
75 C
1 mol e−
1 mol Cu
3
...
5 g Cu
= 11
...
(a) AuCl4 − (aq) + 3 e− → Au(s) + 4 Cl− (aq)

3 mol e−
96,500 C
1 mol Au
= 213,000 C
145 g Au
−
197 g Au
1 mol Au
1 mol e

60 s
= 13,500 s
225
...
8 A
13,500 s

(b) Since 3 hours is 180 minutes, 3 hours and 45
...
0 minutes, so this problem is exactly the same as that in
part (a)
...
771 V
3
...
44 V
Fe2+ (aq) + 2 e− → Fe(s)
Reversing the second of these equations yields
Fe(s) → Fe2+ (aq) + 2 e−

+0
...
21 V

Since the potential is positive, the reaction goes spontaneously as
written; iron(II) is produced
...
2
...
303)(8
...
0591 V
(The constant in the Nernst equation is essentially equal to this
ratio
...
Cu+ is not stable in aqueous solution
...

0
...
Ag+ (aq) + e− → Ag(s)
2+
−
− 0
...
80 V

Zn(s) → Zn2+ (aq) + 2 e−

+0
...
56 V

7
...

8
...
80 V − 0
...
46 V
= o −

[Cu2+ ]
0
...
00 = 0
...
0592
log(1
...
5
[Ag+ ]2 = 3 × 10−16
+

[Ag ] = 2 × 10

−8

(Watch the signs!)

M

9
...
0592 log

P H0
...
040 = 0
...
0592 log{1
...
000/[H+ ]} =

−0
...
68
0
...
68

(Watch the signs!)

[H+ ] = 4
...
Cu2+ (aq) + 2 e− → Cu(s)
Ag+ (aq) + e− → Ag(s)

Since the same number of moles of electrons pass through each
one,

1 mol Cu
108 g Ag
2 mol e−
1 mol Ag
10
...
5 g Cu
1 mol Cu
1 mol Ag
1 mol e−

11
...
0 g Sn2+

1 mol Sn
119 g Sn

2+

2+

2 mol e

−

1 mol Sn

2+

96,500 C
1 mol e−

= 34
...
45 C
= 6200 s

12
...

140

Chapter 10

Equilibrium

Before studying this chapter, review Section 5
...
2, Acid-Base Equilibrium, review net ionic equations from the textbook
...
1 Equilibrium Constant Calculations
Chemical equilibrium is a state in which two exactly opposite
chemical reactions occur in the same system at the same rate
...

Equilibrium constant calculations are rather straightforward,
but we must keep several things in mind
...
(2)
The equilibrium constant expression has no addition or subtraction
within it
...
(4) Pure solids and liquids, and the
solvents for dilute solutions, do not appear in the equilibrium constant expression
...

The equilibrium constant expression for a general equilibrium reaction
a A+bB
c C+d D
is
K =

[C]c [D]d
[A]a [B]b

The square brackets mean “concentration of,” so [A] means the concentration of A, and so on
...
Equilibrium constant problems are ordinarily done without expressly writing the units; only occasionally is it
necessary to specify the units
...
)
[O2 ]
[HCOOCH3 ][H2 O]
(Water is included because the
(c) K =
[CH3 OH][HCOOH]
reaction is not in aqueous solution
...

(a) Write equilibrium constant expressions for equations 1 to 3 below
...

EXAMPLE 2

Z
1
...
Z
3
...
In multiplying the values of K 1 times K 2 , the [Z] term
142

cancels out and the product is equal to K 3
...

Calculation of Values of K
EXAMPLE 3

Calculate the value of the equilibrium constant for the

reaction
A+2B
C
if the concentrations at equilibrium are [A] = 2
...
5 M,
and [C] = 0
...

Solution The equilibrium concentrations are given, so all we
need do is write the equilibrium constant expression and substitute
these values into it:
[C]
(0
...
0022

[A][B]2
(2
...
5)2
EXAMPLE 4

(a) Using the data of Example 3, calculate the value of the equilibrium constant for the reaction
C
A+2B
(b) What is the relationship between the value of K in Example 3
and the value of this K ?
Solution

(a) Again, the equilibrium concentrations are given, so all we need
do is write the equilibrium constant expression and substitute
the values into it:
(2
...
5)2
[A][B]2
=
= 4
...
010)
(b) The two equilibrium constants are reciprocals of each other
...

Much more often, initial concentrations for the reactants are given
along with the equilibrium concentration of one substance, so we
must calculate the equilibrium concentrations for all the other substances from the data given
...
20 mol of A and 1
...
00 L of solution,
whereupon 0
...

Solution Because all the substances are dissolved in the same
solution, they all have the same volume, and therefore their numbers of moles are proportional to their molarities
...

First we write the balanced chemical equation and put under it rows
for “Initial concentrations,” “Change due to reaction,” and “Equilibrium concentrations
...
We assume that there is no C or D present initially,
since none was mentioned
...
20

+

2B
1
...
000

+

2D
0
...
100

It is obvious that the 0
...

As soon as we place that 0
...
The ratio of coefﬁcients in this problem is
1 : 2 : 1 : 2, so the ratio of changes of concentrations in this problem
is 0
...
200 : 0
...
200
...
20
1
...
000
0
...
100
0
...
100
0
...
100
The equilibrium concentration of each reactant is its initial concentration minus the change, since the reactants are used up by the
reaction
...
We now have all the equilibrium concentrations:
144

+ 2D
A
+ 2B
C
Initial
concentration (M) 1
...
70
0
...
000
Change due to
reaction (M)
0
...
200
0
...
200
Equilibrium
concentration (M) 1
...
50
0
...
200
We put the equilibrium concentrations into the equilibrium constant expression and solve:
K =

[C][D]2
(0
...
200)2
=
= 0
...
10)(1
...
Note also that we got those
concentrations from the stoichiometry of the problem, and we did
not double the B or D concentration
...
Please also note that the “Change due to reaction” row
in the ﬁnal table is the only row that is in the ratio of the balanced
chemical equation
...
20 mol of Z and 1
...
00-L solution and
allowed to come to equilibrium, at which point the concentration
of E is 0
...

Solution Note speciﬁcally that the volume is 2
...
600 M and 0
...
The equilibrium concentration of E was given in the problem,
so we don’t have to calculate that using the volume of the solution
...

2Z
+
Initial concentration (M)
0
...
400
Equilibrium concentration (M) 0
...
850
0
...
650

[E]2
(0
...
15
2
[Z] [Q]
(0
...
650)

2E
0
...
400
0
...
We let one of the
equilibrium concentrations be represented by an unknown variable,
say x, and solve the equilibrium constant expression algebraically
...
0 × 10−4 , calculate
the equilibrium concentration of Z if 0
...
500 mol
of B are dissolved in 1
...

EXAMPLE 7

Solution

2Z
A
+ B

Initial concentration
0
...
500
0
...
500−x
0
...
0 × 10−4
[A][B]
(0
...
020
(0
...
500 − x)(0
...
010 − 0
...
020x = 0
...
0050
2x = [Z] = 0
...
0050)]2
=
= 4
...
500 − 0
...
1 × 10−4 (to two signiﬁcant digits), so the process works
...
(It happens to be twice something else, but that
does not matter
...
(We did
that so that the change in concentration of A and B would not be
146

fractional
...
) Also note that a check is a very good
idea to see that our calculation is correct
...

However, we are not always so lucky to have an equation that we
can take the square root of
...
We will neglect very small quantities when added to or
subtracted from larger quantities, but not when the small quantity
is multiplied or divided
...
An error less than 5%
(or even somewhat more) is generally acceptable
...
00 × 10−8
...
50 mol of A and 2
...
00 L of
solution and allowed to come to equilibrium
...
50
x
1
...
50
2x
2
...
00
x
x

[C]
x
=
= 1
...
50 − x)(2
...
That is, x, and
even 2x, should be negligible compared to 1
...
50, so we will
neglect them and get
K =

[C]
x
=
= 1
...
50)(2
...
38 × 10−8 M = [C]
[A] = 1
...
38 × 10−8 = 1
...
50 − 2(9
...
50 M
147

The value of x (and even 2x) is certainly negligible when subtracted
from the larger quantities
...
38 × 10−8
=
2
[A][B]
[1
...
38 × 10−8 ][2
...
38 × 10−8 )]2
= 1
...

10
...

