Download The science of chemistry is concerned

61
Chapter three
USING CHEMICAL EQUATIONS
IN CALCULATIONS
There are a great many circumstances in which you may need to use a balanced equation.
For example, you might want to know how much air pollution would occur when 100
metric tons of coal were burned in an electric power plant, how much heat could be
obtained from a kilogram of natural gas, or how much vitamin C is really present in a 300mg tablet. In each instance someone else would probably have determined what reaction
takes place, but you would need to use the balanced equation to get the desired information.
3.1 EQUATIONS AND MASS RELATIONSHIPS
A balanced chemical equation such as
4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g)
(3.1)
not only tells how many molecules of each kind are involved in a reaction, it also indicates
the amount of each substance that is involved. Equation (3.1) says that 4NH3 molecules can
react with 5O2 molecules to give 4 NO molecules and 6H2O molecules. It also says that 4
mol NH3 would react with 5 mol O2 yielding 4 mol NO and 6 mol H2O.
The balanced equation does more than this, though. It also tells us that 2 × 4 = 8 mol NH3
will react with 2 × 5 = 10 mol O2, and that
62
½ × 4 = 2 mol NH3 requires only ½ × 5 = 2.5 mol O2. In other words, the equation indicates
that exactly 5 mol O2 must react for every 4 mol NH3 consumed. For the purpose of
calculating how much O2 is required to react with a certain amount of NH3 therefore, the
significant information contained in Eq. (3.1) is the ratio
5 mol O 2
4 mol NH 3
We shall call such a ratio derived from a balanced chemical equation a stoichiometric
ratio and give it the symbol S. Thus, for Eq. (3.1),
 O2  5 mol O2

NH
4 mol NH3
3 

S
(3.2)
The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron,
“measure.“ Hence the stoichiometric ratio measures one element (or compound) against
another.
EXAMPLE 3.1 Derive all possible stoichiometric ratios from Eq. (3.1)
Solution Any ratio of amounts of substance given by coefficients in the equation may be
used:
 NH3  4 mol NH3

O
 2  5 mol O2
S
 NH3  4 mol NH3

 NO  4 mol NO
S
 NH3  4 mol NH3

 H2O  6 mol H2 O
S
 O 2  5 mol O 2

 NH  4 mol NH
S
 O2  5 mol O2

 H 2 O  6 mol H 2 O
S
S
 NO  4 mol NO

 H 2 O  6 mol H 2 O
There are six more stoichiometric ratios, each of which is the reciprocal of one of these.
[Eq. (3.2) gives one of them.]
When any chemical reaction occurs, the amounts of substances consumed or produced are
related by the appropriate stoichiometric ratios. Using Eq. (3.1) as an example, this means
that the ratio of the amount of O2 consumed to the amount of NH3 consumed must be the
stoichiometric ratio S(O2/NH3):
nO consumed
 NH3  4 mol NH3
 S

nNH consumed
 O2  5 mol O2
2
3
Similarly, the ratio of the amount of H 2 O produced to the amount of NH 3 consumed must be
S(H2O/NH3):
nH O produced
2
nNH
3
consumed
 H 2 O  6 mol H 2 O

 NH3  4 mol NH3
 S
63
In general we can say that
 X  amount of X consumed or produced

 Y  amount of Y consumed or produced
Stoichiometric ratio 
(3.3a)
or, in symbols,
 X  nX consumed or produced

 Y  nY consumed or produced
S
(3.3.b)
Note that in the word Eq. (3.3a) and the symbolic Eq. (3.3b), X and Y may represent any
reactant or any product in the balanced chemical equation from which the stoichiometric
ratio was derived. No matter how much of each reactant we have, the amounts of reactants
consumed and the amounts of products produced will be in appropriate stoichiometric
ratios.
EXAMPLE 3.2 Find the amount of water produced when 3.68 mol NH3 is consumed
according to Eq. (3.1).
Solution The amount of water produced must be in the stoichiometric ratio S(H2O/NH3) to
the amount of ammonia consumed:
 H 2 O  nH O produced

 NH3  nNH consumed
S
2
3
Multiplying both sides nNH3 consumed, by we have
nH O produced  nNH
2
3
consumed
 H2O 

 NH 3 
 S
 3.68 mol NH 3 
6 mol H 2 O
4 mol NH 3
 5.52 mol H 2 O
This is a typical illustration of the use of a stoichiometric ratio as a conversion factor.
Example 3.2 is analogous to Examples 1.10 and 1.11, where density was employed as a
conversion factor between mass and volume. Example 3.2 is also analogous to Examples
2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion
factors. As in these previous cases, there is no need to memorize or do algebraic
manipulations with Eq. (3.3) when using the stoichiometric ratio. Simply remember that the
coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper
choice results in cancellation of units. In road-map form
stoichiometric
ratio X/Y
amount of X consumed 
 amount of Y consumed
or produced
or produced
64
or symbolically.
When using stoichiometric ratios, be sure you always indicate moles of what. You can only
cancel moles of the same substance. In other words, 1 mol NH3 cancels 1 mol NH2 but does
not cancel 1 mol H2O.
The next example shows that stoichiometric ratios are also useful in problems involving the
mass of a reactant or product.
EXAMPLE 3.3 Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is
reacted with FeS2 according to the equation
4FeS2 + 11O2 → 2Fe2O3 + 8SO2
Solution The problem asks that we calculate the mass of SO2 produced. As we learned in
Sec. 2.8, Example 2.6, the molar mass can be used to convert from the amount of SO2 to
the mass of SO2. Therefore this problem in effect is asking that we calculate the amount of
SO2 produced from the amount of O2 consumed. This is the same problem as in Example
3.2. It requires the stoichiometric ratio
 SO2  8 mol SO2

