Chemistry

The Complete Guide to GCSE Quantitative Chemistry — Moles Made Simple

Moles, reacting masses, concentration, percentage yield and atom economy — every calculation with worked examples.

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Quantitative chemistry is worth a significant number of marks on every GCSE Chemistry paper, and it's one of the most learnable areas — the same calculation types appear repeatedly, and once you know the method for each, you can score consistently. This guide covers every quantitative chemistry skill at GCSE with a clear method and worked example for each.

Relative Atomic Mass and Relative Formula Mass

The relative atomic mass (Ar) of an element is found on the periodic table — it's the larger of the two numbers shown for each element. For most elements at GCSE you use a rounded whole number: carbon = 12, hydrogen = 1, oxygen = 16, nitrogen = 14, chlorine = 35.5, sodium = 23, and so on.

The relative formula mass (Mr) of a compound is found by adding up the Ar values of all atoms in the formula.

Mr of H₂O: (2 × 1) + 16 = 18
Mr of NaCl: 23 + 35.5 = 58.5
Mr of CaCO₃: 40 + 12 + (3 × 16) = 100
Mr of H₂SO₄: (2 × 1) + 32 + (4 × 16) = 98

Get comfortable calculating Mr quickly — it's the foundation of every other quantitative calculation.

The Mole

A mole is simply a specific number of particles — 6.02 × 10²³ (Avogadro's number). The mole connects mass (which we can measure) to the number of atoms or molecules (which we can't directly measure).

One mole of any substance has a mass in grams equal to its relative formula mass. So one mole of water (Mr = 18) has a mass of 18 g. One mole of carbon (Ar = 12) has a mass of 12 g.

Moles = mass (g) ÷ Mr
Mass = moles × Mr
Mr = mass ÷ moles

The Mole Triangle

Draw a triangle with mass at the top, moles (n) at bottom left and Mr at bottom right. Cover the quantity you want to find: mass = n × Mr. Moles = mass ÷ Mr. Mr = mass ÷ n. This triangle works exactly like the speed-distance-time triangle and covers every moles calculation at Foundation level.

Reacting Masses — Using the Balanced Equation

This is the most commonly tested quantitative skill. Given a balanced equation, you can calculate the mass of any reactant or product if you know the mass of one other substance.

Method — always follow these steps:

  1. Write the balanced equation.
  2. Write the Mr values below the relevant substances.
  3. Multiply Mr by the stoichiometric coefficient (the big number in front of the formula) to get the ratio mass.
  4. Scale up or down proportionally to find the unknown mass.
Example: What mass of water is produced when 8 g of hydrogen burns?
2H₂ + O₂ → 2H₂O
Mr: (2×2)=4 — (2×18)=36
So 4 g H₂ produces 36 g H₂O
8 g H₂ produces 72 g H₂O (×2)

Concentration Calculations

Concentration is the amount of solute dissolved per unit volume of solution. At GCSE it is measured in g/dm³ (grams per decimetre cubed) or mol/dm³ (moles per decimetre cubed).

Concentration (g/dm³) = mass (g) ÷ volume (dm³)
Concentration (mol/dm³) = moles ÷ volume (dm³)

❌ Unit trap: volume must be in dm³, not cm³. To convert cm³ to dm³, divide by 1000. 250 cm³ = 0.25 dm³. Forgetting this conversion is the single most common error in concentration calculations.

Titration Calculations

Titrations use concentration and volume data to find an unknown concentration. The method:

  1. Calculate moles of the solution whose concentration you know: moles = concentration × volume (in dm³).
  2. Use the molar ratio from the balanced equation to find moles of the other substance.
  3. Calculate concentration of the unknown: concentration = moles ÷ volume.

Percentage Yield

In practice, reactions rarely produce as much product as theoretically predicted. Percentage yield compares the actual yield (what you actually got) to the theoretical yield (the maximum calculated from the equation).

% yield = (actual yield ÷ theoretical yield) × 100

Reasons for yield being less than 100%: the reaction may be reversible and not go to completion; some product may be lost during filtration, evaporation or transfer between containers; side reactions may produce different products.

Atom Economy (Higher Tier)

Atom economy measures how much of the mass of the reactants ends up in the desired product. A high atom economy means less waste is produced — which is important for both economic and environmental reasons.

% atom economy = (Mr of desired product ÷ sum of Mr of all products) × 100

Addition reactions have 100% atom economy — all the atoms from the reactants end up in the single product. Substitution reactions have lower atom economy — some atoms end up in unwanted byproducts.

The AQA quantitative chemistry specification is at the AQA GCSE Chemistry specification page. Edexcel's is at the Edexcel GCSE Chemistry page.

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