Weak acids and weak bases undergo equilibrium reactions with
water
...
(It is a hydrogen ion
attached to a water molecule, and is more real than a free hydrogen
ion in aqueous solution, because the hydrogen ion has no electrons
...
Use whichever form the textbook uses
...
Most acids are weak;
the only common strong acids are HCl, HBr, HI, HClO3 , HClO4 ,
HNO3 , and H2 SO4
...
Constants are not given for strong
acids because they can be regarded as 100% ionized
...
That is not to say that
the reverse reactions cannot proceed; indeed they proceed to a far
greater extent than the forward reaction, but the tabulated values
are for the un-ionized molecules reacting to form ions
...
The expressions for
148

the two forms of the equations for the ionization of acetic acid are
Ka =

[C2 H3 O2 − ][H3 O+ ]
[HC2 H3 O2 ]

or

Ka =

[C2 H3 O2 − ][H+ ]
[HC2 H3 O2 ]

Note that neither expression has the concentration of water in it
...
6 M, and in a dilute solution
that might drop to 55
...
Thus the concentration of water might
be thought to be a constant that has been incorporated into the
equilibrium constant
...

A weak base like ammonia in water also is in equilibrium to a
slight extent:
−
NH3 (aq) + H2 O(l)
NH4 + (aq) + OH (aq)

Its equilibrium constant, the base dissociation constant, is symbolized K b , with the subscript denoting base
...
(It turns out that
these equilibria are inherently easier because there are few coefﬁcients different from 1 in the balanced equations
...
100 M acetic acid solution is 1
...

Solution

HC2 H3 O2 (aq)
C2 H3 O2 − (aq) + H+ (aq)
Initial
concentrations (M)
Change due
to reaction (M)
Equilibrium
concentrations (M)

Ka =

0
...
000

0
...
35 × 10−3

1
...
35 × 10−3

0
...
35 × 10−3

1
...
35 × 10−3 )2
[C2 H3 O2 − ][H+ ]
=
= 1
...
099

The accepted value of K a for acetic acid is 1
...

The equilibria of weak bases can be handled in the same way
...
100 M NH3 has a hydroxide ion
concentration of 1
...

Solution
H2 O(l) + NH3 (aq)
Initial
concentrations (M)
Change due
to reaction (M)
Equilibrium
concentrations (M)

Kb =

NH4 + (aq)

+ OH− (aq)

0
...
000

0
...
35 × 10−3

1
...
35 × 10−3

0
...
35 × 10−3

1
...
35 × 10−3 )2
[NH4 + ][OH− ]
=
= 1
...
099

Just by coincidence, acetic acid and ammonia have the same values

for their dissociation constants, 1
...

Calculate the hydroxide ion concentration of a
0
...
8 × 10−5
...
350

0
...
000

x

x

x

0
...
8 × 10−5
[NH3 ]
0
...
8 × 10−5
0
...
3 × 10−6

and

x = 2
...

150

What would LeChatelier’s principle predict about adding
0
...
The ammonium ions ought
to shift the equilibrium of the ammonia ionization to the left,
decreasing the hydroxide ion concentration
...
350 M solution of ammonia that also has 0
...
For NH3 , K b = 1
...

Solution The equilibrium that we are considering is still the
ionization of the ammonia
...

H2 O(l) + NH3 (aq)
NH4 + (aq) + OH− (aq)
Initial
concentrations (M)
Change due
to reaction (M)
Equilibrium
concentrations (M)

0
...
200

0
...
350 − x

0
...
200)
[NH4 + ][OH− ]
=
= 1
...
350

x = [OH− ] = 3
...
Just as predicted by
LeChatelier’s principle, the hydroxide ion concentration dropped from 1
...
2 × 10−5 M by addition of the
ammonium chloride
...
The cation of the salt repressed the ionization of the base
...
3 Water Autoionization
The Bronsted theory designates water as an acid in its reactions
with bases and as a base in its reactions with acids
...
Its equilibrium constant expression, with its value, is
K w = [H3 O+ ][OH− ] = 1
...
This equation means
that in every dilute aqueous solution, there is at least some hydronium ion and some hydroxide ion
...
If the hydroxide ion concentration is greater than the hydrogen ion concentration, the solution is basic
...

EXAMPLE 13

Calculate the hydronium ion concentration in pure

water
...
Let each equal x:

K w = [H3 O+ ][OH− ] = x2 = 1
...
0 × 10−7 M

EXAMPLE 14 (a) Calculate the hydrogen ion concentration in
0
...
(b) Calculate the hydroxide ion concentration in
0
...

Solution (a) Since NaOH is a strong base, 0
...
100 M in OH−
...
100) = 1
...
0 × 10−13 M
(b) Since HCl is a strong acid, 0
...
100 M in H+
...
100)[OH− ] = 1
...
0 × 10−13 M
EXAMPLE 15

Calculate the hydronium ion concentration in

0
...

Solution NH3 is a weak base, so we have to calculate the hydroxide ion concentration using K b
...
350 M NH3
152

in Example 11, and found [OH− ] to be 2
...
Then,
K w = [H3 O+ ][OH− ] = [H3 O+ ](2
...
0 × 10−14
[H3 O+ ] = 4
...
100 M NaCl?
Since neither the sodium ion nor the chloride ion appears in the K w
expression, the hydrogen ion concentration and the hydroxide ion
concentration are equal, and each is equal to 1
...

10
...
Also note that the pH, since
it is a logarithm, has as many decimal place digits as the hydrogen
ion concentration has signiﬁcant digits
...
100 M H+ ; (b) 0
...
100 M NaOH
...
100) = 1
...
100 and the pH is the same
as in part (a)
...
0 × 10−13 M
...
00
...
50; (b) 2
...
70
...
(See Section 1
...
)

(a) pH = 7
...
50; [H+ ] = 3
...
6 × 10−3 ; (c) 2
...
5 Buffer Solutions
A buffer solution is a solution of an un-ionized weak acid and its
conjugate base or an un-ionized weak base and its conjugate acid
...

1
...

2
...
They can also be
prepared by carrying out a limiting quantities reaction in solution
to result in one of the same combinations of reagents
...
The buffer solution maintains a relatively constant pH by shifting
according to LeChatelier’s principle:

HC2 H3 O2 (aq) + H2 O(l)

Large
Huge
concentration
concentration

C2 H3 O2 − (aq) + H3 O+ (aq)
Large
Tiny
concentration
concentration

If H3 O+ is added from a strong acid, there is enough acetate ion
present to cause this equilibrium to shift to the left, using up
most of the added H3 O+ and keeping the pH relatively constant
...
This lowering of the H3 O+ concentration causes
the equilibrium to shift to the right, replacing H3 O+ used up and
again maintaining a relatively constant pH
...

What compounds remain in solution after 0
...
100 mol of NaOH?

EXAMPLE 18

154

Solution

HC2 H3 O2 (aq) + NaOH(aq) → NaC2 H3 O2 (aq) + H2 O(l)
Since the reagents react in a 1 : 1 ratio, the 0
...
100 mol of HC2 H3 O2 leaving 0
...
100 mol of NaC2 H3 O2
...

EXAMPLE 19 (a) Calculate the pH of 1
...
125 mol of NH3 and 0
...
(b) Recalculate the pH after 0
...

Assume no change in volume
...
125

0
...
000

x

x

x

0
...
125 + x

x

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

[NH4 + ][OH− ]
x(0
...
8 × 10−5
[NH3 ]
0
...
8 × 10−5 M
pH = 9
...
The easiest way to do a problem like this is to assume
that the acid and base react completely, and then do the equilibrium
problem with the resulting concentrations as initial concentrations
for that problem:
NH4 + (aq) + Cl− (aq)
HCl(aq) + NH3 (aq)
Beginning
concentrations (M)
0
...
125
0
...
010
0
...
010
End of acid-base reaction
concentrations (M)
0
...
115
0
...
115

0
...
000

x

x

x

0
...
135 + x

x

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

[NH4+ ][OH− ]
x(0
...
8 × 10−5
[NH3 ]
0
...
5 × 10−5 M
pH = 9
...
26 to 9
...

10
...
For example, the acetate ion reacts with water according
to the following equilibrium:
C2 H3 O2 − (aq) + H2 O(l)
HC2 H3 O2 (aq) + OH− (aq)
The equilibrium constant expression for this reaction is written as
usual for an ionic equilibrium:
Kh =

[HC2 H3 O2 ][OH− ]
[C2 H3 O2 − ]

where K h is the hydrolysis constant
...
The K h for the hydrolysis reaction is equal
to K w /K a
...
For the hydrolysis of the conjugate acid
of a weak base,
K h = K w /K b
How do we recognize a hydrolysis problem? It is a problem in
which there is no molecular weak acid or weak base present initially
(just a “salt”)
...
)
EXAMPLE 20 Calculate the hydroxide ion concentration in a solution of 0
...

