A-Level Chemistry: Amount of Substance

Amount of substance is the calculation engine of A-Level Chemistry: nearly every quantitative mark in the qualification routes through moles. This quiz drills the full toolkit with concrete numbers.

The mole itself is tested conceptually through its relationship to the Avogadro constant, then put to work immediately. The ideal gas equation question is a full pV = nRT calculation with the unit conversions that decide the mark — cm³ to m³, Celsius to Kelvin — where most errors are actually made. Empirical formula determination works from percentage composition through the divide-by-Ar, find-the-ratio routine, and a molecular formula question pins down the relationship between empirical formula and molar mass.

The titration question is exam-standard: sodium hydroxide against sulfuric acid with a 2:1 stoichiometry that must be applied, not just noticed. A dilution question exercises C₁V₁ = C₂V₂, and the atom economy question computes a deliberately unglamorous 1.46% for hydrogen production — making the point that atom economy is about the equation, not the effort.

Percentage yield appears twice: once conceptually (which factors genuinely lower yield — reversibility, losses on transfer, side reactions) and once numerically, in a calcium carbonate decomposition worked through theoretical mass to a clean 80.0%.

Explanations show every step of every calculation, making the set equally useful as diagnosis or as worked-example study.

  • Use the mole and the Avogadro constant fluently
  • Apply pV = nRT with correct unit conversions
  • Determine empirical and molecular formulae from composition data
  • Perform titration and dilution calculations with non-1:1 stoichiometry
  • Calculate atom economy and percentage yield and explain what lowers yield

Aligned to the quantitative chemistry core of the DfE A level science subject content: the mole, the ideal gas equation, empirical and molecular formulae, volumetric analysis, atom economy and percentage yield.

Sample question

Titration set-up: a graduated burette clamped to a stand holding the titrant, above a conical flask of the solution being titrated, with one drop falling.

In a titration, 25.0 cm3 of 0.100 mol dm-3 NaOH reacts exactly with 20.0 cm3 of H2SO4. What is the concentration of the H2SO4 solution? The equation is H2SO4 + 2NaOH -> Na2SO4 + 2H2O.

See the answer

0.0625 mol dm-3

Moles of NaOH = $0.025 \cdot 0.100 = 0.0025$ mol. From the stoichiometry, moles of H2SO4 = $0.0025 / 2 = 0.00125$ mol. Concentration = $0.00125 / 0.020 = 0.0625 \text{ mol dm}^{-3}$.

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