The mole and stoichiometry (Higher tier)
The mole is the chemist's way of counting particles. One mole of any substance contains 6.02 × 10²³ particles (Avogadro's number) and has a mass equal to its relative formula mass (Mr) in grams.
Three core formulas
- n = m / Mr — moles from mass.
- n = c × V — moles from concentration (mol/dm³) and volume (dm³).
- n = V / 24 — moles of a gas at room temperature and pressure (24 dm³/mol).
Reading a balanced equation
The numbers in front of the formulas (the stoichiometric coefficients) tell you the mole ratio, not the gram ratio. For 2H₂ + O₂ → 2H₂O, two moles of hydrogen react with one mole of oxygen to give two moles of water.
Worked calculation — mass to mass
What mass of magnesium oxide is formed when 4.8 g of magnesium burns in excess oxygen? 2Mg + O₂ → 2MgO. Mr(Mg) = 24, Mr(MgO) = 40.
- n(Mg) = 4.8 / 24 = 0.20 mol.
- Mole ratio Mg:MgO = 2:2 = 1:1, so n(MgO) = 0.20 mol.
- m(MgO) = n × Mr = 0.20 × 40 = 8.0 g.
Percentage yield
% yield = (actual yield / theoretical yield) × 100. A reaction that should give 8.0 g but actually gives 6.4 g has a yield of 80 %. Yields fall short because of incomplete reactions, side reactions, or losses during transfer/filtration.
CCEA tip
Always show the moles step on a stoichiometry calculation. Even if the final answer is wrong, showing n = m / Mr correctly secures the M1 method mark.
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