Relative formula mass and moles

GCSE Chemistry revision notes, key terms and practice questions.

The mole (Higher)

  • Chemical amounts are measured in moles (mol).
  • One mole of a substance contains the same number of particles as one mole of any other substance: the Avogadro constant, 6.02 × 10²³ per mole.
  • The mass of one mole of a substance in grams equals its relative formula mass (Mr) or relative atomic mass (Ar). For example, one mole of CO₂ has a mass of 44 g.
  • Number of moles = mass (g) ÷ Mr.

Reacting masses (Higher)

  • A balanced equation shows how many moles react. 2Mg + O₂ → 2MgO means 2 mol of Mg react with 1 mol of O₂ to make 2 mol of MgO.
  • To find a mass: work out moles of the known substance, use the equation's ratio to find moles of the other, then multiply by its Mr.
  • Example: 12 g of Mg is 12 ÷ 24 = 0.5 mol, which makes 0.5 mol of MgO = 0.5 × 40 = 20 g.

Balancing equations from masses (Higher)

  • Convert the masses of reactants and products to moles, then divide by the smallest number of moles to get the simplest whole-number ratio.
  • Example: 4.8 g Mg, 3.2 g O₂ and 8.0 g MgO are 0.2, 0.1 and 0.2 mol, a ratio of 2 : 1 : 2, so 2Mg + O₂ → 2MgO.

Limiting reactants (Higher)

  • In a reaction, one reactant may be completely used up. It is the limiting reactant, because it limits the amount of product that can form.
  • The other reactant is in excess: some of it is left over.
  • The mass of product depends on the amount of the limiting reactant.

Volumes of gases (Higher, separate sciences)

  • One mole of any gas occupies 24 dm³ at room temperature and pressure (20 °C and 1 atmosphere).
  • Volume of gas (dm³) = moles × 24. For example, 0.25 mol of any gas occupies 6 dm³ at RTP.

Empirical formulae (Edexcel and OCR)

  • The empirical formula is the simplest whole-number ratio of atoms of each element in a compound. For example, the empirical formula of glucose, C₆H₁₂O₆, is CH₂O.
  • To find it from masses: divide the mass of each element by its Ar to get moles, then divide by the smallest number of moles to get the simplest whole-number ratio.
  • Example: 2.4 g of magnesium reacts with 1.6 g of oxygen. Mg: 2.4 ÷ 24 = 0.1 mol; O: 1.6 ÷ 16 = 0.1 mol. The ratio is 1 : 1, so the empirical formula is MgO.
  • Core practical (Edexcel): heat a weighed piece of magnesium ribbon in a crucible, lifting the lid now and then, until it stops glowing. Reweigh to find the mass of oxygen that combined, then work out the empirical formula.
  • The molecular formula is a whole-number multiple of the empirical formula: divide the Mr by the mass of the empirical formula to find the multiple.

Key terms

Mole
The unit for amount of substance; one mole contains 6.02 × 10²³ particles.
Avogadro constant
The number of particles in one mole: 6.02 × 10²³.
Molar mass
The mass of one mole of a substance in grams, equal to its Mr.
Limiting reactant
The reactant that is completely used up, limiting the amount of product.
In excess
More than enough of a reactant; some is left over.
RTP
Room temperature and pressure: 20 °C and 1 atmosphere.
Molar gas volume
The volume of one mole of any gas at RTP: 24 dm³.
Empirical formula
The simplest whole-number ratio of atoms of each element in a compound.

Practise Relative formula mass and moles: 13 questions