Relative formula masses can be calculated and used in conservation of mass calculations. Calculations can be carried out to find out concentrations of solutions.

A balanced equation shows the amounts in moles of reactants that react and the amounts of products that are made. From these amounts, the masses of reactants and products can be calculated.

In a balanced equation, the coefficients (the numbers in front of the formulae) show the amounts of the reactants and products. If there is no coefficient, the amount is one mole.

For example, the equation below shows that one mole of zinc reacts with two moles of hydrogen acid to make one mole of zinc chloride and one mole of hydrogen molecules.

Zn(s) + 2HCl(aq) → ZnCl_{2}(aq)+ H_{2}(g)

Use the equation below to calculate the amount in moles of oxygen molecules that reacts with 2 mol of magnesium metal.

2Mg(s) + O_{2}(g) → 2MgO(s)

The equation shows that 2 mol of magnesium metal reacts with 1 mol of oxygen molecules.

The masses of substances shown in a balanced equation can be calculated using the equation:

mass = relative formula mass × amount

For example, in the equation 2Mg(s) + O_{2}(g) → 2MgO(s) the masses of the substances shown are:

Magnesium, Mg | 24 × 2 = 48 g |

Oxygen, O_{2} | 2 × 16 = 32 g |

Magnesium oxide, MgO | (24 + 16) × 2 = 80 g |

Balanced equations and relative formula mass values can be used to calculate:

- the mass of product made from a given mass of reactant
- the mass of reactant needed to make a given mass of product

In the reaction shown by the equation below, what mass of sulfur dioxide can be made from 16 g of sulfur? (*M*_{r} of SO_{2} = 64)

S(s) + O_{2}(g) → SO_{2}(g)

= 0.5 mol

The equation shows that 1 mol of sulfur reacts with 1 mol of oxygen molecules to make 1 mol of sulfur dioxide. This means that 0.5 mol of sulfur makes 0.5 mol of sulfur dioxide.

mass of SO_{2} = relative formula mass × amount

= 64 × 0.5

= 32 g

- Question
In the reaction shown by the equation below, what mass of nitrogen, N

_{2}, is needed to make 120 g of nitrogen monoxide, NO? (*M*_{r}of NO = 30 and*M*_{r}of N_{2}= 28)N

_{2}(g) + O_{2}(g) → 2NO(g)= 4.0 mol

1 mol of N

_{2}makes 2 mol of NO.This means that 2 mol of N

_{2}makes 4.0 mol of NO.Mass of 2 mol of N

_{2}= 2 × 28= 56 g