Solution The sodium ion is a spectator ion
...
100

0
...
000

x

x

x

0
...
0 × 10−14
[HC2 H3 O2 ][OH− ]
=
=
= 5
...
100)
1
...
6 × 10−11
x = 7
...
100 M NH4 Cl
...
The ammonium
ion hydrolyzes according to the equation
H2 O(l) + NH4+ (aq)
NH3 (aq) + H3 O+ (aq)
Initial
concentrations (M)
Change due
to reaction (M)
Equilibrium
concentrations (M)

Ka =

0
...
000

0
...
100 − x

x

x

x2
1
...
6 × 10−10
+
[NH4 ]
(0
...
8 × 10−5

x2 = 5
...
5 × 10−6 M = [H3 O+ ]

pH = 5
...
The equilibrium constant for a certain reaction has a value of
2
...
What is the value of the constant for the reverse
reaction?
2
...
For which kind of calculation do we use an algebraic variable like
x —calculation of the value of K or calculation using the value of K ?
4
...
100 M acetic acid solution, the acid is 1
...
7% in
the un-ionized form
...
100 M
before any reaction occurs:
158

(a) C2 H3 O2− (aq) + H+ (aq)
HC2 H3 O2 (aq)
(b) NaC2 H3 O2 (aq) + HCl(aq)
HC2 H3 O2 (aq) + NaCl(aq)
5
...
The value of this K is 1/(2
...
0 × 1012
...
Whenever water is involved in a reaction but is not the solvent for
the reaction
...

3
...
(Remember, we calculated values
of K before we used x
...
(a) The ions react to form 98
...
3% in
the ionic form
...
(b) The net ionic
equation for this reaction is that given in part (a), so the answer is
the same
...
(a) The ﬁrst equation represents an equilibrium reaction of a weak
acid, that proceeds to the right only a tiny percentage
...

(b) The ﬁrst equation represents an equilibrium reaction of the
conjugate of a weak acid with water that proceeds to the right
only a tiny percentage
...
) The second
equation proceeds extensively, since its reverse is the ionization
of a weak acid in water, which we know proceeds only slightly
...
From the value of the equilibrium constant for
A+B
2C

K = 16
...
Write the equilibrium constant expression for the following
reaction:
CaCO3 (s)
CaO(s) + CO2 (g)

3
...
20 mol of A and 1
...
00 L of
solution, whereupon 0
...

4
...
220 M, [B] = 0
...
22 M
...
02 M
...
(b) Calculate the
value of the new equilibrium constant
...

5
...
100 M formic acid solution is 4
...

6
...
250 M solution of
HC2 H3 O2 that also has 0
...
K a = 1
...

7
...
Calculate the pH of a solution containing 0
...
150 M
NH4 Cl
...
8 × 10−5
...
Calculate the hydrogen ion concentration of each of the following
solutions: (a) pH = 7
...
90
...
Calculate the pH of 0
...

K a = 7
...

160

11
...
500 mol HCl + 1
...
00 L of solution
0
...
00 mol NaCl in 1
...
500 mol NaCl + 1
...
00 L of solution
0
...
00 mol NaC2 H3 O2 in 1
...
The hydroxide ion
concentration in 0
...
6 × 10−3 M
...
00 L of solution, has the highest pH:
(i) 0
...
250 mol NaCHO2
...
480 mol HCHO2 and 0
...

(iii) 0
...
290 mol NaCHO2
...
440 mol HCHO2 and 0
...

(b) How much NaOH do we need to convert solution (i) to
solution (ii)? solution (ii) to solution (iii)? solution (iii) to
solution (iv)?
(c) If that much NaOH were added to 1
...
6 × 10−5 M HCl, what would be the ﬁnal pH?
Calculate the value of K b for a base (B) if a 0
...
28
...
100 M HA
and 0
...
22
...
100 M HC2 H3 O2 and 0
...
K a = 1
...

When 0
...
150 mol of NH4 Cl are dissolved in
enough water to make 1
...
010 M concentration? (c) What other ions are present? (d ) What
is the principal equilibrium reaction? (e) What effect does each of
the ions of part (b) have on the equilibrium of part (d )? ( f ) What is
the hydroxide ion concentration of the solution? (g ) What is the
pH of the solution?
Calculate the pH of the solution that results after 0
...
Assume no
change in volume
...
00-L solutions involve the use of a
hydrolysis constant for the determination of equilibrium
concentrations?

(a)
(b)
(c)
(d )
12
...

14
...

16
...

18
...

161

(a) 0
...
100 mol NH4 Cl
(c) 0
...
100 mol NaOH + 0
...
100 mol HCl + 0
...
Calculate the hydroxide ion concentration of a solution prepared by
adding 0
...
200 M NaOH to 0
...
200 M
HC2 H3 O2
...
Calculate the pH of a solution prepared by adding 0
...
200 M HCl to 0
...
200 M NH3
...
For the given reaction, the equilibrium constant expression is
[C]2
= 16
...
00
...
0) = 0
...

1

(c)

K =

1

[A] 2 [B] 2
[C]

which is the square root of the value in part (b), or 0
...
(Note
that the square root of a number smaller than 1 is larger than
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
(Chapter 6); be sure to use the correct unit with each one
...

11
...
If the solute is nonvolatile (nonevaporating), the solution
has a lower vapor pressure than the pure solvent does
...
) From a molecular view, the solute
particles at the surface of the liquid inhibit the movement of solvent molecules from going into the vapor phase, but do not inhibit
solvent molecules in the vapor phase from returning to the liquid
phase, so the rate of evaporation is lower than the rate of condensation until there are fewer solvent molecules in the vapor phase
...
Raoult’s law is approximate for many solutions, and is exact only for “ideal solutions
...

169

Solvents generally are volatile (evaporate easily) whereas solutes may
be volatile or nonvolatile
...

(a) Calculate the vapor pressure of benzene in a solution of naphthalene (a nonvolatile solute) in benzene at 21
...
900
...
7 kPa
...
(c) What is the vapor pressure of the solution?

EXAMPLE 1

Solution

(a) Psolvent = Xsolvent P osolvent = (0
...
7 kPa) = 9
...
7 kPa − 9
...
1 kPa
(c) 9
...

(Since the solute is nonvolatile, it has no
vapor pressure
...
These equations look
very similar; do not confuse them
...
The new equation can give a
more precise answer than the ﬁrst
...
200 mol of naphthalene in 1
...
3◦ C
...
7 kPa
...
200 mol
= 0
...
00 mol total

o
Pbenzene = Xnaphthalene P benzene

= (0
...
7 kPa) = 1
...

170

11
...
The freezing-point depression is directly proportional to
the molality of the solute particles:
t f = k f m
where t f is the freezing-point depression, k f is the freezing-pointdepression constant, and m is the molality of the solution
...
It is customary, however, to report both as positive
values
...
A set
of freezing-point-depression constants along with freezing points is
given in Table 11-1
...
100 m
solution of naphthalene in benzene
...

EXAMPLE 3

Solution The freezing point of benzene (the solvent) and its
freezing-point constant are taken from Table 11-1
...
12◦ C/m)(0
...
512◦ C
(b) t = 5
...
512◦ C = 5
...

EXAMPLE 4 Which of the following has (a) the greatest freezingpoint depression? (b) the highest freezing point?

1 m aqueous CH3 OH, 2 m aqueous CH3 OH,
or 3 m aqueous CH3 OH
Table 11-1

Freezing-Point Data

Solvent

Freezing Point (◦ C)

k f (◦ C/m)

Benzene
Bromoform
Cyclohexane
Naphthalene
Water

5
...
8
6
...
2
0
...
12
14
...
0
6
...
86
171

Solution (a) 3 m CH3 OH
...
(b) 1 m CH3 OH
...

The boiling point of a substance is the temperature at which
its vapor pressure is equal to that of the surroundings
...

This boiling-point elevation, like the freezing-point depression, is
directly proportional to the molality of the solute particles:
tb = kb m
where tb is the boiling-point elevation, kb is the boiling-pointelevation constant, and m is the molality
...

Determine the boiling point of a 0
...
00 atm pressure
...
512◦ C/m)(0
...
0512◦ C
t = 100
...
0512◦ C = 100
...

Note that here the change in temperature was added to the normal
boiling point (it is an elevation) whereas the change was subtracted
in Example 3 (where a depression was observed)
...
7◦ C
...
12
155
...
88
218
100
...
53
6
...
79
5
...
512

172

Solution Using data from Table 11-1, we calculate that the
freezing-point depression is

80
...
7◦ C = 2
...
9◦ C/m)(m) = 2
...
36 m

EXAMPLE 7 Calculate the value of the boiling-point-elevation constant for a solvent that boils at 53
...
359 m
solution boils at 55
...

