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EXAMPLE 10.7
Gasoline is a mixture of hydrocarbons, including octane (C 8 H 18 ).
Combustion of octane produces CO 2 and H 2 O, with the release of en-
ergy. Write a balanced equation for this reaction.

SOLUTION
Step 1: Write the correct formulas in an unbalanced equation.

C 8 H 18 ( g) + O 2 ( g) → CO 2 ( g) + H 2 O( g)
(not balanced)
Step 2: Take an inventory.
Reactants: 8 C Products: 1 C
18 H 2 H
2 O 3 O FIGURE 10.8 Hydrocarbons are composed of the elements
hydrogen and carbon. Propane (C 3 H 8 ) and gasoline, which contain
(not balanced) octane (C 8 H 18 ), are examples of hydrocarbons. Carbohydrates are
composed of the elements hydrogen, carbon, and oxygen. Table
Step 3: Start with the compound with the most atoms
sugar, for example, is the carbohydrate C 12 H 22 O 11 . Generalizing, all
(generalization 1) and place coefficients to balance these
hydrocarbons and carbohydrates react completely with oxygen to
atoms.
yield CO 2 and H 2 O.
C 8 H 18 ( g) + O 2 ( g) → 8 CO 2 ( g) + 9 H 2 O( g)
(not balanced)
Redo the inventory.
predict what will happen in similar reactions. For example, you
Reactants: 8 C Products: 8 C have studied equations in the “Balancing Equations” section
18 H 18 H describing the combustion of methane (CH 4 ), propane (C 3 H 8 ),
and octane (C 8 H 18 ). Each of these reactions involves a hydro-
2 O 25 O
carbon, a compound of the elements hydrogen and carbon. Each
The O atoms are still unbalanced. There are 2 O atoms in the reactants hydrocarbon reacted with O 2 , yielding CO 2 and releasing the
but 25 O atoms in the products. Since the subscript cannot be changed,
energy of combustion. Generalizing from these reactions, you
it will take 12.5 O 2 to produce 25 oxygen atoms (generalization 3).
could predict that the combustion of any hydrocarbon would
involve the combination of atoms of the hydrocarbon molecule
C 8 H 18 ( g) + 12.5 O 2 ( g) → 8 CO 2 ( g) + 9 H 2 O( g)
with O 2 to produce CO 2 and H 2 O with the release of energy.
(balanced)
Such reactions could be analyzed by chemical experiments, and
Step 4: (a) An inventory will show that the atoms balance, the products could be identified by their physical and chemical
properties. You would find your predictions based on similar
Reactants: 8 C Products: 8 C
reactions to be correct, thus justifying predictions from such
18 H 18 H
generalizations. Butane, for example, is a hydrocarbon with the
25 O 25 O
formula C 4 H 10 . The balanced equation for the combustion of
(b) The coefficients are not in the lowest whole number butane is
ratio (one-half an O 2 does not exist). To make the lowest
possible whole number ratio, all coefficients are multi- 2 C 4 H 10 ( g) + 13 O 2 ( g) → 8 CO 2 ( g) + 10 H 2 O( g)
plied by 2. This results in a correct balanced equation of You could extend the generalization further, noting that the
combustion of compounds containing oxygen as well as carbon
and hydrogen also produces CO 2 and H 2 O (Figure 10.8). These
2 C 8 H 18 ( g) + 25 O 2 ( g) → 16 CO 2 ( g) + 18 H 2 O( g)
compounds are carbohydrates, composed of carbon and water.
(balanced)
Glucose, for example, was identified earlier as a compound
with the formula C 6 H 12 O 6 . Glucose combines with oxygen to
produce CO 2 and H 2 O, and the balanced equation is
GENERALIZING EQUATIONS C 6 H 12 O 6 (s) + 6 O 2 ( g) → 6 CO 2 ( g) + 6 H 2 O( g)
In the previous chapters, you learned that the act of classifying, Note that three molecules of oxygen were not needed from the
or grouping, something according to some property makes the O 2 reactant because the other reactant, glucose, contains six ox-
study of a large body of information less difficult. Generaliz- ygen atoms per molecule. An inventory of atoms will show that
ing from groups of chemical reactions also makes it possible to the equation is thus balanced.

10-9 CHAPTER 10 Chemical Reactions 259