The Mole & Stoichiometry

🔬 Science Years 10–11 Higher

Counting particles by weighing — using the mole to link masses, gas volumes, concentrations, yields and formulae.

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Chemists can't count atoms one by one — they are far too small and far too many. Instead they count by weighing. The clever trick is the mole: a fixed, enormous number of particles (6.02 × 1023, the Avogadro constant) chosen so that the mass of one mole of a substance in grams equals its relative formula mass. Master the mole and almost every calculation in chemistry becomes the same three-step routine.

Why this matters Stoichiometry — working out how much reacts and how much you get — underpins the whole of quantitative chemistry: making medicines to the right dose, mixing concrete, refining fuel. In Paper 4 (Extended) these calculations are guaranteed marks, and the method is always the same.

Relative masses: Ar and Mr

The relative atomic mass (Ar) compares the average mass of an atom to 1⁄12 of a carbon-12 atom; it has no units. The relative molecular / formula mass (Mr) is just the Ar values added up for the formula.

FormulaWorkingMr
H₂O(2×1) + 1618
CO₂12 + (2×16)44
CaCO₃40 + 12 + (3×16)100
Ca(OH)₂40 + 2×(16+1)74

The mole triangle

The single most useful relationship is moles = mass ÷ Mr. Rearranged: mass = moles × Mr, and Mr = mass ÷ moles. Cover the quantity you want in the triangle to read off the formula.

massmolM₁mass = mol × M₁
Cover the unknown: moles = mass ÷ Mr.
Worked: moles from mass. How many moles are in 25 g of calcium carbonate, CaCO₃ (Mr = 100)?
moles = mass ÷ Mr = 25 ÷ 100 = 0.25 mol.

Reacting masses from balanced equations

A balanced equation gives the mole ratio of reactants to products. The routine: (1) write the balanced equation, (2) turn the known mass into moles, (3) use the ratio, (4) turn moles back into a mass.

2Mg + O₂ → 2MgO2 mol Mg2 mol MgO= 48 g= 80 gratio 2 : 2 = 1 : 1
Set the known mass over the ratio, then scale.
Worked: reacting mass. What mass of magnesium oxide forms when 12 g of magnesium burns? 2Mg + O₂ → 2MgO (Ar: Mg = 24, MgO Mr = 40).
moles Mg = 12 ÷ 24 = 0.5 mol. Ratio Mg : MgO = 1 : 1, so moles MgO = 0.5 mol. mass MgO = 0.5 × 40 = 20 g.

Concentration, gas volume, yield and formulae

Worked: empirical formula. A compound is 4.8 g magnesium combined with 3.2 g oxygen.
Mg: 4.8 ÷ 24 = 0.2 mol. O: 3.2 ÷ 16 = 0.2 mol. Ratio 0.2 : 0.2 = 1 : 1, so the empirical formula is MgO.
Common exam mistake: forgetting to convert cm³ to dm³. Concentration must use dm³, so divide any volume in cm³ by 1000 before using moles = concentration × volume. And always balance the equation first — the mole ratio comes from the big numbers in front, not the small subscripts.

Exam technique — for “Calculate” command words, always show working: write the equation, then mass → moles → ratio → moles → answer. Method marks are awarded even if the final number slips. Memory hook: “Mr divides, Mr multiplies” — divide by Mr to get moles, multiply by Mr to get mass.

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