 O2  11 mol O2
S
The amount of SO2 produced is then
nSO  nSO
2
2
consumed
× conversion factor
 3.84 mol O2 
8 mol SO 2
11 mol O 2
 2.79 mol SO 2
The mass of SO2 is
mSO  2.79 mol SO 2 
2
64.06 g SO 2
1 mol SO 2
 179 g SO 2
With practice this kind of problem can be solved in one step by concentrating on the units.
The appropriate stoichiometric ratio will convert moles of O2 to moles of SO2 and the
molar mass will convert moles of SO2 to grams of SO2. A schematic road map for the onestep calculation can be written as
Thus
65
The chemical reaction in this example is of environmental interest. Iron pyrite (FeS2 ) is
often an impurity in coal, and so burning this fuel in a power plant produces sulfur dioxide
( SO 2 ), a major air pollutant.
Our next example also involves burning a fuel and its effect on the atmosphere.
EXAMPLE 3.4 What mass of oxygen would be consumed when 3.3 × 1015 g, 3.3 Pg
(petagrams), of octane (C8H18) is burned to produce CO2 and H2O?
Solution First, write a balanced equation
2C8H18 + 25O2 → 16CO2 + 18H2O
The problem gives the mass of C8H18 burned and asks for the mass of O2 required to
combine with it. Thinking the problem through before trying to solve it, we realize that the
molar mass of octane could be used to calculate the amount of octane consumed. Then we
need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass
of O 2 permits calculation of the mass of O2. Symbolically
Thus 12 Pg (petagrams) of O2 would be needed.
The large mass of oxygen obtained in this example is an estimate of how much O2 is
removed from the earth’s atmosphere each year by human activities. Octane, a component
of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total
mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we
were to go on burning fuel at the present rate, it would take about 100 000 years to use up
all the O2. Actually we will consume the fossil fuels long before that! One of the least of
our environmental worries is running out of atmospheric oxygen.
The Limiting Reagent
Example 3.4 also illustrates the idea that one reactant in a chemical equation may be
completely consumed without using up all of another. In the laboratory as well as the
environment, inexpensive reagents like atmospheric O2 are often supplied in excess. Some
portion of such a reagent will be left unchanged after the reaction. Conversely, at least one
reagent is
66
usually completely consumed. When it is gone, the other excess reactants have nothing to
react with and they cannot be converted to products. The substance which is used up first is
the limiting reagent.
EXAMPLE 3.5 When 100.0 g mercury is reacted with 100.0 g bromine to form mercuric
bromide, which is the limiting reagent?
Solution The balanced equation
Hg + Br2 → HgBr2
tells us that according to the atomic theory, 1 mol Hg is required for each mole of Br2. That
is, the stoichiometric ratio S(Hg/Br2) + 1 mol Hg/ 1 mol Br2. Let us see how many moles of
each we actually have
1 mol Hg
nHg  100.0 g 
 0.4985 mol Hg
200.59 g
nBr  100.0 g 
2
1 mol Br2
159.80 g
 0.6258 mol Br2
When the reaction ends, 0.4985 mol Hg will have reacted with 0.4985 mol Br2 and there
will be
(0.6258 – 0.4985) mol Br2 = 0.1273 mol Br2
left over. Mercury is therefore the limiting reagent.
From this example yo6 can begin to see what needs to be done to determine which of two
reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the
actual ratio of amounts of X and Y which were initially mixed together. In Example 3.5 this
ratio of initial amounts
nHg (initial) 0.4985 mol Hg 0.7966 mol Hg


nBr (initial) 0.6258 mol Br2
1 mol Br2
2
was less than the stoichiometric ratio
 Hg  1 mol Hg

 Br2  1 mol Br2
S
This indicated that there was not enough Hg to react with all the Br2 and mercury was the
limiting reagent. The corresponding general rule, for any reagents X and Y, is
If
If
nX (initial)
nY (initial)
nX (initial)
nY (initial)
X
 , then X is limiting.
Y
is less than S 
X
 , then Y is limiting.
Y
is greater than S 
67
(Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both
reagents will be completely consumed at the same time, and neither is in excess.). This
general rule for determining the limiting reagent is applied in the next example.
EXAMPLE 3.6 Iron can be obtained by reacting the ore hematite (Fe2O3) with coke (C).
The latter is converted to CO2. As manager of a blast furnace you are told that you have
20.5 Mg (megagrams) of Fe2O3 and 2.84 Mg of coke on hand. (a) Which should you order
first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can
you make with the materials you have?
Solution
a) Write a balanced equation
2Fe2O3 + 3C → 3CO2 + 4Fe
The stoichiometric ratio connecting C and Fe2O3 is


3 mol C
1.5 mol C


 Fe2 O3  2 mol Fe2 O3 1 mol Fe2O3
C
S
The initial amounts of C and Fe2O3 are calculated using appropriate molar masses
1 mol C
nC (initial)  2.84  106 g 
 2.36  105 mol C
12.01 g
nFe O (initial)  20.5  106 g 
2
1 mol Fe 2 O3
159.69 g
3
 1.28  105 mol Fe 2O 3
Their ratio is
nC (initial)
nFe O (initial)
2
3

2.36  105 mol C
1.28 10 mol Fe2 O3
5

1.84 mol C
1 mol Fe2 O3
Since this ratio is larger than the stoichiometric ratio, you have more than enough C to react
with all the Fe2O3. Fe2O3 is the limiting reagent, and you will want to order more of it first
since it will be consumed first.
b) The amount of product formed in a reaction may be calculated via an appropriate
stoichiometric ratio from the amount of a reactant which was consumed. Some of the
excess reactant C will be left over, but all the initial amount of Fe2O3 will be consumed.
Therefore we use nFe2O3 (initial) to calculate how much Fe can be obtained
This is 1.43 × 106 g, or 14.3 Mg, Fe.
68
As you can see from the example, in a case where there is a limiting reagent, the initial
amount of the limiting reagent must be used to calculate the amount of product formed.
Using the initial amount of a reagent present in excess would be incorrect, because such a
reagent is not entirely consumed.
The concept of a limiting reagent was used by the nineteenth century German chemist
Justus von Liebig (1807 to 1873) to derive an important biological and ecological law.
Liebig’s law of the minimum states that the essential substance available in the smallest
amount relative to some critical minimum will control growth and reproduction of any
species of plant or animal life. When a group of organisms runs out of that essential
limiting reagent, the chemical reactions needed for growth and reproduction must stop.
Vitamins, protein, and other nutrients are essential for growth of the human body and of
human populations. Similarly, the growth of algae in natural bodies of water such as Lake
Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of
phosphates. It is for this reason that many states have regulated or banned the use of
phosphates in detergents and are constructing treatment plants which can remove
phosphates from municipal sewage before they enter lakes or streams.
Percent Yield
Not all chemical reactions are as simple as the ones we have considered, so far. Quite often
a mixture of two or more products containing the same element is formed. For example,
when octane (or gasoline in general) burns in an excess of air, the reaction is
2C8H18 + 25O2 → 16CO2 + 18H2O
If oxygen is the limiting reagent, however, the reaction does not necessarily stop short of
consuming all the octane available. Instead, some carbon monoxide (CO) forms:
2C8H18 + 24O2 → 14CO2 + 2CO + 18H2O
Burning gasoline in an automobile engine, where the supply of oxygen is not always as
great as that demanded by the stoichiometric ratio, often produces carbon monoxide, a
poisonous substance and a major source of air pollution.
In other cases, even though none of the reactants is completely consumed, no further
increase in the amounts of the products occurs. We say that such a reaction does not go to
completion. When a mixture of products is produced or a reaction does not go to
completion, the effectiveness of the reaction is usually evaluated in terms of percent yield
of the desired product. A theoretical yield is calculated by assuming that all the limiting
reagent is converted to product. The experimentally determined mass of product is then
compared to the theoretical yield and expressed as a percentage:
Percent yield 
actual yield
theoretical yield
100 percent
69
EXAMPLE 3.7 When 100.0 g N2 gas and 25.0 g H2 gas are mixed at 350°C and a high
pressure, they react to form 28.96 g NH3 (ammonia) gas. Calculate the percent yield.
Calculate the percent yield.
Solution We must calculate the theoretical yield of NH3, and to do this, we must first
discover whether N2 or H2 is the limiting reagent. For the balanced equation
N2 + 3H2 → 2NH3
the stoichiometric ratio of the reactants is
 H2  3 mol H2

 N2  1 mol N2
S
Now, the initial amounts of the two reagents are
nH (initial)  25.0 g H 2 
2
1 mol H 2
2.016 g H 2
nN (initial)  100.0 g N 2 
and
2
 12.4 mol H 2
1 mol N 2
28.02 g N 2
 3.569 mol N 2
The ratio of initial amounts is thus
nH (initial)
2
nN (initial)

