IGCSE Chemistry Mole Calculations, Step by Step (Every Question Type)

Ask any IGCSE Chemistry class which topic decides who's "good at chemistry" and you'll hear the same word: moles. It's the topic that turns a memorisation subject into a maths subject mid-year — and the one where students who learn the system overtake students who memorise examples. Here's the whole of mole calculations as one repeatable method, with worked examples for every question type the exam actually asks.

The only three formulas you need

The universal 4-step method

1. Balance the equation
   The coefficients are the ratio — everything depends on them. If the question gives the equation, check it's balanced anyway; if not, write and balance it first.
2. Convert the known substance to moles
   Take whatever the question gave you (grams, cm³ of gas, cm³ of solution) and use the matching formula from the table.
3. Apply the mole ratio
   Read the coefficients between your known and your target substance and scale: if the ratio is 2:1, halve; 1:3, triple.
4. Convert to the units the question wants
   Back to grams (× M_r), gas volume (× 24 dm³), or concentration (÷ volume in dm³). Then round sensibly and attach the unit.

Worked examples — the four exam species

1. Mass → mass. What mass of magnesium oxide forms when 4.8 g of magnesium burns completely? (2Mg + O₂ → 2MgO)
Moles of Mg = 4.8 ÷ 24 = 0.2 mol → ratio Mg:MgO is 2:2 = 1:1 → 0.2 mol MgO → mass = 0.2 × 40 = 8 g.

2. Mass → gas volume. What volume of CO₂ (at r.t.p.) is produced when 25 g of calcium carbonate decomposes? (CaCO₃ → CaO + CO₂)
M_r of CaCO₃ = 40 + 12 + 48 = 100 → moles = 25 ÷ 100 = 0.25 mol → ratio 1:1 → 0.25 mol CO₂ → volume = 0.25 × 24 = 6 dm³.

3. Solution → solution (titration). 25.0 cm³ of NaOH is neutralised by 20.0 cm³ of 0.10 mol/dm³ HCl. Find the concentration of the NaOH. (HCl + NaOH → NaCl + H₂O)
Moles of HCl = 0.10 × (20.0 ÷ 1000) = 0.002 mol → ratio 1:1 → 0.002 mol NaOH → concentration = 0.002 ÷ (25.0 ÷ 1000) = 0.08 mol/dm³.

4. Limiting reactant & percentage yield. When the question gives amounts of two reactants, convert both to moles, divide each by its coefficient — the smaller result is the limiting reactant, and every later step uses it. Percentage yield = (actual ÷ theoretical) × 100; the theoretical value is just the answer to a normal mole calculation.

Make it automatic

Moles yield to volume of practice like nothing else in chemistry: the method never changes, only the costume. Drill mole questions in the 0620 past-paper library with instant AI marking — the marking shows exactly which of the four steps you fumbled — and slot the topic into a proper revision timetable early, because the rest of quantitative chemistry (titrations, energetics, electrolysis calculations) is built on it. The full grade plan lives in the Chemistry 0620 A* guide.

FAQ

Moles stop being the hard topic the day you stop treating each question as new: balance, convert, ratio, convert back — four moves, every time, forever. Run the loop on real marked questions until it bores you, and you've just converted chemistry's most feared topic into its most reliable marks.