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 reaction finishes when one of the reactants is all used up. The other reactant has nothing left to react with, so some of it is left over:

- the reactant that is all used up is called the limiting reactant
- the reactant that is left over is described as being in excess

The mass of product formed in a reaction depends upon the mass of the limiting reactant. This is because no more product can form when the limiting reactant is all used up.

The maximum mass of product formed in a reaction can be calculated using:

- the balanced equation
- the mass of the limiting reactant, and
- the
*A*_{r}(relative atomic mass) or*M*_{r}(relative formula mass) values of the limiting reactant and the product

12 g of magnesium reacts completely with excess hydrochloric acid to form magnesium chloride and hydrogen:

Mg(s) + 2HCl(aq) → MgCl_{2}(aq) + H_{2}(g)

Calculate the maximum mass of hydrogen that can be produced. (*A*_{r} of Mg = 24, *M*_{r} of H_{2} = 2)

Amount of magnesium =

Amount of magnesium =

= 0.5 mol

Looking at the equation, 1 mol of Mg forms 1 mol of H_{2}, so 0.5 mol of Mg forms 0.5 mol of H_{2}.

Mass of H_{2} = *M*_{r} × amount

= 2 × 0.5

= 1 g

- Question
1.0 g of calcium carbonate decomposes to form calcium oxide and carbon dioxide:

CaCO

_{3}(g) → CaO(s) + CO_{2}(g)Calculate the maximum mass of carbon dioxide that can be produced. (

*M*_{r}of CaCO_{3}= 100,*M*_{r}of CO_{2}= 44)Amount of calcium carbonate =

= 0.01 mol

Looking at the equation, 1 mol of CaCO

_{3}forms 1 mol of CO_{2}, so 0.01 mol of CaCO_{3}forms 0.01 mol of CO_{2}.Mass of CO

_{2}= relative formula mass × amount= 44 × 0.01

= 0.44 g

The balancing numbers in an equation can be worked out using masses found by experiment.

6.0 g of magnesium reacts with 4.0 g oxygen to produce magnesium oxide, MgO. Deduce the balanced equation for the reaction. (*A*_{r} of Mg = 24, *M*_{r} of O_{2} = 32)

Step | Action | Result | Result |
---|---|---|---|

1 | Write the formulae of the substances | Mg | O_{2} |

2 | Calculate the amounts | = 0.25 mol | = 0.125 mol |

3 | Divide both by the smaller amount | = 2 | = 1 |

This means that 2 mol of Mg reacts with 1 mol of O_{2}, so the left-hand side of the equation is:

2Mg + O_{2}

Then balancing in the normal way gives: 2Mg + O_{2} → 2MgO