2
12.4 mol H 2

3.569 mol N 2
3.47 mol H 2
1 mol N 2
 H2 
 , there is an excess of H2. N2 is the limiting reagent.
 N2 
Since this ratio is greater than S 
Accordingly we must use 3.569 mol N2 (rather than 12.4 mol H2) to calculate the
theoretical yield of NH3. We then have
nNH (theoretical)  3.569 mol N 2 
2 mol NH 3
3
1 mol N 2
 7.138 mol NH3
so that
m NH (theoretical)  7.138 mol NH 3 
3
17.03 g NH 3
1 mol NH 3
 121.6 g NH3
The percent yield is then
actual yield
Percent yield 
 100 percent
theoretical yield

28.96 g
121.6 g
 100 percent  23.81 percent
70
Combination of nitrogen and hydrogen to form ammonia is a classic example of a reaction
which does not go to completion. Commercial production of ammonia is accomplished
using this reaction in what is called the Haber process. Even at the rather unusual
temperatures and pressures used for this industrial synthesis, only about one-quarter of the
reactants can be converted to the desired product. This is unfortunate because nearly all
nitrogen fertilizers are derived from ammonia and the world has come to rely on them in
order to produce enough food for its rapidly increasing population. Ammonia ranks third
[after sulfuric acid (H2SO4) and oxygen (O2)] in the list of most-produced chemicals,
worldwide. It might rank even higher if the reaction by which it is made went to
completion. Certainly ammonia and the food it helps to grow would be less expensive and
would require much less energy to produce if this were the case.
3.2 ANALYSIS OF COMPOUNDS
Up to this point we have obtained all stoichiometric ratios from the coefficients of balanced
chemical equations. Chemical formulas also indicate relative amounts of substance,
however, and stoichiometric ratios may be derived from them, too. For example, the
formula HgBr2 tells us that no matter how large a sample of mercuric bromide we have,
there will always be 2 mol of bromine atoms for each mole of mercury atoms. That is, from
the formula HgBr2 we have the stoichiometric ratio
 Br  2 mol Br

 Hg  1 mol Hg
S
We could also determine that for HgBr2
(The reciprocals of these stoichiometric ratios are also valid for HgBr2.)
Stoichiometric ratios derived from formulas instead of equations are involved in the most
common procedure for determining the empirical formulas of compounds which contain
only C, H, and O. A weighed quantity of the substance to be analyzed is placed in a
combustion train (Fig. 3.1) and
Figure 3.1 A combustion train. H2O and C 2O, produced by
combination of O2 with H and C in the sample, are selectively
absorbed by tubes containing Dehydrite [Mg(ClO 4) 2•3H2O]
and ascarite (NaOH on asbestos).
71
heated in a stream of dry O2. All the H in the compound is converted to H2O(g) which is
trapped selectively in a previously weighed absorption tube. All the C is converted to
CO2(g) and this is absorbed selectively in a second tube. The increase of mass of each tube
tells, respectively, how much H2O and CO2 were produced by combustion of the sample.
EXAMPLE 3.8 A 6.49-mg sample of ascorbic acid (vitamin C) was burned in a
combustion train. 9.74 mg CO2 and 2.64 mg H2O were formed. Determine the empirical
formula of ascorbic acid.
Solution We need to know the amount of C, the amount of H, and the amount of O in the
sample. The ratio of these gives the subscripts in the formula. The first two may be
obtained from the masses of CO2 and H2O using the molar masses and the stoichiometric
ratios
 C  1 mol C

 CO2  1 mol CO2
 H  2 mol H

 H 2 O  1 mol H 2 O
S
S
Thus
nC  9.74  10-3g CO 2 
1 mol CO 2
44.01 g CO 2

1 mol C
1 mol CO 2
 2.21  10-4 mol C
nH  2.64  10-3g H 2 O 
1 mol H 2 O

2 mol H
18.02 g H 2 O 1 mol H 2 O
 2.93  10-4 mol H
The compound may also have contained oxygen. To see if it does, calculate the masses of
C and H and subtract from the total mass of sample
Thus we have
6.49 mg sample – 2.65 mg C – 0.295 mg H = 3.54 mg O
and
nO  3.54  10-3g O 
1 mol O
16.00 g O
 2.21  10-4 mol O
72
The ratios of the amounts of the elements in ascorbic acid are therefore
nH
nC


