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Step 2: Inventory the number of each kind of atom on both Reaction:
sides of the unbalanced equation. In the example Methane reacts with oxygen to yield
there are carbon dioxide and water
Reactants: 1 C Products: 1 C Balanced equation:
CO 2 + 2 H 2 O
CH 4 + 2 O 2
4 H 2 H
2 O 3 O Sketches representing molecules:

This shows that the H and O are unbalanced.
+ +
Step 3: Determine where to place coefficients in front of for-
mulas to balance the equation. It is often best to focus
on the simplest thing you can do with whole number
ratios. The H and the O are unbalanced, for example,
Meaning:
and there are 4 H atoms on the left and 2 H atoms on
1 molecule 2 molecules 1 molecule 2 molecules
the right. Placing a coefficient 2 in front of H 2 O will of methane + of oxygen of carbon + of water
balance the H atoms: dioxide
CH 4 + O 2 → CO 2 + 2 H 2 O FIGURE 10.6 Compare the numbers of each kind of atom in
the balanced equation with the numbers of each kind of atom in
(not balanced)
the sketched representation. Both the equation and the sketch
Now take a second inventory: have the same number of atoms in the reactants and in the
products.
Reactants: 1 C Products: 1 C
4 H 4 H

2 O 4 O (O 2 + 2 O) 2. A correct formula of a compound cannot be changed by
altering the number or placement of subscripts. Changing
This shows the O atoms are still unbalanced with 2 on the left
subscripts changes the identity of a compound and the
and 4 on the right. Placing a coefficient of 2 in front of O 2 will
meaning of the entire equation.
balance the O atoms.

3. A coefficient in front of a formula multiplies everything in

CH 4 + 2 O 2 → CO 2 + 2 H 2 O the formula by that number.
(balanced) There are also a few generalizations that can be helpful for
success in balancing equations:
Step 4: Take another inventory to determine if the numbers

of atoms on both sides are now equal. If they are, 1. Look first to formulas of compounds with the most atoms,
determine if the coefficients are in the lowest possible and try to balance the atoms or compounds they were
whole number ratio. The inventory is now formed from or decomposed to.
2. Polyatomic ions that appear on both sides of the
Reactants: 1 C Products: 1 C
equation should be treated as independent units with

4 H 4 H a charge. That is, consider the polyatomic ion as a unit
while taking an inventory rather than the individual
4 O 4 O

atoms making up the polyatomic ion. This will save time
The number of each kind of atom on each side of the equa- and simplify the procedure.
tion is the same, and the ratio of 1:2 → 1:2 is the lowest possible 3. Both the crossover technique and the use of
whole number ratio. The equation is balanced, which is illus- fractional coeffi cients can be useful in fi nding the
trated with sketches of molecules in Figure 10.6. least common multiple to balance an equation. All
Balancing chemical equations is mostly a trial-and-error of these generalizations are illustrated in examples 10.5,
procedure. But with practice, you will find there are a few 10.6, and 10.7.
generalized “role models” that can be useful in balancing
The physical state of reactants and products in a reaction is
equations for many simple reactions. The key to success at
often identified by the symbols (g) for gas, (l) for liquid, (s) for
balancing equations is to think it out step by step while re-
solid, and (aq) for an aqueous solution (aqueous means “water”).
membering the following:
If a gas escapes, this is identified with an arrow pointing up (↑).
1. Atoms are neither lost nor gained, nor do they change A solid formed from a solution is identified with an arrow point-

their identity in a chemical reaction. The same kind and ing down (↓). The Greek symbol delta (Δ) is often used under or
number of atoms in the reactants must appear in the over the yield sign to indicate a change of temperature or other
products, meaning atoms are conserved. physical values.

10-7 CHAPTER 10 Chemical Reactions 257