Solution

tb = 55
...
8◦ C = 2
...
359 m) = 2
...
6◦ C/m

Since these equations can get us molalities, we can use them
with other data to determine molar masses
...
50 g
of the solute in 55
...
3◦ C
...
5◦ C − 4
...
2◦ C
...
12◦ C/m)(m) = 1
...
23 m
That means that there is 0
...
00 kg of benzene
...
00 kg of benzene:

1
...
2 g solute
=
55
...
23 mol is 27
...
2 g)/(0
...
2 × 102 g/mol

11
...
11-1), an osmotic pressure
173

Semipermeable membrane
Pure solvent

h

Solution

Fig
...

develops
...
The solute particles partially block passage of solvent molecules from the solution
into the pure solvent, but not the other way
...
An
external piston can be used to keep the levels the same and prevent
dilution of the solution (Fig
...
This added pressure is the osmotic pressure of the solution
...

Calculate the osmotic pressure of a 0
...

EXAMPLE 9

Solution

πV = nRT
Dividing both sides of this equation by V yields an equation
with molarity as one of the variables:
n
π=
(RT) = (0
...
31 L·kPa/mol·K)(298 K) = 248 kPa
V
Pressure
Piston

Pure solvent

Solution

Fig
...

174

This is a very large pressure indeed! Compared to other colligative
properties, osmotic pressure measurements allow very precise determinations of molar masses
...
00 g in
100
...
77 kPa
...
77 kPa
=
=M=
= 0
...
31 L·kPa/mol·K)(298 K)
There is 0
...
0 g per liter:

2
...
0 g
=
100
...
0 g
= 8580 g/mol = 8
...
00233 mol
It would be impossible to determine the molar mass of this solute
with a freezing-point depression experiment
...
)

11
...

Thus a 1 m aqueous solution of NaCl contains 1 mol of Na+ ions
and 1 mol of Cl− ions per kilogram of solvent
...
In very dilute solutions, the colligative properties of
solutions of ionic solutes are a multiple of the analogous properties
of nonionic solutes
...
of Particles
Formula Unit

CH3 OH
NaCl
MgCl2
AlCl3
MgSO4

1 molecule/formula unit
2 ions/formula unit
3 ions/formula unit
4 ions/formula unit
2 ions/formula unit

1
2
3
4
2
175

EXAMPLE 11 Which of the following 0
...
00100 m AlCl3 , which has four ions per formula unit, has the greatest freezing-point depression
...
00400 m
in particles
...
00100 m NaCl has about twice the colligative
properties as a solution of 0
...
However, as
the concentration of the ionic solution increases, the interionic attractions increase markedly, and the particles become less independent of each other
...
Thus NaCl might have 1
...

Leading Questions
1
...
State the difference between each pair:
(a) Vapor pressure and vapor-pressure lowering
(b) Freezing point and freezing-point depression
(c) Boiling point and boiling-point elevation
3
...

P solvent = Xsolvent P osolvent
P solvent = Xsolute P osolvent

2
...

176

3
...
(This
fact might be useful to remind us which phase change is lowered and
which is raised
...
(a) Calculate the vapor pressure of ethyl alcohol containing glucose
(a sugar) at 25◦ C in which the mole fraction of glucose is 0
...

The vapor pressure of pure ethyl alcohol at 25◦ C is 59
...

(b) Calculate the vapor-pressure lowering
...
(a) Calculate the freezing-point depression of a 0
...
(b) Calculate the freezing point of
the solution
...
Calculate the molality of a solution of a nonionic solute if its
solution in cyclohexane freezes at 3
...

4
...
225 mol of
sucrose (sugar) in 1550 mL of aqueous solution at 25◦ C
...
Determine the molar mass of a solute if 1
...
145 atm
...
Calculate the freezing point of an aqueous solution of a nonvolatile
solute with a boiling point of 101
...
000 atm pressure
...
Calculate the molar mass of a nonionic solute if 4
...
7◦ C
...
Calculate the molar mass of a nonvolatile solute if a solution of
1
...
0 g of cyclohexane, C6 H12 , has a vapor
pressure of 0
...
7◦ C
...
106 atm
...
Calculate the freezing point of a solution of glucose (a simple sugar)
in water with mole fraction of glucose equal to 0
...

10
...
00 m
aqueous sucrose (table sugar) at 25◦ C
...
20 kPa
...
Calculate the freezing point of a solution of sucrose (C12 H22 O11 ) in
water if the density of the solution is 1
...
5 kPa
...
Calculate the vapor pressure of a solution of sucrose (table sugar)
in 1550 mL of aqueous solution at 25◦ C if its freezing point could
be measured to be 0
...
The vapor pressure of pure water at
that temperature is 24
...
(Assume that the density of the
solution is 1
...
)
177

13
...
00200 mol of
NaCl in 345 mL of aqueous solution at 25◦ C
...
Calculate the osmotic pressure of a solution of 0
...
550 L of aqueous solution at 25◦ C
...
Calculate the freezing point of a solution of 0
...
500 kg of water at 25◦ C
...
A solution of 4
...
7◦ C
...
0% carbon, 6
...
3% oxygen
...
A solution of 4
...
7◦ C
...
0% carbon, 6
...
3% oxygen
...
(b) Calculate
the molar mass of the sample
...
(d ) Calculate the molecular formula of the sample
...
Calculate the molecular formula of 5
...
1◦ C
...
0%
carbon, 6
...
3% oxygen
...
Calculate the freezing-point depression of a solution containing
2
...
0 g (a) of water;
(b) of bromoform (Table 11-1)
...
Calculate the molality of 0
...

(Assume that the density of the solution is 1
...
)

Solutions to Supplementary Problems
1
...
0000 − 0
...
9500
...
9500)(59
...
1 torr

(b) 59
...
1 torr = 2
...
0500(59
...
95 torr
2
...
86◦ C/m)(0
...
279◦ 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 ◦ = − ◦ n F = −(0
...
31 J/K)(298 K) ln K
ln K = 10
...
(a) 6 (equal to the total change in oxidation number of M or Au)
[Cl− ]8 [M2+ ]3
0
...
250 = ◦ −

(1
...
500)3
0
...
110)2

= ◦ − (0
...
01)
◦ = −0
...
240 V)(6 mol e− )(96,500 C/mol e− ) = 139 kJ

(d)

G ◦ = −RT ln K
ln K = −G ◦ /RT
= − (139,000 J)/(8
...
1
K = 4 × 10−25

13
...

RT
14
...

196

G ◦ = −817
...

G ◦ = −RT ln K
ln K = −

(−817,900 J)
= 330
(8
...
303) = 143
K = 10143

This huge value of K corresponds to the very large magnitude of G ◦
...
G ◦ = −RT ln K = − ◦ n F
ln K =

(−0
...
0779
RT
(8
...
925

17
...
2 kJ) + 4(−237
...
8 kJ) − 3(0 kJ)
= −1121
...
31 J/K)(298 K)

log K = (ln K )/(ln 10) = 453/2
...
5 O2 (g) → CO(g) + 2 H2 O(l)
◦

G = G f◦ (products) − G f◦ (reactants)
= G f◦ (CO) + 2G f◦ (H2 O) − G f◦ (CH4 ) + 1
...
2 kJ) + 2(−237
...
8 kJ) − 1
...
7 kJ
197

G ◦ = −R T ln K
ln K =

560,700 J
= 226
(8
...
303 = 98
...
Within rounding error, the answer of part (b) is the square root of
that in part (a), as predictable from the equilibrium constant
expressions:
(a)

198

K =

[CO]2
[CH4 ]2 [O2 ]3

(b) K =

[CO]
[CH4 ][O2 ]1
...

1
...
What are the units of (a) molar
mass? (b) density? (c) molarity? (d) speciﬁc heat? (e) mole fraction? ( f ) vapor pressure of water? (g) entropy? (h) freezing-point
lowering constant? (i) R, the ideal gas law constant?
2
...

(a) Density of a gas at a given temperature and pressure
(b) Molarity of a solution and its volume
(c) Mass of each reactant in a reaction and mass of all but one
product
(d) Density of a solution and its mass
(e) Mass and volume of a sample
(f) Formula of a compound and the atomic masses of its elements
(g) Percent composition of a compound and the atomic masses
of its elements
(h) Empirical formula of a compound and its molar mass
(i) Concentration and volume of one reactant in a titration and
volume of the other
(j) Concentration and volume of both reactants in a reaction
3
...
Calculate the density of ammonia gas at 373 K and 1
...