nO
nC

2.93  10-4 mol H
2.21 10 mol C
-4
1 13 mol H
1 mol C

1.33 mol H
1 mol C
4 mol H
3 mol C
2.21  10-4 mol O
2.21  10 mol C
-4


1 mol O
1 mol C

3 mol O
3 mol C
Since nC:nH:nO is 3 mol C:4 mol H:3 mol O, the empirical formula is C3H4O3.
A drawing of a molecule of ascorbic acid is shown in Fig. 3.2. You can determine by
counting the atoms that the molecular formula is C6H8O6—exactly double the empirical
formula. It is also evident that there is more to know about a molecule than just how many
atoms of each kind are present.
In ascorbic acid, as in other molecules, the way the atoms are connected together and their
arrangement in three-dimensional space are quite important. A picture like Fig. 3.2,
showing which atoms are connected to which, is called a structural formula. Empirical
formulas may be obtained from percent composition or combustion-train experiments, and,
if the molecular weight is known, molecular formulas may be determined from the same
data. More complicated experiments are required to find structural formulas.
In Example 3.8 we obtained the mass of O by subtracting the masses of C and H from the
total mass of sample. This assumed that only C, H, and O were present. Sometimes such an
assumption may be incorrect. When penicillin was first isolated and analyzed, the fact that
it contained sulfur
Figure 3.2 The structural formula of
ascorbic acid (vitamin C), C 6H8O6. Carbon
atoms are dark gray, hydrogen atoms are
light red, and oxygen atoms are dark red.
(Computer-generated.) (Copyright@1976 by
W.G.Davis and J.W.Moore.)
73
was missed. This mistake was not discovered for some time because the atomic weight of
sulfur is almost exactly twice that of oxygen. Two atoms
of oxygen were substituted in place of one sulfur atom in the formula.
3.3 THERMOCHEMISTRY
When a chemical reaction occurs, there is usually a change in temperature of the chemicals
themselves and of the beaker or flask in which the reaction is carried out. If the temperature
increases, the reaction is exothermic—energy is given off as heat when the container and
its contents cool back to room temperature. (Heat is energy transferred from one place to
another solely because of a difference in temperature.) An endothermic reaction produces a
decrease in temperature. In this case heat is absorbed from the surroundings to return the
reaction products to room temperature. Thermochemistry, a word derived from the Greek
thermé “heat,” is the measurement and study of energy transferred as heat when chemical
reactions take place. It is extremely important in a technological world where a great deal
of work is accomplished by transforming and harnessing heat given off during combustion
of coal, oil, and natural gas.
Energy
Energy is usually defined as the capability for doing work. For example, a billiard ball can
collide with a second ball, changing the direction or speed of motion of the latter. In such a
process the motion of the first ball would also be altered. We would say that one billiard
ball did work on (transferred energy to) the other. Energy due to motion is called kinetic
energy and is represented by Ek. For an object moving in a straight line, the kinetic energy
is one-half the product of the mass and the square of the speed:
Ek = ½mu2
(3.4)
Where m = mass of the object
u = speed of object
If the two billiard balls mentioned above were studied in outer space, where friction due to
their collisions with air molecules or the surface of a pool table would be negligible, careful
measurements would reveal that their total kinetic energy would be the same before and
after they collided. This is an example of the law of conservation of energy, which states
that energy cannot be created or destroyed under the usual conditions of everyday life.
Whenever there appears to be a decrease in energy somewhere, there is a corresponding
increase somewhere else.
Even when there is a great deal of friction, the law of conservation of energy still applies. If
you put a milkshake on a mixer and leave it there for 10 min, you will have a warm, rather
unappetizing drink. The whirling
74
mixer blades do work on (transfer energy to) the milkshake, raising its temperature. The
same effect could be produced by heating the milkshake, a fact which suggests that heating
also involves a transfer of energy. The first careful experiments to determine how much
work was equivalent to a given quantity of heat were done by the English physicist James
Joule (1818 to 1889) in the 1840s. In an experiment very similar to our milkshake example,
Joule connected falling weights through a pulley system to a paddle wheel immersed in an
insulated container of water. This allowed him to compare the temperature rise which
resulted from the work done by the weights with that which resulted from heating.
Units with which to measure energy may be derived from the SI base units of Table 1.2 by
using Eq. (3.4).
EXAMPLE 3.9 Calculate the kinetic energy of a Volkswagen Beetle of mass 844 kg (1860
lb) which is moving at 13.4 m s–1 (30 miles per hour).
Solution
Ek = ½mu2
= ½ × 8.44 kg × (13.4 m s–1) 2 = 7.58 × 104 kg m2 s–2
In other words the units for energy are derived from the SI base units kilogram for mass,
meter for length, and second for time. A quantity of heat or any other form of energy may
be expressed in kilogram meter squared per second squared. In honor of Joule’s pioneering
work this derived unit 1 kg m2 s–2 called the joule, abbreviated J. The Volkswagen in
question could do nearly 76 000 J of work on anything it happened to run into.
Another unit of energy still widely used by chemists is the calorie. The calorie used to be
defined as the energy needed to raise the temperature of one gram of water from 14.5°C to
15.5°C but now it is defined as exactly 4.184 J. The Calorie used by dieticians and others
for measuring the energy values of foods is actually a kilocalorie, i.e., 4.184 kJ. This
nutritional Calorie is distinguished by a capital C in its name.
Thermochemical Equations
Energy changes which accompany chemical reactions are almost always expressed by
thermochemical equations, such as
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
(25°C, 1 atm pressure)
ΔHm = –890 kJ mol–1
(3.5)
75
Here the ΔHm (delta H subscript m) tells us whether heat energy is released or absorbed
when the reaction occurs and also enables us to find the actual quantity of energy involved.
By convention, if ΔHm is positive, heat is absorbed by the reaction; i.e., it is endothermic.
More commonly, ΔHm is negative as in Eq. (3.5), indicating that heat energy is released
rather than absorbed by the reaction, and that the reaction is exothermic. This convention
as to whether ΔHm is positive or negative looks at the heat change in terms of the matter
actually involved in the reaction rather than its surroundings. In the reaction in Eq. (3.5),
the C, H, and O atoms have collectively lost energy and it is this loss which is indicated by
a negative value of ΔHm.
It is important to notice that ΔHm is not an energy but rather an energy per unit amount. Its
units are not kilojoules but kilojoules per mole (hence the m subscript in ΔHm). This is
necessary because the quantity of heat released or absorbed by a reaction is proportional to
the amount of each substance consumed or produced by the reaction. Thus Eq. (3.5) tells us
that 890.4 kJ of heat energy is given off for every mole of CH4 which is consumed.
Alternatively, it tells us that 890.4 kJ is released for every mole of 2H2O produced, i.e., for
every 2 mol H2O produced. Seen in this way, ΔHm is a conversion factor enabling us to
calculate the heat absorbed when a given amount of substance is consumed or produced. If
q is the quantity of heat absorbed and n is the amount of substance involved, then
H m 
q
(3.6)
n
EXAMPLE 3.10 How much heat energy is obtained when 1 kg of ethane gas, C2H6, is
burned in oxygen according to the equation:
2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l)
ΔHm = –3083 kJ mol–1
(3.7)
Solution The mass of C2H6 is easily converted to the amount of C2H6 from which the heat
energy q is easily calculated by means of Eq. (3.6). The value of ΔHm is –3083 kJ per mole
of 2 mole of 2 C2H6, i.e., per 2 mol C2H6. The road map is
Note: By convention a negative value of q corresponds to a release of heat energy by the
matter involved in the reaction.
76
The quantity ΔHm is usually referred to as an enthalpy change per mole. In this context
the symbol Δ (delta) signifies change in” while H is the symbol for the quantity being
changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15.
For the moment we can think of it as a property of matter which increases when matter
absorbs energy and decreases when matter releases energy.
It is important to realize that the value of ΔHm given in thermochemical equations like (3.5)
or (3.7) depends on the physical state of both the reactants and the products. Thus, if water
were obtained as a liquid instead of a gas in the reaction in Eq. (3.5), the value of ΔHm
would be different from -890.4 kJ mol–1. It is also necessary to specify both the temperature
and pressure since the value of ΔHm depends very slightly on these variables. If these are
not specified [as in Eq. (3.7)] they usually refer to 25°C and to normal atmospheric
pressure.
Two more characteristics of thermochemical equations arise from the law of conservation
of energy. The first is that writing an equation in the reverse direction changes the sign of
the enthalpy change. For example,
H2O(l) → H2O(g)
ΔHm = 44 kJ mol–1
(3.8a)
tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This
corresponds to the fact that heat is absorbed from your skin when perspiration evaporates,
and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off
exactly the same quantity of heat.
H2O(g) → H2O(l)
ΔHm = –44 kJ mol–1
(3.8b)
To see why this must be true, suppose that ΔHm [Eq. (3.8a)] = 44 kJ mol–1 while ΔHm [Eq.
(3.8b)] = –50.0 kJ mol–1. If we took 1 mol of liquid water and allowed it to evaporate, 44
kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be
given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ
of heat which had been created from nowhere! This would violate the law of conservation
of energy. The only way the problem can he avoided is for ΔHm of the reverse reaction to
be equal in magnitude but opposite in sign from ΔHm of the forward reaction. That is,
ΔHm forward = –ΔHm reverse
Hess’ Law
Perhaps the most useful feature of thermochemical equations is that they can be combined
to determine ΔHm values for other chemical reactions. Consider, for example, the following
two-step sequence. Step 1 is reaction of 1 mol C(s) and 0.5 mol O2(g) to form 1 mol CO(g):
C(s) + ½O2(g) → CO(g)
ΔHm = –110.5 kJ mol–1 = ΔH1
(Note that since the equation refers to moles, not molecules, fractional coefficients are
permissible.) In step 2 the mole of CO reads with an additional 0.5 mol O2 yielding 1 mol
CO2:
CO(g) + ½O2(g) → CO2(g) ΔHm = –283.0 kJ mol–1 = ΔH2
77
The net result of this two-step process is production of 1 mol CO2 from the original 1 mol
C and 1 mol O2 (0.5 mol in each step). All the CO produced in step 1 is used up in step 2.
On paper this net result can be obtained by adding the two chemical equations as though
they were algebraic equations. The CO produced is canceled by the CO consumed since it
is both a reactant and a product of the overall reaction
Experimentally it is found that the enthalpy change for the net reaction is the sum of the
enthalpy changes for steps 1 and 2:
ΔHnet = –110.5 kJ mol–1 + (–283.0 kJ mol–1) + –393.5 kJ mol–1 = ΔH1 + ΔH2
That is, the thermochemical equation
