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How to Solve
Word Problems in Chemistry

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

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


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

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


vi

Chapter 1

Introduction

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

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

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

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

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

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

4
...

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

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

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

must memorize
...

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

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

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

2
...

3
...

4
...

5
...

4

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

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

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

EXAMPLE 1

Convert 2
...

Solution




100 cents
✥
✥
✥
= 225 cents
2
...

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



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


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



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

EXAMPLE 3

Change 1
...


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



60 minutes
1
...
60 minutes
1 hour



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



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

6

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


EXAMPLE 4

Solution

45 voters




100 people total
75 voters


= 60 people




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

=

66
...
0 feet/second
1 second

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


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

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

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

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

6 × 3 ÷ 2 ÷ 4 =
or

6 × 3 ÷ ( 2 × 4 ) =

(c) 512
...

(d) −9
...

9

(e) +9
...

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

EXAMPLE 7

Solution

(a) Enter 1


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



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

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

Alternatively, we could have determined:



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

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

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

13

Chapter 2

Measurement

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

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

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

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

Table 2-1

Most Important Metric Terms and Abbreviations

Unit

Abbreviation

Prefix

Abbreviation

meter
gram
liter

m
g
L

kilodecicentimilli-

k
d
c
m

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

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

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

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

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

∗

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

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



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



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

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

EXAMPLE 2

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


Solution




(a) 1
...
00149 km

(substitute 1000 for the k)

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

Metric

Equivalent

1 m3
1 dm3
1 cm3
1 mm3

1 kL
1L
1 mL
1 µL

1000 L
1L
0
...
000001 L
17

Cubic meter

Cubic decimeter
Liter

Cubic centimeter
Milliliter

100 cm
1m

10 cm

100 cm
1m

1 cm

10 cm
100 cm
1m

10 cm
1 dm

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

(10 cm)3
1,000 cm3

1 cm
1 cm
(1 cm)3
1 cm3

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



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

1,000,000 cm3
1 m3


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

1000 L
1 m3


= 2500 L




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

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

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


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

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

19

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

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

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

Solution

(a) 0
...
4 kg

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



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

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

20

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

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



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

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

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

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

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

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

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

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

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

22

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


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


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




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


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

exponent
\ /
1
...

EXAMPLE 10

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

tific notation?
23

(a) 1
...
246 × 103
...
00 × 10−3

(b) 0
...
246 × 100

(c) 10
...
0 × 10−3

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

cm

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


24

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

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



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




1 hour

60 minutes
1 hour




15 minutes




60 seconds
1 minute

60 seconds
1 minute


= 3600 seconds

= 900 seconds




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

Table 2-4

Temperature Scales

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

Freezing Point
of Water

Normal Boiling
Point of Water

32◦ F
0◦ C
273 K

212◦ F
100◦ C
373 K

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

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

g/mL

mg/mL

kg/m3

kg/L

kg/dm3

g/cm

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

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

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

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

5
...

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

7
...
55 cm2 times 2
...

8
...
0 g by 23
...

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

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

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

29

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

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


Solutions to Supplementary Problems



1
...


3
...


5
...
44 m
= 244 cm

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



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







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





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

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

8
...

10
...

12
...

14
...


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

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







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



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





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



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

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





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



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



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





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

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




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

33

Chapter 3

Classical Laws of
Chemical Combination

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

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

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



34

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

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

100
...
0906


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




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

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



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




6
...
20 g oxygen


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

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

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

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



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

EXAMPLE 9

Compound 1
Compound 2

Element 1
29
...
4%

Element 2
40
...
6%

Element 3
30
...
0%

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

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

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

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

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

1
5

0
...
4

2
5

0
...
6

3
5

0
...
8

4
5

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

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


EXAMPLE 10

Solution

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

1
2
3
2

(c) (2
...
)
2 23

=

8
3

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


= 29
...


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

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

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

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

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

10
...

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

Element 3
1
...
4%

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

Element 3
1
...
4%

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

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

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

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

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

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

40

3
...
3%
...
)



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

Element 1
Element 2
Element 3

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

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

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

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

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

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

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

3
4

so multiply numerator and denominator

(19
...
00 g B)

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

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

(d ) No
...




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

Compound 1
Compound 2

Element 1
7
...
0 g

Element 2
0
...
56 g

Element 3
1
...
4 g

Compound 1
Compound 2

Element 1
12
...
99 g

Element 2
1
...
00 g

Element 3
2
...
99 g

(c)

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

13
...




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

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

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

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

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


43

Chapter 4

Formula Calculations

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

44

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

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

Solution

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



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


EXAMPLE 2
24

45

Table 4-1

Types of Formula Masses

Formula Unit

Name

Example

Atom
Molecule
Molecule

Atomic mass
Molecular mass
Molecular mass

Hg
NH3
H2

Formula unit of an
ionic compound

Formula mass

MgCl2

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

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


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

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

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




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

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

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

Solution


(a) 3
...
50 mol

12 lemons
1 dozen


= 42 lemons

6
...
11 × 1024 atoms




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

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

Solution





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

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


= 85
...
00 mol NH3
1 mol NH3

(a) 5
...

EXAMPLE 8

Calculate the number of molecules in 56
...


Solution




56
...
34 mol NH3

1 mol NH3
17
...
34 mol NH3

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





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




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

1
...
20 mol K



49

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


EXAMPLE 10

Solution

The masses are calculated as shown above:




14
...
02 g N
1 mol N



1
...
032 g H
1 mol H



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

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

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



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

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

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

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


empirical formula to be CH2 O
...
Instead of masses of the elements, a problem is usually stated in terms of percentages of the
elements
...
0-g sample, the percentages are equal to the masses in grams
...
The
second complication is the division by the magnitude of the smallest number of moles may not give all integers, but some decimal
fractions in addition to integers
...
(We had the
same 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

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