Balancing Equations

The goal of balancing equations is to ensure that you have the same number of atoms for each element on both sides of the reaction. This ensures that we have written out the correct reaction and that our reaction formula will accurately predict the stoichiometric ratios of reactants and products. Remember that mass and charge are conserved, so however many atoms of a particular element we begin with we must also end with.

As an example, let’s consider the following unbalanced reaction, where we have propane reacting with oxygen to form carbon dioxide and water.

C₃H₈ + O₂ → CO₂ +H₂O

To balance the reaction, let’s begin by tallying up the number of atoms of each element we have in our reactants, in this case three carbon atoms, eight hydrogen atoms, and two oxygen atoms. Next, repeat the same process for the product compounds, on the right side of the equation. The products contain one carbon atom, two hydrogen atoms, and three oxygen atoms, an unbalanced equation. We will first pick one atom to balance. The easiest atom to balance will typically be an atom that is present in its elemental state (in this case, oxygen) so we will save that as the last atom to balance. Let’s begin with carbon. The imbalance of carbon is weighted to the left side of the reaction. Therefore, to balance it, we must add a coefficient to the product containing carbon, CO₂. If we had a ‘3’ coefficient before carbon dioxide, carbon will now be balanced, giving us this intermediate equation:

C₃H₈ + O₂ → 3CO₂ +H₂O

Remembering that we’ll save oxygen for last, let’s now balance hydrogen. Hydrogen is also imbalanced to the left side of the equation, meaning a coefficient will need to be added to the products. There are two hydrogen atoms in water and eight in propane, so balancing hydrogen will require adding a ‘4’ coefficient before water. Our not-yet balanced equation will now look like this:

C₃H₈ + O₂ → 3CO₂ +4H₂O

Our final step will be to balance oxygen. There are currently two oxygens on the left side of the equation and, with our new balancing coefficients, ten oxygens on the right side of the equation. Therefore, adding a ‘5’ coefficient before elemental oxygen will balance the oxygens across the reaction. This will give us a final balanced equation of:

C₃H₈ + 5O₂ → 3CO₂ +4H₂O

We can see that there are now three carbons, eight hydrogens, and ten oxygens on the reactant side, as well as three carbons, eight hydrogens, and ten oxygens on the product side. Reflecting back on our discussion of stoichiometric ratios, this means that the combustion of one mole of propane will require five moles of oxygen, and will produce as products three moles of carbon dioxide and four moles of water (and quite a bit of heat, not shown in the reaction).