C(s) + O2(g) → CO2(g)
ΔHm = –393.5 kJ mol–1
is the correct one for the overall reaction.
In the general case it is always true that whenever two or more chemical equations can be
added algebraically to give a net reaction, their enthalpy changes may also be added to
give the enthalpy change of the net reaction.
This principle is known as Hess' law. If it were not true, it would be possible to think up a
series of reactions in which energy would be created but which would end up with exactly
the same substances we started with. This would contradict the law of conservation of
energy.
Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out
experimentally, as the next example shows.
EXAMPLE 3.11 Acetylene (C2H2) cannot be prepared directly from its elements
according to the equation
2C(s) + H2(g) → C2H2(g)
(3.9)
Calculate ΔHm for this reaction from the following thermochemical equations, all of which
can be determined experimentally:
C(s) + O2(g) → CO2(g)
H2(g) + ½O2(g) → H2O(l)
C2H2(g) + 52 O2(g) → 2CO2(g) + H2O(l)
ΔHm = –393.5 kJ mol–1 (3.10a)
ΔHm = –285.8 kJ mol–1 (3.10b)
ΔHm = –1299.8 kJ mol–1 (3.10c)
Solution We use the following strategy to manipulate the three experimental equations so
that when added they yield Eq. (3.9):
a) Since Eq. (3.9) has 2 mol C on the left, we multiply Eq. (3.10a) by 2.
78
b) Since Eq. (3.9) has 1 mol H2 on the left, we leave Eq. (3.10b) unchanged.
c) Since Eq. (3.9) has 1 mol C2H2 on the right, whereas there is 1 mol C2H2 on the left of
Eq. (3.10c) we write Eq. (3.10c) in reverse.
We then have
2 C(s) + O 2 (g)  CO 2 (g)
H 2 (g) +
1
2
O 2 (g)  H 2 O(l )
H m  2 (  393.5) kJ mol-1
H m  285.8 kJ mol-1
2 CO 2 (g) + H 2 O(l )  C 2 H 2 (g) + 52 O 2 (g) H m  (  1299.8 kJ) mol-1
2 C(s) + H 2 (g) + 2 12 O 2 (g)  C 2 H 2 (g) + 52 O 2 (g)
H m  (  787.0  285.8 + 1299.8) kJ mol-1  227.0 kJ mol-1
Thus the desired result is
2C(s) + H2(g) → C2H2(g)
ΔHm = 227.0 kJ mol–1
3.4 STANDARD ENTHALPIES OF FORMATION
By now chemists have measured the enthalpy changes for so many reactions that it would
take several large volumes to list all the thermochemical equations. Fortunately Hess’ law
makes it possible to list a single value, the standard enthalpy of formation ΔHf, for each
compound. The standard enthalpy of formation is the enthalpy change when 1 mol of a
pure substance is formed from its elements. Each element must be in the physical and
chemical form which is most stable at normal atmospheric pressure and a specified
temperature (usually 25°C).
For example, if we know that ΔHf[H2O(l)] = –285.8 kJ mol–1, we can immediately write the
thermochemical equation
H2(g) + ½O2(g) → H2O(l)
ΔHm = –285.8 kJ mol–1
(3.11)
The elements H and O appear as diatomic molecules and in gaseous form because these are
their most stable chemical and physical states. Note also that 285.8 kJ are given off per
mole of H2O(l) formed. Equation (3.11) must specify formation of 1 mol H2O(l), and so the
coefficient of O2 must be ½.
In some cases, such as that of water, the elements will react directly to form a compound,
and measurement of the heat absorbed serves to determine ΔHf. Quite often, however,
elements do not react directly with each other to form the desired compound, and ΔHf must
be calculated by combining the enthalpy changes for other reactions. A case in point is the
gas acetylene, C2H2. In Example 3.11 we used Hess’ law to show that the thermochemical
equation
2C(s) + H2(g) → C2H2(g)
ΔHm = 227.0 kJ mol–1
79
TABLE 3.1 Some Standard Enthalpies of Formation at 25°C.
H f kcal
Compound
AgCl(s)
AgN 3 (s)
Ag 2 O(s)
Al 2 O 3 (s)
Br2 (l)
Br2 (g)
C(s), graphite
C(s), diamond
CH 4 (g)
CO(g)
CO 2 (g)
C 2 H 2 (g)
C 2 H 4 (g)
C 2 H 6 (g)
C 6 H 6 (l)
CaO(s)
CaCO 3 (s)
CuO(s)
Fe 2 O 3 (s)
HBr(g)
HCl(g)
HI(g)
mol
-1
- 127.0
+ 310.3
- 31.0
- 1675.7
0.0
+ 31.0
0.0
+ 1.9
- 74.8
- 110.5
- 393.5
+ 226.9
+ 52.6
- 84.5
+ 49.1
- 635.5
- 1207.8
- 157.3
- 822.2
- 36.4
- 92.3
+ 26.4
H f kcal
mol
-1
- 30.35
+ 66.79
- 7.41
- 400.50
0.00
+ 7.41
0.00
+ 0.45
- 17.88
- 26.41
- 94.05
+ 54.23
+ 12.57
- 20.20
+ 11.74
- 151.89
- 288.67
- 37.60
- 196.51
- 8.70
- 22.06
+ 6.31
H f kcal
Compound
mol
-1
- 241.8
H 2 O(g)
- 285.8
H 2 O(l)
H 2 O 2 (l) - 187.8
- 20.2
H 2S(g)
- 90.8
HgO(s)
0.0
I 2 (s)
+ 62.3
I 2 (g)
KCl(s)
- 435.9
KBr(s)
- 392.0
MgO(s) - 601.8
NH 3 (g)
- 46.1
NO(g)
+ 90.3
NO 2 (g)
+ 33.8
N 2 O 4 (g)
+ 9.2
NF3 (g)
- 124.7
NaBr(s) - 359.9
NaCl(s) - 411.0
O 3 (g)
+ 142.7
SO 2 (g)
- 296.8
SO 3 (g)
- 395.8
ZnO(s)
- 348.1
H f kcal
mol
-1
- 57.79
- 68.31
- 44.84
- 4.83
- 21.70
0.0
+ 14.89
- 104.18
- 93.69
- 143.83
- 11.02
+ 21.58
+ 8.08
+ 2.20
- 29.80
- 86.02
- 98.23
+ 34.11
- 70.94
- 94.60
- 83.20
was valid. Since it involves 1 mol C2H2 and the elements are in their most stable forms, we
can say that ΔHf.[C2H2(g)] = 227.0 kJ mol–1.
One further point arises from the definition of ΔHf. The standard enthalpy of formation for
an element in its most stable state must be zero. If we form mercury from its elements, for
example, we are talking about the reaction
Hg(l) → Hg(l)
Since the mercury is unchanged, there can be no enthalpy change, and ΔHf = 0 kJ mol–1.
Standard enthalpies of formation for some common compounds are given in Table 3.1.
These values may be used to calculate ΔHm for any chemical reaction so long as all the
compounds involved appear in the tables. To see how and why this may be done, consider
the following example.
EXAMPLE 3.12 Use standard enthalpies of formation to calculate ΔHm for the reaction
2CO(g) + O2(g) → 2CO2(g)
80
Solution We can imagine that the reaction takes place in two steps, each of which involves
only a standard enthalpy of formation. In the first step CO (carbon monoxide) is
decomposed to its elements:
2CO(g) → 2C(s) + O2(g)
ΔHm = ΔHf
(3.12)
Since this is the reverse of formation of 2 mol CO from its elements, the enthalpy change is
ΔH1 = 2 × {–ΔHf [CO(g)]} = 2 × [– (–110.5 kJ mol–1)] = +221.0 kJ mol–1
In the second step the elements are combined to give 2 mol CO 2 (carbon dioxide):
2C(s) → 2O2(g) + CO2(g)
ΔHm = ΔH2
(3.13)
In this case
ΔH2 = 2 × ΔHf [CO(g)] = 2 × (393.5 kJ mol–1) = 787.0 kJ mol–1
You can easily verify that the sum of Eqs. (3.12) and (3.13) is
2CO(g) + 2 O2(g) → CO2(g) ΔHm = ΔHnet
Therefore
ΔHnet = ΔH1 + ΔH2 = 221.0 kJ mol–1 – 393.5 kJ mol–1 = – 172.5 kJ mol–1
Note carefully how Example 3.12 was solved. In step 1 the reactant compound CO(g) was
hypothetically decomposed to its elements. This equation was the reverse of formation of
the compound, and so ΔH1 was opposite in sign from ΔHf. Step 1 also involved 2 mol
CO(g) and so the enthalpy change had to be doubled. In step 2 we had the hypothetical
formation of the product CO2(g) from its elements. Since 2 mol were obtained, the enthalpy
change was doubled but its sign remained the same.
Any chemical reaction can be approached similarly. To calculate ΔHm we add all the ΔHf
values for the products, multiplying each by the appropriate coefficient, as in step 2 above.
Since the signs of ΔHf for the reactants had to be reversed in step 1, we subtract them,
again multiplying by appropriate coefficients. This can he summarized by the equation
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
(3.14)
The symbol Σ means “the sum of.” Since ΔHf values are given per mole of compound, you
must be sure to multiply each ΔHf by an appropriate coefficient derived from the equation
for which ΔHm is being calculated.
81
EXAMPLE 3.13 Use Table 3.1 to calculate ΔHm for the reaction
4NH3(g) 5O2(g) → 6H2O(g) + 4NO(g)
Solution Using Eq. (3.14), we have
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
= [6 ΔHf (H2O) + 4 ΔHf (NO)] – [4 ΔHf (NH3) + 5 ΔHf (O2)]
= 6(–241.8) kJ mol–1 + 4(90.3) kJ mol–1 – 4(–46.1 kJ mol–1) – 5 × 0
= –1450.8 kJ mol–1 + 361.2 kJ mol–1 + 184.4 kJ mol–1
= –905.2 kJ mol–1
Note that we were careful to use ΔHf [H2O(g)] not ΔHf [H2O(l)]. Even though water vapor
is not the most stable form of water at 25°C, we can still use its ΔHf value. Also the
standard enthalpy of formation of the element O2(g) is zero by definition. Obviously it
would be a waste of space to include it in Table 3.1.
3.5 SOLUTIONS
In the laboratory, in your body, and in the outside environment, the majority of chemical
reactions take place in solutions. Macroscopically a solution is defined as a homogeneous
mixture of two or more substances, that is, a mixture which appears to be uniform
throughout. On the microscopic scale a solution involves the random arrangement of one
kind of atom or molecule with respect to another.
There are a number of reasons why solutions are so often encountered both in nature and in
the laboratory. The most common type of solution involves a liquid solvent which
dissolves a solid solute. (The term solvent usually refers to the substance present in greatest
amount. There may be more than one solute dissolved in it.) Because a liquid adopts the
shape of its container but does not expand to fill all space available to it, liquid solutions
are convenient to handle. You can easily pour them from one container to another, and their
volumes are readily measured using graduated cylinders, pipets, burets, volumetric flasks,
or other laboratory glass-ware. Moreover, atoms or molecules of solids dissolved in a liquid
are close together but still able to move past one another. They contact each other more
frequently than if two solids were placed next to each other. This “intimacy” in liquid
solutions often facilitates chemical reactions.
Concentration
Since solutions offer a convenient medium for carrying out chemical reactions, it is often
necessary to know how much of one solution will react
82
with a given quantity of another. Examples earlier in this chapter have shown that the
amount of substance is the quantity which determines how much of one material will react
with another. The ease with which solution volumes may be measured suggests that it
would be very convenient to know the amount of substance dissolved per unit volume of
solution. Then by measuring a certain volume of solution, we would also be measuring a
certain amount of substance.
The concentration c of a substance in a solution (often called molarity) is the amount of
the substance per unit volume of solution:
Concentration of solute 
amount of solute
volume of solute
csolute 
nsolute
Vsolute
Usually the units moles per cubic decimeter (mol dm–3) or moles per liter (mol liter–1) are
used to express concentration.
If a pure substance is soluble in water, it is easy to prepare a solution of known
concentration. A container with a sample of the substance is weighed accurately, and an
appropriate mass of sample is poured through a funnel into a volumetric flask, as shown in
Fig. 3.3. The container is then reweighed. Any solid adhering to the funnel is rinsed into the
flask, and water is added until the flask is about three-quarters full. After swirling the flask
to dissolve the solid, water is added carefully until the bottom of the meniscus coincides
with the calibration mark on the neck of the flash.
EXAMPLE 3.14 A solution of KI was prepared as described above. The initial mass of the
container plus KI was 43.2874 g, and the final mass after pouring was 30.1544 g. The
volume of the flask was 250.00 ml. What is the concentration of the solution?
Solution The concentration can be calculated by dividing the amount of solute by the
volume of solution [Eq. (3.15)]:
n
cKI  KI
V
We obtain nKI from the mass of KI added to the flask:
m KI  43.2874 g  30,1544 g  13.1330 g
nKI  13.1330 g 
1 mol
-2
166.00 g
 7.9114  10 mol
The volume of solution is 250.00 ml, or
Vsolution  250.00 cm3 
1 dm3
3
10 cm
-1
3
 2.5000  10 dm3
Thus
cKI 
nKI
Vsolution

