OxfordAQA IGCSE Chemistry: Quantitative Chemistry — Revision Notes

Condensed for the final weeks. For the full explanation, use the Quantitative Chemistry study guide.

Relative masses

Relative atomic mass (Aᵣ) — the weighted mean mass of an atom compared with 1/12 the mass of a carbon-12 atom.

Relative formula mass (Mᵣ) — the sum of the Aᵣ values of all atoms in the formula.

Relative atomic masses are not whole numbers because they are weighted averages of isotopes. This specification does not require calculating Ar from isotopic abundances — the weighted-mean formula (Ar = sum of isotope mass x abundance, divided by 100) is A-level material.

The mole

moles = mass / Mr
mass  = moles x Mr
number of particles = moles x 6.02 x 10^23

One mole is the amount containing 6.02 × 10²³ particles — the Avogadro constant. Its mass in grams equals the relative formula mass.

Conservation of mass

Mass is conserved in a chemical reaction because atoms are rearranged, not created or destroyed.

Apparent mass changes are always explained by a gas:

• Mass decreases in an open container when a gas escapes — for example carbonate decomposition releasing CO₂.
• Mass increases when a gas from the air is a reactant — for example a metal oxidising.

In a closed system, mass is always unchanged. Being able to explain both directions is a standard question.

Empirical and molecular formulae

Method: divide each mass or percentage by the relative atomic mass, divide all results by the smallest, then scale to whole numbers.

molecular formula = empirical formula x (Mr / empirical formula mass)

Worked example. A compound is 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass (A_r: C = 12, H = 1, O = 16). Divide by A_r: C = 40.0 ÷ 12 = 3.33, H = 6.7 ÷ 1 = 6.7, O = 53.3 ÷ 16 = 3.33. Divide by the smallest (3.33): C = 1, H = 2.01 ≈ 2, O = 1 — empirical formula CH₂O (mass 30). If the relative molecular mass is 180, the scaling factor is 180 ÷ 30 = 6, giving the molecular formula C₆H₁₂O₆.

Reacting masses

Balanced equations give the ratio in moles, never directly in grams. The reliable route:

1. Convert the known mass to moles.
2. Use the equation’s ratio to find the moles of the substance you want.
3. Convert back to mass.

Skipping step 1 and working with the mass ratio directly is the commonest error in the topic.

Even though no atoms are gained or lost in a chemical reaction, it is not always possible to obtain the calculated amount of a product, because the reaction may not go to completion (if it is reversible), some of the product may be lost during separation and purification, or some of the reactants may react in ways other than the expected reaction.

Gas volumes

volume of gas (dm3) = moles x 24        at room temperature and pressure

Equal volumes of any gases at the same temperature and pressure contain equal numbers of molecules, which is why gas volume ratios equal the mole ratios in the equation.

Concentration

concentration (g/dm3)     = mass / volume
concentration (mol/dm3)   = moles / volume
g/dm3 = mol/dm3 x Mr

Volume must be in dm³ — divide cm³ by 1000. This conversion accounts for more lost marks than any concept in the topic.

Worked example: reacting masses

4.8 g of magnesium is burned in excess oxygen (2Mg + O₂ → 2MgO, A_r: Mg = 24, O = 16): moles Mg = 4.8 ÷ 24 = 0.2 mol; moles MgO formed = 0.2 mol (1:1 ratio); mass MgO = 0.2 × 40 = 8.0 g. If a student actually collects less than 8.0 g, this is because the reaction did not go to completion, some product was lost during separation, or some of the reactants reacted in an unexpected way — not because atoms were gained or lost.

Titrations

c1 V1 / n1 = c2 V2 / n2

Use only concordant titres — those within 0.10 cm³ of each other — and exclude the rough titre from the average.

Worked example. 25.0 cm³ of 0.0500 mol dm⁻³ sulfuric acid is exactly neutralised by 20.0 cm³ of sodium hydroxide solution (H₂SO₄ + 2NaOH → Na₂SO₄ + 2H₂O). Moles H₂SO₄ = 0.0500 × (25.0 ÷ 1000) = 1.25 × 10⁻³ mol. Moles NaOH = 2 × 1.25 × 10⁻³ = 2.50 × 10⁻³ mol (mole ratio 1:2). Concentration NaOH = 2.50 × 10⁻³ ÷ (20.0 ÷ 1000) = 0.125 mol dm⁻³.

Exam traps

• Working from mass ratios instead of converting to moles first.
• Forgetting to divide cm³ by 1000.
• Explaining a mass change without identifying the gas involved.
• Assuming that failing to obtain the calculated mass of product means atoms were lost.
• Rounding partway through instead of at the end.

Self-test

1. Why are relative atomic masses not whole numbers?
2. Explain why mass appears to decrease when a metal carbonate is heated in an open crucible.
3. Give the three steps of a reacting-mass calculation.
4. Give three reasons why the calculated mass of a product might not be obtained in practice, even though no atoms are gained or lost.

Answers: 1. They are weighted averages of the masses of an element’s isotopes, taking their relative abundances into account. 2. Carbon dioxide gas is produced and escapes from the open container, so the mass remaining falls; in a closed system the total mass would be unchanged. 3. Convert the known mass to moles, use the balanced equation’s mole ratio to find the moles of the target substance, then convert back to mass. 4. The reaction may not go to completion because it is reversible; some of the product may be lost when it is separated from the reaction mixture; some of the reactants may react in ways different from the expected reaction — any three.