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Chemistry

Mole calculations

GCSE & A-level

Choose AQA GCSE 8462 or A-level 7405 before revising. The GCSE view contains only GCSE equations; the A-level view adds the ideal-gas equation. Flip on Test yourself to check you can recall each one under pressure.

AQA GCSE Chemistry only — the ideal-gas equation is deliberately excluded.

Moles, from a massGCSEEquationn =mMUnits / notesm in g; M = Mr in g mol−1
Mass, from molesGCSEEquationm = n × MUnits / notesrearrangement of the above
Moles, in a solutionGCSEEquationn = c × VUnits / notesV in dm3; c in mol dm−3
ConcentrationGCSEEquationc =nVUnits / notesmol dm−3 (× M to get g dm−3)
Moles of a gas (at RTP)GCSEEquationn =V24Units / notesV in dm3 at RTP (÷ 24000 if V in cm3)
Number of particlesGCSEEquationN = n × LUnits / notesL = 6.02×1023 mol−1 (Avogadro)
Percentage yieldGCSEEquationactualtheoretical× 100Units / notescompare moles (or mass) of product
Atom economyGCSEEquationMr of desired productΣMr of reactants× 100Units / noteshow much of the reactant mass ends up wanted
% by mass of an elementGCSEEquationn × ArMr× 100Units / notesn = number of that atom in the formula

Unit conversions worth memorising

  • 1 dm3 = 1000 cm3 — divide cm3 by 1000 to get dm3.
  • For volume conversions: 1 m3 = 1000 dm3.
  • 1 tonne = 1×10⁶ g; 1 kg = 1000 g.

The method examiners reward

  1. Work out moles of what you're given.
  2. Use the balanced equation's ratio to get moles of the target.
  3. Convert those moles to the quantity asked for.

Now put them to work

Reading the equations isn't the same as landing the marks — drill them until the method is automatic.

Free revision tool by Oscar Song Tutoring

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