17.9114  10-2 mol
250.00  10 dm
-1
3
-1
 3.1645  10 mol dm-3
83
Figure 3.3 The preparation of a sodium chloride solution of concentration 1.000 mol dm–3. (a)
A sample of pure solid NaCl is firs weighed with its container. (b) Most of the sample is now
transferred into the volumetric flask. (c) The container and remaining solid NaCl are again
weighed. Subtraction yields mNaCl = 58.44 g. Thus n NaCl = 1.000 mol. (d) Any solid remaining
in the funnel is washed down into the body of the flask. (e) The flask is now filled to about 80
percent capacity and the solid allowed to dissolve. The flask is shaken to achieve a uniform
solution. (f) Solvent is now added carefully until the bottom of the me- niscus coincides with
the mark on the flask. The flask is again shaken to achieve a uniform solution. Since the
volume of the solution is 1.000 dm3, the concentration is 1.000 mol/1.000 dm3 = 1.000 mol dm–
3.
Note that the definition of concentration is entirely analogous to the definitions of density,
molar mass, and stoichiometric ratio that we have previously encountered. Concentration
will serve as a conversion factor relating the volume of solution to the amount of dissolved
solute.
concentration
c
Volume of solution 
 amount of solute V 
n
Because the volume of a liquid can be measured quickly and easily, concentration is a
much-used quantity. The next two examples show how this conversion factor may be
applied to commonly encountered solutions in which water is the solvent (aqueous
solutions).
EXAMPLE 3.15 An aqueous solution of HCl [represented or written HCl(aq)] has a
concentration of 0.1396 mol dm–3. If 24.71 cm³ (24.71 ml) of this solution is delivered from
a buret, what amount of HCl has been delivered?
84
Solution Using concentration as a conversion factor, we have
c
V 
n
nHCl  24.71 cm 3 
0.1396 mol
1 dm 3
The volume units will cancel if we supply a unity factor to convert cubic centimeters to
cubic decimeters:
3
0.1396 mol  1 dm 
3
nHCl  24.71 cm 


