# Balancing Chemical Equations with Fractions

Sometimes it is easier to balance chemical equations by using a fraction as a coefficient.

The is especially useful when you have a diatomic element like O_{2} and N_{2} on one side of the chemical equation, and you find yourself "stuck" trying to balance it.

Let's examine this technique by trying to balance the equation for the combustion of hexane.

The products of the combustion of hexane in the presence of oxygen are carbon dioxide and water.

C_{6}H_{14} + O_{2} ------> CO_{2} + H_{2}O

1. First we will balance the carbon atoms by placing a 6 in front of CO_{2} in the products.

C_{6}H_{14} + O_{2} ------> 6 CO_{2} + H_{2}O

2. Balance the hydrogen atoms by placing a 7 in front of H_{2}O in the products.

C_{6}H_{14} + O_{2} ------> 6 CO_{2} + 7 H_{2}O

3. Now we get "stuck" balancing the oxygen atoms.

Because oxygen is diatomic in the reactants , and there are an odd total number of oxygen atoms in the products , we cannot balance the equation at this point by simply using whole number coefficients. We can start over , or use a fraction.

There are 2 oxygen atoms on the reactant (left) side of the equation

There are 19 oxygen atoms on the product (right) side of the equation

In order to balance the number of atoms we can multiply the 2 atoms of oxygen in the reactants by 19/2.

2 x 19/2 = 19

Applying this to our nearly balanced equation we get:

C_{6}H_{14} + 19/2 O_{2} ------> 6 CO_{2} + 7 H_{2}O

The equation is now balanced.

Since we only want whole numbers as coefficients , we can now multiply the all the coefficients in the entire chemical equation by 2,

2 x (C_{6}H_{14} + 19/2 O_{2} ------> 6 CO_{2} + 7 H_{2}O)

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