The Law of Conservation of Mass states that matter can neither be created nor destroyed. Therefore, the # of atoms of each element must be equal on both sides of the equation.

The left half of the equation is called the “reactants” and the right side are the “products.” So we have 1 A on each side of equation and 2 B atoms on each side.

Coefficients are used in front of each formula to change the # of atoms. A coefficient is multiplied or distributed to all atoms that follow it.

Here are some other examples:

**Guidelines to Balance Chemical Equations:**

- The # of all atoms must be the same on both sides of the arrow.
- Begin with all NON oxygen and hydrogen atoms…those will almost always work themselves out. Also begin with the element which appears in only ONE reactant or product (this is usually the NON oxygen and hydrogen element(s)
- Use only COEFFICIENTS to balance the equation. DO NOT CHANGE SUBSCRIPTS.
- It is easier to balance polyatomic ions as units rather than as individual atoms.
- Make sure that you use the simplest combination of coefficients and double check that all # on both sides of the equation balance.

Let’s try another example:

C_{2}H_{6} + O_{2} → CO_{2} + H_{2}O

Notice that the # of carbon atoms in the reactants is 2 but there is only 1 C in the products. The # of atoms of oxygen and hydrogen are also uneven; so we have to balance the #s by adding coefficients.

First inclination is to look at the carbon atom

Add a coefficient of **2** in front of CO_{2} to balance the carbon atoms

Next turn to hydrogen atoms

Add coefficient of **3** in front of H2O to balance the hydrogen atoms

Now balance the remaining oxygen atoms

There is already a **2** in front of CO_{2} so that cannot change now. The only places where a coefficient can be added is in front of O_{2} and H_{2}O. However, notice how there is no value that when placed in front of H_{2}O and that when added to the 4 oxygen atoms in CO_{2}, would yield an even number. Why does it have to be even? Because any value placed in front of O_{2} would yield an even number. Additionally, adding a coefficient in front of H_{2}O would also disrupt the balance of H atoms.

Back to the drawing board! In the first trial, we left C_{2}H_{6} alone. Let’s try a **2** in front of C_{2}H_{6} and follow the same process to see if we get a better result.

Add a **4** in front of CO_{2} to balance carbon

Now balance the H in the products

And finally the O

Success! All the atoms are balanced. Double check each side

Reactant Atoms | Product Atoms | |

C | 4 | 4 |

H | 12 | 12 |

O | 14 | 14 |

It is important to apply the moto : “if at first you don’t succeed, right, right again” when balancing chemical equations because the first try might not work.

Also note that coefficients placed in front of a formula apply to and change the # of atoms in the *entire* formula, including polyatomic ions!