1 dm3
 10 cm 
0.1396 mol
 24.71 cm3 
1 dm3

1 dm3
103 cm3
 0.003 450 mol
The concentration units of moles per cubic decimeter are often abbreviated M, pronounced
molar. That is, a 0.1-M (one-tenth molar) solution contains 0.1 mol solute per cubic
decimeter of solution. This abbreviation is very convenient for labeling laboratory bottles
and for writing textbook problems; however, when doing calculations, it is difficult to see
that
1 dm3 × 1 M = 1 mol
Therefore we recommend that you always write the units in full when doing any
calculations involving solution concentrations. That is,
1 dm 3  1
mol
dm 3
 1mol
Problems such as Example 3.15 are easier for some persons to solve if the solution
concentration is expressed in millimoles per cubic centimeter (mmol cm–3) instead of moles
per cubic decimeter. Since the SI prefix m means 10–3, 1 mmol = 10–3 mol, and
1 mol
1 dm
3

1 mol
1 dm
3

1 dm3
3
10 cm
3

1 mmol
-3
10 mol

1 mmol
1 cm3
Thus a concentration of 0.1396 mol dm–3 (0.1396 M) can also be expressed as 0.1396 mmol
cm–3. Expressing the concentration this way is very convenient when dealing with
laboratory glassware calibrated in milliliters or cubic centimeters.
EXAMPLE 3.16 Exactly 25.0 ml NaOH solution whose concentration is 0.0974 M was
delivered from a pipet. (a) What amount of NaOH was present? (b) What mass of NaOH
would remain if all the water evaporated?
85
Solution
a) Since 0.0974 M means 0.0974 mol dm–3, or 0.0974 mmol cm–3, we choose the latter,
more convenient quantity as a conversion factor:
0.0974 mol
nNaOH  25.0 cm3 
 2.44 mmol
1 cm3
 2.44  10-3 mol
b) Using molar mass, we obtain
40.01 g
m NaOH  2.44  10-3 mol 
 9.76  10-2 g
1 mol
Note: The symbols nNaOH and mNaOH refer to the amount and mass of the solute NaOH,
respectively. They do not refer to the solution. If we wanted to specify the mass of aqueous
NaOH solution, the symbol mNaOH(aq) could be used.
Diluting and Mixing Solutions
Often it is convenient to prepare a series of solutions of known concentrations by first
preparing a single stock solution as described in Example 3.14. Aliquots (carefully
measured volumes) of the stock solution can then be diluted to any desired volume. In other
cases it may be inconvenient to weigh accurately a small enough mass of sample to prepare
a small volume of a dilute solution. Each of these situations requires that a solution be diluted to obtain the desired concentration.
EXAMPLE 3.17 A pipet is used to measure 50.0 ml of 0.1027 M HCl into a 250.00-ml
volumetric flask. Distilled water is carefully added up to the mark on the flask. What is the
concentration of the diluted solution?
Solution To calculate concentration, we first obtain the amount of HCI in the 50.0 ml (50.0
cm3) of solution added to the volumetric flask:
Dividing by the new volume gives the concentration
cHCl 
nHCl
V

5.14 mol
250.00 cm
3
 0.0205 mmol cm -3
Thus the new solution is 0.0205 M.
EXAMPLE 3.18 What volume of the solution prepared in Example 3.14 would be
required to make 50.00 ml of 0.0500 M KI?
86
Solution Using the volume and concentration of the desired solution, we can calculate the
amount of KI required. Then the concentration of the original solution (0.316 46 M) can be
used to convert that amount of KI to the necessary volume. Schematically
Thus we should dilute a 7.90-ml aliquot of the stock solution to 50.00 ml. This could be
done by measuring 7.90 ml from a buret into a 50.00-ml volumetric flask and adding water
up to the mark.
Titrations
When solutions are used quantitatively in the laboratory, titration is usually involved.
Titration is a technique used to determine the volume of one solution necessary to
consume exactly some reactant in another solu-
Figure 3.4 The technique of titration.
87
tion. As shown in Fig. 3.4, a measured volume of the solution to be titrated is placed in a
flask or other container. The titrant (the solution to be added) is placed in a buret. The
volume of titrant added can be determined by reading the level of liquid in the buret before
and after titration. This reading can usually be estimated to the nearest hundredth of a
milliliter, for example, 25.62 ml.
In Fig. 3.4 the solution to be titrated is colorless aqueous hydrogen peroxide, H2O2(aq),
which contains excess sulfuric acid H2SO4(aq). The titrant is purple-colored potassium
permanganate solution KMnO4(aq). The reaction which occurs is
2KMnO4(aq) + 5H2O2 + 3H2SO4 → 2MnSO4(aq) + 5O2(g) + K2SO4(aq) + 8H2O(l)
(3.16)
As the first few cubic centimeters of titrant flow into the flask, there is a large excess of
..
H2O2. The limiting reagent KMnO4 is entirely consumed, and its purple color disappears
almost as soon as it is added. Eventually, though, all the H2O2 is consumed. Addition of
just one more drop of titrant produces a lasting pink color due to unreacted KMnO4 in the
flask. This indicates that all the H2O2 has been consumed and is called the endpoint of the
titration. If more KMnO4 solution were added, there would be an excess of KMnO4 and the
color of the solution in the flask would get much darker. The darker color would show that
we had overtitrated, or overshot the endpoint, by adding more than enough KMnO4 to react
with all the H2O2.
In the titration we have just described, the intense purple color of permanganate indicates
the endpoint. Usually, however, it is necessary to add an indicator—a substance which
combines with excess titrant to produce a visible color or to form an insoluble substance
which would precipitate from solution. No matter what type of reaction occurs or how the
endpoint is detected, however, the object of a titration is always to add just the amount of
titrant needed to consume exactly the amount of substance being titrated. In the KMnO4—
H2O2 reaction [Eq. (3.16)], the endpoint occurs when exactly 2 mol KMnO4 have been
added from the buret for every 5 mol H2O2 originally in the titration flask. That is, at the
endpoint the ratio of the amount of KMnO4, added to the amount of H2O2 consumed must
equal the stoichiometric ratio
nKMnO (added from buret)
4
nH O (initially in flask)
2
2
 KMnO4  2 mol KMnO4