5
...
76 g
of a solute (MM = 133 g/mol) in 50
...
0◦ C/m, f
...
= 6
...

6
...
30 g neutralizes
25
...
000 M NaOH
...
(a) Calculate the number of grams of magnesium in 14
...
(b) Calculate the percentage of magnesium in
Mg(ClO3 )2
...
Calculate the number of liters in 2
...

9
...
00 mL of 2
...
00 mL of 2
...

10
...
0200 mol of K2 SO4
in 0
...

11
...
0 kPa barometric pressure prepared by thermal decomposition of 1
...
(PH2 O = 3
...
Calculate the volume of 12
...
00 atm
pressure
...
Calculate the mass of KClO3 that must be decomposed to produce 3
...

14
...
1-g sample of a gas that occupies 10
...

15
...
The ﬁrst emits 2
...
0 × 10−14 J/photon, and the second 1
...
0 × 10−14 J/photon
...
Calculate the mass of K2 SO4 that will be produced by the
reaction of 156 g of (aqueous) KOH and 262 mL of 3
...

17
...
00 mL of the liquid has a mass of
107
...
00 g
...
Calculate the mass of Li2 CO3 that will be produced by the reaction of 38
...
00 M LiOH and 1
...
00 atm pressure
...
Calculate the mass of copper metal produced from CuCl4 2
solution by passage of a 4
...
0 hours
...
State in another way the fact that red light consists of photons
200

21
...

23
...

25
...

27
...

29
...

31
...

33
...

35
...

having less energy than photons of violet light
...
”
Can all light be detected by the human eye? Explain
...
00 ×
10−6
...
20 mol of A and 2
...
00 L of
solution and allowed to come to equilibrium
...
6 g of CO
with oxygen to yield carbon dioxide: H f (CO) = −110 kJ/mol;
H f (CO2 ) = −393 kJ/mol
Calculate the molar mass of ammonia at 373 K and 1
...

Determine the new volume of a 1
...
0%
...

Calculate the ﬁnal temperature after 43
...
451 J/g·◦ C) at 59
...
5◦ C
...
34 V) is combined with the
iron(III)/iron(II) half-cell ( ◦ = 0
...

Calculate the number of mercury atoms in 1
...
6 g/mL)
...
7 g of liquid water at 50
...
4◦ C : cliquid = 4
...
089 J/g·◦ C; Hfusion = 335 J/g
...
44 at 25◦ C
...
100 M H+ , 0
...
11 M MnO4 − ( ◦ = 1
...

Calculate the pH of 1
...
175 mol of HC2 H3 O2 and 0
...
010 mol of HCl is added to the solution
...

A 1
...
0200 mol of a weak acid, HA
...
528 atm at 25◦ C
...

Calculate the free energy change for a reaction at 80◦ C in which
the enthalpy change is 14
...
44 J/K
...
96; (b) 13
...
55
...
00 mL of 2
...
00 mL
...
Show that the following data are in accord with the law of multiple proportions:
Compound A Compound B
Element 1 40
...
2%
Element 2 6
...
0%
Element 3 53
...
8%
38
...
0% carbon, 6
...
3% oxygen if 10
...
0 g of water freezes at −1
...

39
...
20 minutes to diffuse there under the
same conditions
...
Calculate the osmotic pressure at 25◦ C of an aqueous solution
of a nonionic solute with molar mass 80
...
11◦ C
...
11 g/mL
...
Calculate the value of K for the reaction A + B
C, in which
G is −25
...
00 M, [B] = 4
...
100 M
...
Which of the following solutions require the use of a hydrolysis constant to calculate its pH? (a) 0
...
100 mol HC2 H3 O2
...
100 mol HCl + 0
...

(c) 0
...
100 mol NH3
...
100 mol NaOH +
0
...
(e) 0
...
200 mol NaC2 H3 O2
...
100 mol HCl + 0
...

43
...
03 g of element A and 11
...
03 g of element A and
15
...
Calculate the mass of magnesium chloride produced and the
mass of excess reactant left unreacted after 92
...
1 g of hydrochloric acid (in water)
...
Calculate the mole fraction of 1
...
21 g/mL
...
After a 2
...
500 mol of solute was added
to a 1
...
95 m solution was produced
...
(Hint: Let x equal the mass of solvent in the ﬁnal
solution
...
Calculate the average kinetic energy of a gas molecule in a sample at 25◦ C
...
Calculate the root mean square speed (u) of the nitrogen
molecules in a sample of the gas at 298 K
...
Calculate the freezing point of 0
...

50
...

51
...
2 kJ and a value of G =
−16
...

Solutions to Miscellaneous Problems
1
...

2
...
Assume a volume, calculate the number of
moles of the gas (with the ideal gas law, for example), and divide
that into the number of grams in the selected volume of
gas
...
Deﬁnition of molarity
...
The law of conservation of mass
...
Deﬁnition of density
...
Deﬁnition of density
...
Multiply the numbers of atoms of each element in
a formula unit by the atomic mass, sum these values to get the
molecular mass
...

(g) The empirical formula
...

(h) The molecular formula
...

(i) The concentration of the other reactant, assuming that a
balanced chemical equation is available
...
For example, it
might be necessary to do this type of calculation to determine if
one or two hydrogen atoms in H2 SO4 have been replaced in its
203

reaction with NaOH:
H2 SO4 (aq) + NaOH(aq) → NaHSO4 (aq) + H2 O(l)
H2 SO4 (aq) + 2 NaOH(aq) → Na2 SO4 (aq) + 2 H2 O(l)

3
...

(b) Divide the mass by the number of moles
...

(c) Add the atomic masses for each atom in the empirical formula
and divide the sum into the molecular mass
...
Multiply
each subscript in the empirical formula (including the
assumed values of one) by that number to get the molecular
formula
...

(e) Assume a volume (like 1
...

4
...
00 L of gas
...
00 atm)(1
...
0327 mol
RT
(0
...
0 g
0
...
556 g
1 mol
The density of ammonia under these conditions is therefore
0
...

5
...
76 g
= 0
...
0132 mol)/(0
...
264 m
t f = k f m = (20
...
264 m) = 5
...
5◦ C − 5
...
2◦ C
204

6
...
000 mmol
25
...
9 mmol
1 mL
Since they react in a 1 : 1 mol ratio, that is also the number of
millimoles of HA
...
30 g)/(0
...
8 g/mol

7
...
0 g Mg(ClO3 )2

1 mol Mg(ClO3 )2
191 g Mg(ClO3 )2

1 mol Mg
×
1 mol Mg(ClO3 )2

24
...
78 g Mg
1 mol Mg

1
...
7%
14
...
50 × 103 L
8
...
50 m3
1 m3
9
...
110 mmol
50
...
5 mmol H3 PO4
1 mL

2
...
00 mL
= 105
...
The K2 SO4 dissociates into three moles of ions (2 mol K+ and
1 mol SO42− ) for each mole of salt
...
0600 mol)/(0
...
171 m
...
86◦ C/m)(0
...
318◦ C

The freezing point is −0
...

11
...
0 kPa − 3
...
8 kPa
205

The number of moles of oxygen gas is determined:
2 HgO(s) → 2 Hg(l) + O2 (g)

1 mol HgO
1 mol O2
1
...
00392 mol O2
216
...
00392 mol)(8
...
0909 L
P
106
...
Be careful not to use a gas law for water at 25◦ C and 1
...
The volume is found from the mass and density!
(We must not only know the rules, but when to use each one!)

1 mL
V = 12
...
5 mL
1
...
997 g/mL, but 1
...
)
13
...
00 L O2

1 mol O2 (at STP)
22
...
9 g KClO3

14
...
Then divide the mass by
that number of moles
...
7 L)
n=
=
= 0
...
31 L·kPa/mol·K)(298 K)
MM = (21
...
523 mol) = 40
...
(a) The total energy of each source is the product of the number of
photons times the energy of each:
First source: E 1 = (2
...
0 × 10−14 J/photon)
= 2
...
0 × 106 photons)(2
...
0 × 10−8 J
Both sources emit the same total energy
...

206

16
...
79 mol KOH
56
...
00 mol H2 SO4
= 0
...
262 L H2 SO4
1 L H2 SO4
Less than 1 mol of H2 SO4 will react with less than 2 mol of KOH,
but there is more than 2 mol of KOH present, so the H2 SO4 is
limiting
...
786 mol H2 SO4
1 mol H2 SO4
1 mol K2 SO4
17
...
4 g − 52
...
4 g
...
4 g)/(25
...
22 g/mL
...

