# Question #c5c15

Jan 16, 2014

It’s a step-by-step process. It can be tricky but, if your instructor allows fractions, here’s how you do it.

EXAMPLE

Balance the following equation:

C₆H₁₄ + O₂ → CO₂ + H₂O

The usual procedure is to balance all atoms other than H and O; then balance O; then balance H. Let’s see how it works with your equation.

Start with most complicated-looking molecule in the equation. This looks like C₆H₁₄.

BALANCE ALL ATOMS OTHER THAN H AND O.
Put a 1 in front of the C₆H₁₄.

1C₆H₁₄ + O₂ → CO₂ + H₂O

Since we have fixed six C atoms on the left, we need six C atoms on the right. We put a 6 in front of the CO₂.

1C₆H₁₄ + O₂ → 6CO₂ + H₂O

BALANCE O.
We can’t, because both O₂ and H₂O need coefficients. Therefore, we BALANCE H instead.

We have fixed 14 H atoms on the left, so we need 14 H atoms on the right. We have 2 H atoms in the H₂O. We put a 7 in front of the H₂O.

1C₆H₁₄ + O₂ → 6CO₂ + 7H₂O

BALANCE O.
We have 19 O atoms on the right, so we need 19 O atoms on the left. One O atom is half an O₂ molecule, so we need ¹⁹/₂ O₂ molecules. We put ¹⁹/₂ in front of the O₂.

1C₆H₁₄ + ¹⁹/₂O₂ → 6CO₂ + 7H₂O

Our balanced equation is therefore
C₆H₁₄ + ¹⁹/₂O₂ → 6CO₂ + 7H₂O