 H 2 O 2  5 mol H 2 O 2
 S
(3.17)
When the endpoint has been reached in any titration, the ratio of the amounts of substance
of the two reactants is equal to the stoichiometric ratio obtained from the appropriate
balanced chemical equation. Therefore we can use the stoichiometric ratio to convert from
the amount of one substance to the amount of another.
EXAMPLE 3.19 What volume of 0.05386 M KMnO4 would be needed to reach the
endpoint when titrating 25.00 ml of 0.1272 M H2O2.
88
Solution At the endpoint, Eq. (3.17) will apply, and we can use it to calculate the amount
of KMnO4 which must be added:
 KMnO4 

 H 2 O2 
nKMnO (added)  nH O (in flask)  S 
4
2
2
The amount of H2O2 is obtained from the volume and concentration:
mmol
nH O (in flask)  25.00 cm3  0.1272
2
cm3
2
 3.180 mmol H2 O2
Then
nKMnO (added)  3.180 mmol H 2 O 2 
2 mol KMnO 4
4
 3.180 mmol H 2 O 2 
5 mol H 2 O 2

10-3
10-3
2 mmol KMnO 4
5 mmol H 2 O 2
 1.272 mmol KMnO 4
To obtain VKMnO4(aq) we use the concentration as a conversion factor:
VKMnO
4
(aq )
 1.272 mmol KMnO4 
1 cm3
5.386  10 mmol KMnO4
-2
 23.62 cm3
Note that overtitrating [adding more than 23.62 cm 3 of KMnO4(aq) would involve an
excess (more than 1.272 mmol) of KMnO4.
Titration is often used to determine the concentration of a solution. In many cases it is not a
simple matter to obtain a pure substance, weigh it accurately, and dissolve it in a
volumetric flask as was done in Example 3.14. NaOH, for example, combines rapidly with
H2O and CO2 from the air, and so even a freshly prepared sample of solid NaOH will not
be pure. Its weight would change continuously as CO2(g) and H2O(g) were absorbed. Hydrogen chloride (HCl) is a gas at ordinary temperatures and pressures, making it very
difficult to handle or weigh. Aqueous solutions of both of these substances must be
standardized; that is, their concentrations must be determined by titration.
EXAMPLE 3.20 A sample of pure potassium hydrogen phthalate (KHC8H4O4) weighing
0.3421 g is dissolved in distilled water. Titration of the sample requires 27.03 ml
NaOH(aq). The titration reaction is
NaOH(aq) + KHC8H4O4(aq) → NaKC8H4O4(aq) + H2O
What is the concentration of NaOH(aq) ?
89
Solution To calculate concentration, we need to know the amount of NaOH and the
volume of solution in which it is dissolved. The former quantity could be obtained via a
stoichiometric ratio from the amount of KHC8H4O4, and that amount can be obtained from
the mass
KHC8H4O4
8 4 4
mKHC8H4O4 
 nKHC8H4O4 
 nNaOH
M
nNaOH  3.180 g 
S (NaOH/KHC H O )
1 mol KHC8 H 4 O 4
204.22 g

1 mol NaOH
1 mol KHC8 H 4 O 4
 1.674  10-3 mol NaOH  1.675 mmol NaOH
The concentration is
cNaOH 
or
nNaOH

1.675 mmol NaOH
V
0.06197 M.
27.03 cm 3
 0.06197 mmol cm -3
By far the most common use of titrations is in determining unknowns, that is, in
determining the concentration or amount of substance in a sample about which we initially
knew nothing. The next example involves an unknown that many persons encounter every
day.
EXAMPLE 3.21 Vitamin C tablets contain ascorbic acid (C6H8O6) and a starch “filler”
which holds them together. To determine how much vitamin C is present, a tablet can be
dissolved in water and with sodium hydroxide solution, NaOH(aq). The equation is
C6H8O6(aq) + NaOH(aq) → Na C6H7O6(aq) + H2O(l)
If titration of a dissolved vitamin C tablet requires 16.85 cm³ of 0.1038 M NaOH, how
accurate is the claim on the label of the bottle that each tablet contains 300 mg of vitamin
C?
Solution
The known volume and concentration allow us to calculate the amount of
NaOH(aq) which reacted with all the vitamin C. Using the stoichiometric ratio
 C6 H8 O 6
 NaOH
S
 1 mmol C6 H8 O6
  1 mmol NaOH

we can obtain the amount of C6H8O6. The molar mass converts that amount to a mass
which can be compared with the label. Schematically
90
C6H8O6
NaOH
6 8 6
VNaOH 
 nNaOH 
 nC6H8O6 
 m C6 H8 O 6
c
M
S(C H O /NaOH)
m C6 H8O6  16.85 cm3 
0.1038 mmol NaOH

1 cm3
1 mmol C6 H8 O6
1 mmol NaOH

176.1 mg
mmol C6 H8 O6
 308.0 mg
Note that the molar mass of C6H8O6
176.1 g
1 mol C6 H8O6

176.1 g
1 mol C6 H8O6

10-3
10
-3

176.1 g 10-3
-3
10 mol C6 H8O 6

176.1 mg
1 mmol C6 H8O 6
can be expressed in milligrams per millimole as well as in grams per mole.
The 308.0 mg obtained in this example is in reasonably close agreement with the
manufacturer’s claim of 300 mg. The tablets are stamped out by machines, not weighed
individually, and so some variation is expected.
SUMMARY
This chapter has been concerned with the amounts of substances which participate in
chemical reactions, with the quantities of heat given off or absorbed when reactions occur,
and with the volumes of solutions which react exactly with one another. These seemingly
unrelated subjects have been discussed together because many of the calculations
invo1ving them are almost identical in form. The same is true of the density calculations
described in Chap. 1 (Secs. 1.3 and 1.4) and of the calTABLE 3.2 Summary of Related Quantities and Conversion Factors.
Related Quantities
Volume ↔ mass
Conversion Factor
Density, ρ
Amount of substance ↔ mass
Molar Mass, M
Amount of substance ↔
number of particles
Avogadro constant, NA
Amount of X consumed or
produced ↔ amount of Y
consumed or produced
Amount of X consumed or
produced ↔ quantity of heat
absorbed during reaction
Volume of solution ↔
amount of solute
Stoichiometric ratio,
S(Y/X)
Definition
Road Map
m
V
m
M
n
N
NA 
n

V 
 m
n
S (Y/X)  Y
nX
S (Y/X)
nX 
nY

ΔHm for thermochemical equation
H m 
Concentration of
solute, cX
cX 
nX
V
q
nX
M
n 
m
NA
n 

N
H m
nX 
q
cX
V 
 nX
91
culations involving molar mass and the Avogadro constant in Chap. 2 (Secs. 2.4 and 2.5).
In each case one quantity is defined as the ratio of two others. The first quantity serves as a
conversion factor relating the other two. A summary of the relationships and conversion
factors we have encountered so far is given in Table 3.2.
An incredible variety of problems can he solved using the conversion factors in Table 3.2.
Sometimes only one factor is needed, but quite often several are applied in sequence, as in
Example 3.21. In solving such problems, it is necessary first to think your way through,
perhaps by writing down a road map showing the relationships among the quantities given
in the problem. Then you can apply conversion factors, making sure that the units cancel,
and calculate the result.
The examples and end-of-chapter problems in this and the preceding two chapters should
give you some indication of the broad applications of the problem-solving techniques we
have developed here. Once you have mastered these techniques, you will be able to do a
great many useful computations which are related to problems in the chemical laboratory,
in everyday life, and in the general environment. You will find that the same type of
calculations, or more complicated problems based on them, will he encountered again and
again throughout your study of chemistry and other sciences.