18
...
00 mol LiOH
0
...
194 mol LiOH
1 L LiOH
n=

(1
...
46 L)
PV
=
= 0
...
0821 L·atm/mol·K)(298 K)

There is more than twice as much LiOH present as CO2 , so the
CO2 is limiting:

0
...
0 g Li2 CO3
1 mol Li2 CO3

= 4
...
15 C
1 mol e−
1 mol Cu
×
19
...
0 hours
1 hour
1s
96,500 C
2 mol e−

63
...
0 g Cu
1 mol Cu

20
...
”
21
...
The word
“light” is often used to mean the entire electromagnetic spectrum,
so not all light can be detected by the human eye
...

Initial concentration (M)
Change due to reaction (M)
Equilibrium concentrations (M)
K =

A
+
1
...
20 − x

2B

2
...
30 − 2x

2C
0
...
00 × 10−6
[A][B]2
(1
...
30)2

4x 2 = 6
...
26 × 10−3
[C] = 2x = 2
...
20 − 1
...
20 M
[B] = 2
...
26 × 10−3 ) = 2
...

23
...
6 g of CO:

H = 19
...
0 g CO

−566 kJ
2 mol CO

= −198 kJ

24
...
0 g/mol (the sum of the atomic
masses of all four atoms), no matter what the conditions
...

(a)
P
V
1
...
00P 1
V2
2 P 2 = 1
...
00P 1 )(1
...
885 L
P2
1
...
0% did not decrease the
volume 13
...
5%
...
00/1
...
0/100
...
00P 1 )(1
...
500 L
2
...
0% did not decrease
the volume 100
...

26
...
184 J/g·◦ C)(t − 14
...
9 g)(0
...
3◦ C)
◦

t = 15
...
The two reduction half-cells are
Cu2+ (aq) + 2 e− → Cu(s)
Fe3+ (aq) + e− → Fe2+ (aq)

0
...
77 V

We reverse the equation with the lower potential, and change the
sign of the potential:
Cu(s) → Cu2+ (aq) + 2 e−

− 0
...
77 V

All that is left to do is to add these equations, and add the
corresponding potentials:

28
...
43 V

1000 mL
13
...
02 × 1023 atoms
1
...
6 g
1 mol
= 6
...
The total heat involved is the sum of the heats of three steps:
q = mct = (51
...
184 J/g·◦ C)(−50
...
9 kJ

−335 J
51
...
3 kJ
1g
q = mct = (51
...
089 J/g·◦ C)(−5
...
58 kJ

The total heat required is
(−10
...
3 kJ) + (−0
...
8 kJ
30
...
31 J/K)(298 K) ln(4
...
69 × 103 J = −3
...
MnO4− (aq) + 8 H+ (aq) + 5 e− → Mn2+ (aq) + 4 H2 O(l)
= ◦ −

0
...
51 −

(0
...
0592
log
= 1
...
0927 = 1
...
100)8 (1
...
HCl reacts with C2 H3 O2 − to give HC2 H3 O2 (in addition to that
already present)
...
010
Change due to
reaction (mol) 0
...
000

0
...
175

0
...
010

0
...
185

Now the equilibrium reaction is considered:
H2 O(l) + HC2 H3 O2 (aq)
C2 H3 O2 − (aq) + H3 O+ (aq)
Initial
concentrations (M)
Change due
to reaction (M)
Equilibrium
concentrations (M)

0
...
165

0
...
185 − x

0
...
165)x
=
= 1
...
185

x = [H3 O+ ] = 2
...
70

33
...
528 atm)(1
...
0216 mol of particles
(0
...
0200 mol of acid and 0
...

There is 0
...
0200
0
...
0000 0
...
0016
+0
...
0016 +0
...
0184
0
...
0016 0
...
0016)2
= 1
...
0184

It does not matter that H2 O is not included in the equation, since
the H2 O molecules are solvent particles, not solute particles
...
G = H − T S = 14
...
00744 kJ/K) = 12
...
(a) 1
...
9 × 10−14 M (c) 2
...
Number of millimoles of nitrate ion in initial solution:

2
...
00 mL
1 mL
1 mmol Al (NO3 )3
= 90
...
00 mmol NO3 −
= 1
...
00 mL

37
...
00 g
4
...
00 g
1
...
99 g
2
...

211

38
...
0 g C
= 3
...
0 g C

1 mol H
= 6
...
008 g H

1 mol O
53
...
33 mol O
16
...
67 g H

The mole ratio is 1 : 2 : 1, and the empirical formula is CH2 O
...
24◦ C = m(1
...
667 m

In 1
...
667 mol of solute and 100
...
0 g)/(0
...
The empirical formula mass is
30
...

39
...
0 g/mol
= 0
...
0 g/mol

r SO3 = (0
...
461)(4
...
94 minutes

40
...
11◦ C)/(1
...
21 m
Assuming that we have 1
...
21 mol
of solute
...
0 g
2
...
177 kg solute
1 mol
There is 1
...
177 kg
= 1
...
11 kg
212

The osmotic pressure is thus
π=

(2
...
0821 L·atm/mol·K)(298 K)
nRT
=
= 51
...
06 L

41
...
100
= G ◦ + (8
...
00)(4
...
3 kJ = G ◦ − 10
...
4 kJ
ln K =

−G ◦
+14,400 J
=
= 5
...
31 J/K)(298 K)

K = 3
...
(a) 0
...
100 mol HC2 H3 O2 produces 0
...

(b) 0
...
100 mol NaC2 H3 O produces (NaCl plus)
0
...

(c) 0
...
100 mol NH3 produces 0
...

(d) 0
...
200 mol HC2 H3 O2 produces 0
...
100 mol HC2 H3 O2 in excess, a buffer
solution problem
...
100 mol HCl + 0
...
100 mol
HC2 H3 O2 and leaves 0
...

(f) 0
...
200 mol NH3 produces 0
...
100 mol NH3 in excess, a buffer solution
problem
...
(a) 15
...
(b) The law of conservation of mass
...
) (c) 15
...
(d ) The law of deﬁnite
proportions
...
Some of the B is left unreacted
...
MgCO3 + 2 HCl → MgCl2 + CO2 + H2 O

92
...
3 g MgCO3

= 1
...
1 g HCl

1 mol HCl
36
...
36 mol HCl

The quantity of HCl required to react with 1
...
10 mol MgCO3

2 mol HCl
1 mol MgCO3

= 2
...

1
...
2 g MgCl2
1 mol MgCl2

= 105 g MgCl2

2
...
20 mol reacting = 0
...
5 g HCl
0
...
8 g HCl excess
1 mol HCl

45
...
00 L of solution
...
62 mol of sucrose and
1210 g of solution
...
62 mol C12 H22 O11

342 g C12 H22 O11
1 mol C12 H22 O11

= 554 g C12 H22 O11

The mass of the water is 1210 g − 554 g = 660 g = 0
...

The molality of the sucrose is (1
...
66 kg) = 2
...
The ﬁrst solution contained 0
...
500 mol solute

1 kg solvent
2
...
200 kg solvent

The ﬁnal solution contains x kg of solvent and

1
...
95x mol solute
1 kg solvent

The second solution thus contained (x − 0
...
95x − 0
...
Its molality is
1
...
95x − 0
...
200) kg solvent

1
...
500 = 1
...
200)
x = 0
...
40 kg − 0
...
20 kg
...
31 J/K)(298 K)
3RT
47
...
17 × 10−21 J
2N
2(6
...
The average kinetic energy is obtained from the prior problem
...
17 × 10−21 J) 6
...
0 amu)
1 kg

49
...
0200 m in ions
...

t f = k f m = (1
...
0200 m) = 0
...

