Chemistry
Mole calculations
GCSE & A-level
The one skill that turns up on every chemistry paper. Every equation you need — mass, solutions, gases, yield and atom economy — in one place. Flip on Test yourself to check you can recall each one under pressure.
Reading mode — know which equation each quantity needs.
| To find… | Use | Units / notes |
|---|---|---|
| Moles, from a massGCSE | n = m ÷ M | m in g; M = Mr in g mol−1 |
| Mass, from molesGCSE | m = n × M | rearrangement of the above |
| Moles, in a solutionGCSE | n = c × V | V in dm3; c in mol dm−3 |
| ConcentrationGCSE | c = n ÷ V | mol dm−3 (× M to get g dm−3) |
| Moles of a gas (at RTP)GCSE | n = V ÷ 24 | V in dm3 at RTP (÷ 24000 if V in cm3) |
| Number of particlesGCSE | N = n × L | L = 6.02×1023 mol−1 (Avogadro) |
| Moles of a gas (any conditions)A-level | n = pV ÷ RT | pV = nRT; p/Pa, V/m3, T/K, R = 8.31 |
| Percentage yieldGCSE | (actual ÷ theoretical) × 100 | compare moles (or mass) of product |
| Atom economyGCSE | (Mr of desired product ÷ ΣMr of reactants) × 100 | how much of the reactant mass ends up wanted |
| % by mass of an elementGCSE | (n × Ar ÷ Mr) × 100 | n = number of that atom in the formula |
Unit conversions worth memorising
- 1 dm3 = 1000 cm3 — divide cm3 by 1000 to get dm3.
- For the ideal-gas equation: 1 m3 = 1000 dm3, and °C + 273 = K.
- 1 tonne = 1×10⁶ g; 1 kg = 1000 g.
The method examiners reward
- Work out moles of what you're given.
- Use the balanced equation's ratio to get moles of the target.
- 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.
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