When balancing equations, start with the atoms that show up in only one compound on each side. Figure out those coefficients first, and then work back and forth from the reactants side to the products side to find the coefficients of each of the other compounds.

Example 1: Balance the following equation:

… C_{5}H_{12} + … O_{2} → … CO_{2} + … H_{2}O

Solution: This reaction showing the combustion of pentane is not too complicated. Notice that carbon appears in only one compound on each side, so it’s a good place to start. The five carbon atoms on the left need to be balanced by five carbon atoms on the right:

1 C_{5}H_{12} + … O_{2} → 5 CO_{2} + … H_{2}O

There are 12 hydrogen atoms on the left, so balance these on the right:

1 C_{5}H_{12} + … O_{2} → 5 CO_{2} + 6 H_{2}O

Finally, count up all the oxygen atoms on the right (there are 16 of them!) and balance with the O_{2} on the left:

1 C_{5}H_{12} + 8 O_{2} → 5 CO_{2} + 6 H_{2}O

Example 2: Balance the following equation:

… C_{6}H_{14} + … O_{2} → … CO_{2} + … H_{2}O

Solution: Changing the pentane to hexane has made this equation more complicated. Let’s see why. First, let’s balance the carbon and hydrogen as before:

1 C_{6}H_{14} + … O_{2} → 6 CO_{2} + 7 H_{2}O

There are now 19 oxygen atoms on the right side. How do you balance this on the left side? Some teachers will let you use fractions for your coefficients, and 19/2 is the correct fraction in this case. But most teachers want you to use only whole numbers for the coefficients. If you encounter a fraction, you need to multiply the entire equation by the denominator to finish the problem. Here, we multiply the equation above by 2:

2 C_{6}H_{14} + … O_{2} → 12 CO_{2} + 14 H_{2}O

Notice there are now 38 oxygen atoms on the right side, and the last step is simple:

2 C_{6}H_{14} + 19 O_{2} → 12 CO_{2} + 14 H_{2}O

Here’s a trick you can apply when the equation is a double replacement reaction:

Example 3: Balance the following reaction:

… Na_{2}SO_{4} + … Al(NO_{3})_{3} → … NaNO_{3} + … Al_{2}(SO_{4})_{3}

Solution: Let’s start by balancing the aluminum:

… Na_{2}SO_{4} + 2 Al(NO_{3})_{3} → … NaNO_{3} + 1 Al_{2}(SO_{4})_{3}

Now instead of balancing the sulfur atoms, we can save time by balancing the sulfate ions instead. We treat them as a group rather than as separate atoms. There are three sulfates on the right side, so we need three on the left:

3 Na_{2}SO_{4} + 2 Al(NO_{3})_{3} → … NaNO_{3} + 1 Al_{2}(SO_{4})_{3}

Note that you can only balance polyatomic ions like this when the ions do not change from reactant side to product side. Finally, you can see there are six sodium atoms on the left (or, six nitrate ions), so the final answer is:

3 Na_{2}SO_{4} + 2 Al(NO_{3})_{3} → 6 NaNO_{3} + 1 Al_{2}(SO_{4})_{3}