51
...
0372◦ C

1 mol NH4 NO3
2 mol N atoms
146 g NH4 NO3
= 3
...
0 g NH4 NO3
1 mol NH4 NO3
G = −n F

(The value of G ◦ does not matter
...
172 V

215

List of Important Equations

The student must know the conditions, if any, under which each
equation is applicable
...
8
T = t + 273◦
KE =

1
2

mv 2

Chapter 6
M = (moles of solute)/(liter of solution)
m = (moles of solute)/(kilogram of solvent)
XA = (moles of A)/(total number of moles)

Chapter 7
P1 V1 = P2 V2

(constant T)

V1 /T1 = V2 /T2

(constant P )

P1 V1 /T1 = P2 V2 /T2
P V = nRT
Pi
Ptotal

216

=

ni
ntotal

(constant V and T)

ni
Vi
=
Vtotal
ntotal

r1
MM2
=
r2
MM1

(constant P and T)

KE = 3RT/2N

Chapter 8
q = H = mct
H = H f (products) − H f (reactants)

Chapter 9
= o −

[C]c [D]d
0
...
0 × 10−14
pH = − log[H3 O+ ]

or

pH = − log[H+ ]

Chapter 11
PA = XA PAo
PA = XB PAo

(two components)

t f = k f m
tb = kb m
πV = nRT

217

Chapter 12
S o = S o (products) − S o (reactants)
G = H − TS
G = G f (products) − G f (reactants)
G = G o + RT ln Q
G o = −RT ln K
G = −nF

218

Constants with Values That Must Be Remembered
Constant

Identiﬁcation

Value

dH2 O

density of water (4◦ C)
mass of 12 C atom
Avogadro’s number
ideal gas law constant
ideal gas law constant
ideal gas law constant
standard temperature
standard pressure

1
...
0000 amu
6
...
0821 L·atm/mol·K
8
...
31 J/K
0◦ C = 273 K
1 atm, 101
...
184 J/g·◦ C

N
R
R
R

cH2 O
o
H
2

Kw
pHH2 O

speciﬁc heat of water
potential of standard
hydrogen electrode
Nernst equation
constant
water ionization
constant
pH of pure water

Chapter
Introduced
2
4
4
7
7
12
7
7
8

0
...
0592 V

9

1
...
00

10
10

219

Glossary

Absolute temperature
...

Acid dissociation constant
...

Activity
...

Ampere
...

amu
...

Atomic mass
...

Atomic mass unit
...

Atomic weight
...

Avogadro’s number
...
02 × 1023
...
The equilibrium constant for the reaction of a weak base with water to form its cation and hydroxide
ion
...
The rise of the boiling point of a solution
(compared to the pure solvent) due to the presence of a solute
...
At constant temperature, the volume of a given sample
of gas is inversely proportional to its pressure
...
A theory that deﬁnes acids as proton donors and
bases as proton acceptors
...
A solution of a weak acid or base plus its conjugate;
it resists change in its pH even on addition of strong acid or
base
...
The volume of a given sample of gas at constant pressure is directly proportional to its absolute temperature
...
Properties of a solution that depend on
220

the nature of the solvent and the concentration of solute
particles
...
The volume of a given sample of gas is directly
proportional to its absolute temperature and inversely proportional to its pressure
...
The quantity of solute per unit quantity of solution
or solvent
...
)
Condensation
...

Coulomb
...

Cubic meter
...

Dalton’s law of partial pressures
...

∆ (Greek delta)
...

∆G
...

∆G◦f
...

∆H
...

∆H◦f
...

Density
...

Diffusion
...

Dimensional analysis
...

Effusion
...

Electrode
...
(2) A
half-cell
...
A cell in which electric current is used to cause
chemical reaction
...
The simplest formula for a compound, in
which the subscripts are at their lowest integral ratios
...
The point in a titration at which the indicator signals
an end to the process (at a point as close to the stoichiometric
ratio of reactants as possible to that in the balanced equation)
...
The heat of a process carried out at constant
pressure with no work other than expansion against the atmosphere
...
A measure of the randomness of a system
...
The symbol for the potential of a cell
...
A state in which two exactly opposite processes occur
at equal rates, resulting in no net change
...
The mathematical equation including the constant ratio of concentrations of products to concentrations of reactants, all raised to the appropriate powers,
at equilibrium
...
The quantity of one reagent greater than required
by the balanced chemical equation to react with all of another
reagent
...
Dimensional analysis
...
The sum of the atomic masses of each atom in a
formula
...
The collection of bonded atoms represented by the
formula of a substance
...

Free energy change
...

Free energy of formation
...

Freezing-point depression
...

Fusion
...

Galvanic cell
...

Graham’s law
...

Gram
...

Half-cell
...

Heat capacity
...

Heat of sublimation
...

Hess’s law
...

Hydrogen electrode
...

Hydronium ion
...

222

Ideal gas law
...

Indicator (acid-base)
...

Joule
...

Kelvin
...

Kinetic molecular theory
...

KMT
...

Law of combining volumes
...

Law of conservation of mass
...

Law of deﬁnite proportions
...
For example,
every sample of pure water is 88
...
2% hydrogen by mass
...
For two or more compounds consisting of the same elements, for a given mass of one of the
elements, the masses of the other element(s) are in a small,
integral ratio
...
If a stress is applied to a system at equilibrium, the equilibrium will tend to shift in an effort to reduce
the stress
...
The quantity of one reactant in a chemical reaction that is not sufﬁcient to react with all the other reactant(s)
present
...
The unit of volume in the (older) metric system
...
The basic unit of length in the metric system
...
The system of units used by scientists in which multiples or subdivisions of units are powers of 10 times the unit,
and all such multiples or subdivisions are designated by the
same preﬁx no matter what unit is involved
...
One-thousandth of a mole
...
The unit of molality
...
The number of moles of solute per kilogram of solvent in
a solution
...
The unit of molarity
...
The formula mass of any substance, expressed in grams
per mole
...
The volume of one mole of a gas measured at standard temperature and pressure: 22
...

Molarity
...

Mole
...

Mole fraction
...

Molecular formula
...

Molecular mass
...

Nernst equation
...

Nonvolatile
...

Of
...
(For example, “20 is one-half of 40” means
20 = 12 × 40
...
The pressure exerted by a solution because of
the presence of a solute
...
Divided by
...
−log [H+ ]
...
The solid, liquid, or gaseous state
...
A state in which two opposite physical processes occur at equal rates, resulting in no net change
...
The symbol for osmotic pressure
...
The “driving force” of an electrochemical reaction
...
The required order of operations in a mathematical
expression or equation
...

Proton acceptor
...

Proton donor
...

Raoult’s law
...

S
...

S ◦
...

Scientiﬁc notation
...

SI, system internationale
...
(Differences from the older metric system include the
224

cubic meter as the unit of volume rather than the liter, and
pascals used for pressure
...
A digit in a properly reported value that indicates
the precision with which a measurement was made
...
Signiﬁcant digit
...
The heat required to raise one gram of a substance
1◦ C
...
The deﬁned quantity against which all other quantities
are compared
...

Standard absolute entropy
...

Standard enthalpy of formation
...

Standard exponential notation
...

Standard free energy of formation
...

Standard half-cell
...

Standard reduction potential
...

Standard state
...

Standard temperature and pressure
...

Stoichiometry
...

STP
...

Sublimation
...

System internationale
...

Titration
...

Unit activity
...

Vaporization
...

Vapor-pressure lowering
...

225

Volt
...

Voltaic cell
...

Water dissociation constant
...

Weighted average
...
For example, the
atomic mass of an element is a weighted average of the masses
of its isotopes, taking into account the relative abundance of
each isotope
...

absolute temperature, 86, 99, 220(g)
acid(s):
strong, 148
weak, 148
acid-base equilibrium, 148–151
acid dissociation constant, 149, 220(g)
activity, 220(g)
activity series, 132
algebra, distributive law of, 114
ampere, 128, 220(g)
amu, 44, 220(g)
approximation method, for equilibrium
constant calculations, 147
atmospheres, 85
atomic mass, 44, 46(t), 220(g)
atomic mass unit, 44, 220(g)
atomic weight, 44, 220(g)
average kinetic energy, 99
Avogadro’s number, 47, 220(g)
balanced chemical equation, mole
ratios in, 59–60
bar, 99
base, of exponential number, 23
base, weak, 149
base dissociation constant, 149, 220(g)
boiling-point elevation, 171–173,
172(t), 220(g)
boy scouts, 3
Boyle’s law, 85–87, 220(g)
Bronsted theory, 151–152, 220(g)
buffer solution(s), 154–156, 220(g)
built-up fraction, 8
c, 110–114, 111(t)
calculations, scientiﬁc, 1–5
calculator, scientiﬁc, 7–11
calorimetry, 110–114

cell, electrolysis, 128–130
galvanic, 128
voltaic, 128
Celsius temperature scale, 26–27,
26(t), 85
change in, 110
change sign key, 10
Charles’ law, 85–88, 220(g)
chemical combination, calculations for,
34–43
chemical equilibrium, 141–168
chemical reactions, moles in, 59–60
coefﬁcient, of exponential number, 23
coefﬁcients, in equations, 59
colligative properties, 169–183, 220(g)
combination of equilibrium constants,
142–143, 156–157
combined gas law, 86–88, 221(g)
combining volumes, law of, 96–97
concentration, 221(g)
calculations involving, 73–84
units of, 73, 77, 78
condensation, 115, 115(f), 221(g)
conservation of mass, law of, 34–35
constants, important, 219
conversions, for stoichiometry, 61–62
coulomb, 128, 221(g)
cubic decimeter, 17
cubic meter, 17, 221(g)
Dalton’s law, 93–96, 221(g)
Daniell cell, 131, 131(f)
deﬁned values, signiﬁcant digits and, 21
deﬁnite proportions, law of, 35–36
(Greek delta), 110, 221(g)
G, 186, 221(g)
G ◦ , 186, 221(g)
G f , 186, 221(g)

227

H, 117, 221(g)
H f , 221(g)
H ◦f , 221(g)
tb , 171
t f , 171
density, 25, 221(g)
diffusion, 97–99, 221(g)
dilution problems, 75
dimensional analysis, 5–7, 199,
221(g)
distance, 14–15
distributive law of algebra, 114
EE, 10
effusion, 97–99, 221(g)
electrochemistry, 128–140
electrode, 130–136, 221(g)
electrolysis, potential and, 136
electrolysis cell, 128–130, 221(g)
electronic calculator, signiﬁcant digits
on, 22, 24
empirical formula, 51–52, 221(g)
end point, 75, 221(g)
energy, 26, 27
of phase change, 114–117,
115(t), 115(f)
enthalpy, 117
enthalpy change, 117, 221(g)
enthalpy of combustion, 117
enthalpy of formation, 118, 119(t),
221(g)
of elements, 118
enthalpy of reaction, 117–121
entropy, 184–186, 221(g)
(Greek epsilon), 131, 221(g)
◦ , 134
equations, list of important, 216–218,
equilibrium, 141–168, 222(g)
physical, 95
equilibrium constant(s), 141
combination of, 142–143, 156–157
for hydrolysis, 156–158
equilibrium constant expression,
141–142, 222(g)
G and, 189–191
calculation of, 143–145
calculation using, 146–148
equivalence point, 76
excess quantity, 62–63, 222(g)
EXP, 10
exponent, 23
exponential part, of exponential
number, 23
factor, 5
factor label method, 5–7, 222(g)
Fahrenheit scale, 26–27, 26(t)

228

Faraday’s laws, 128–129
formula calculations, 44–58
formula mass(es), 46–47, 46(t), 222(g)
types of, 46(t)
formula unit(s), 46, 46(t), 61–62, 222(g)
formula weight, 46
free energy change, 186–189, 222(g)
potential and, 191–192
spontaneous reaction and, 188–189
free energy of formation, 186–188,
187(t), 222(g)
freezing-point data, 171(t)
freezing-point depression, 171–173,
222(g)
fusion, 115, 115(f), 222(g)
galvanic cell, 128, 222(g)
gas laws, 85–109
Gibbs free energy change, 186–189
Graham’s law, 97–99, 222(g)
gram, 15, 15(t), 222(g)
half-cell, 130–136, 222(g)
heat, 110–114
heat capacity, 110–114, 111(t), 222(g)
heat of condensation, 115, 115(t)
heat of fusion, 115, 115(t)
heat of sublimation, 115, 222(g)
heat of vaporization, 115, 115(t)
Hess’s law, 187, 120–121, 222(g)
hydrogen electrode, 222(g)
hydrolysis, 156–158
hydronium ion, 148, 222(g)
ideal gas law, 90–92, 223(g)
ideal solution, 169
indicator (acid-base), 75, 223(g)
ionic solutions, colligative properties of,
175–176
ionization of water, 151–153
joule, 27, 128, 223(g)
K , 141
K a , 149
K b , 149
KE, 99
kelvin, 223(g)
kelvin temperature scale, 26–27,
26(t), 85
k f , 171
K h , 156
kilogram, 15–16
kilopascals, 85
kinetic energy, 27
kinetic molecular theory, 99, 223(g)
KMT, 99, 223(g)

law of combining volumes, 96–97,
223(g)
law of conservation of mass, 34–35,
223(g)
law of deﬁnite proportions, 35–36,
223(g)
law of multiple proportions, 36–38,
223(g)
law of partial pressures, 93–96
LeChatelier’s principle, 151, 154,
223(g)
length, 14–15
limiting quantity(ies), 62–63, 223(g)
problems, 62–64
liter, 15(t), 17, 223(g)
logarithm keys, 11
m, 77
m, 77
M, 73
M, 73
mass(es), 15–17
in chemical reactions, 60–61
unit of, 15, 15(t)
mass number, 44
mass spectrometer, 45
measurement, 14–33
meter, 14, 15(t), 223(g)
metric conversions, 17–19
metric preﬁxes, 15, 15(t)
metric system, 14–19, 223(g)
milli-, 14
millimole, 47, 223(g)
molal, 77, 223(g)
molality, 77, 223(g)
molar, 73, 88, 223(g)
molar mass, 47, 223(g)
molar volume, 88–89, 224(g)
molarity, 73–75, 224(g)
mole(s), 47–49, 224(g)
in chemical reactions, 59–60
of gas, 88–93
conversions for, 91(f)
mole fraction, 77–79, 224(g)
mole ratios in chemical reactions,
59–60
molecular formula, 52–53, 224(g)
molecular mass, 46(t), 224(g)
multiple proportions, law of, 36–38
n, 134–135
Nernst equation, 134–136, 192, 224(g)
nonvolatile, 169, 224(g)
of, 224(g)
osmotic pressure, 173–175, 224(g)
ounce, 16

partial pressures, law of, 93–96
per, 7, 224(g)
percent, 2, 6–7
percent composition, 50–51
pH, 153–154, 224(g)
signiﬁcant digits in, 153
phase, 224(g)
phase change(s), energy of, 114–117,
115(t), 115(f)
heat of, 114–117, 115(t)
names of, 115, 115(f)
physical equilibrium, 95, 224(g)
π (osmotic pressure), 174, 224(g)
potential, 131, 224(g)
G and, 191–192
electrolysis and, 136
pound, 16
precedence, 224(g)
precedence rules, 8–10, 9(t)
preﬁxes, metric, 15, 15(t)
proton acceptor, 224(g)
proton donor, 224(g)
q, 110
R, 90
randomness, 184
Raoult’s law, 169, 224(g)
reciprocal key, 10–11
rounding off, 22–23
S, 184, 224(g)
S ◦ , 184, 224(g)
scientiﬁc calculations, 1–5
units in, 19
scientiﬁc calculator, 7–11
scientiﬁc notation, 23–24, 224(g)
semipermeable, 173, 174(f)
SI (system internationale), 14,
224–225(g)
signiﬁcant digit(s), 19–23, 225(g)
in calculations, 20–22
and deﬁned values, 21
electronic calculator and, 22, 24
in exponential numbers, 153
in pH values, 153
signiﬁcant ﬁgure, 19–23, 225(g)
...

solution, ideal, 169
speciﬁc heat (capacity), 110–114, 111(t),
225(g)
spontaneous reaction, 132
G and, 188–189
square brackets, 135, 141
standard, 15–16, 225(g)
standard absolute entropy, 184, 185(t),
225(g)

229

standard enthalpy of formation,
118–120, 119(t)
standard exponential notation, 23–24,
225(g)
standard free energy change, 186
standard free energy of formation,
225(g)
standard half-cell, 132, 225(g)
standard hydrogen electrode, 131–132,
132(f)
standard reduction potential, 132–133,
133(t), 225(g)
standard state, 118, 225(g)
standard temperature and pressure, 88,
225(g)
stoichiometry, 59–72, 225(g)
STP (standard temperature and
pressure), 88, 118, 225(g)
strong acids, 148
sublimation, 115, 115(f), 225(g)
subscripts, 47
surroundings, 110
symbol, 1
system, 110
system internationale, 14, 224–225(g)
temperature, 26–27, 26(t)
temperature change, 111
temperature difference, 111
thermochemistry, 110–127

230

thermodynamics, 110–127, 184–198
time, 25–26
titration, 75–77
torr, 85
unary minus, 9
unit(s), 1
adjustment of, 19
in scientiﬁc calculations, 19
unit activity, 132, 225(g)
vapor pressure of water, 95–96
vapor-pressure lowering, 169–170,
225(g)
vaporization, 115, 115(f), 225(g)
variables, designation of, 4–5
volt, 128, 226(g)
voltaic cell, 128, 130–136, 226(g)
volume, 15(t), 17, 17(t)
SI vs
...
D
...
He is author or coauthor of 15 books, over 30
journal articles, and numerous bookelets for student use

Title: Problem solving medicinal chemistry
Description: It is a problem solving medicinal chemistry in English language in very easy way that can understand to every students and teachers.

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