GCSE Chemistry: Moles in Solution, Titrations and Yield

Quantitative chemistry does not stop at converting mass into moles. In separate GCSE Chemistry the same mole idea is carried into reactions where one reactant runs out first, into solutions measured in mol/dm³, into titrations and into volumes of gas — and most of these calculations are Higher-tier content, so they are marked "(Higher tier)" wherever that applies. This material extends the category's existing quantitative chemistry set (conservation of mass, relative formula mass, moles from mass and g/dm³) rather than repeating it.

The quiz works through the calculation chain one step type at a time. You decide which reactant is limiting by comparing moles against the ratio in the balanced equation, then use the limiting reactant to find the maximum mass of product. Percentage yield appears twice: once from an actual and a theoretical mass, and once where you must first calculate the theoretical mass from the equation. Atom economy is calculated for making hydrogen from methane and steam, and a second question asks why a reaction whose only product is the one you want is preferred in industry.

Solutions follow: moles and a volume in cm³ become a concentration in mol/dm³, and mol/dm³ is converted into g/dm³ using Mr. The titration questions cover the practical side — which apparatus measures a fixed 25.0 cm³ of alkali, how the end point is approached, what concordant titres are — and the calculations for a 1 : 1 reaction and for sulfuric acid reacting with sodium hydroxide in a 1 : 2 ratio. The set closes with the molar volume of a gas: 1 mol of any gas occupies 24 dm³ at room temperature and pressure. Every value you need is given in the question, and each explanation shows the full working with units; the wrong options are built from the slips that cost marks, such as forgetting to divide cm³ by 1000 or ignoring the mole ratio.

The flashcards hold the formulas and definitions to learn by heart: moles, concentration, percentage yield, atom economy, the molar gas volume, limiting reactant, end point, concordant results and the unit conversions.

The printable written work gives eight longer tasks to answer by hand with full working: a limiting-reactant problem taken through to the mass of product, a yield calculation with reasons why yields fall short of 100%, a comparison of two routes to ethanol by atom economy, a titration method, a titration calculation, choosing concordant titres, a gas-volume problem and a concentration conversion. Each task is checked against a model answer.

The content is based on the quantitative chemistry section of the DfE GCSE chemistry subject content; for example, AQA places yield, atom economy, titrations and gas volumes in the chemistry-only part of this section.

  • Identify the limiting reactant from given masses and use it to calculate the maximum mass of product (Higher tier)
  • Calculate percentage yield, including working out the theoretical yield from a balanced equation
  • Calculate atom economy and explain why a high atom economy matters for industry
  • Calculate concentration in mol/dm³ and convert it to g/dm³ (Higher tier)
  • Describe how to carry out an acid–alkali titration and identify concordant titres
  • Use titration results to calculate an unknown concentration for 1 : 1 and 1 : 2 reactions (Higher tier)
  • Use the molar volume of 24 dm³ at room temperature and pressure to calculate gas volumes (Higher tier)

Practice material written by Zestly, based on the DfE GCSE chemistry subject content (quantitative chemistry: limiting reactants, yield and atom economy, concentrations of solutions, titrations and volumes of gases); board placement given as an example from the AQA GCSE Chemistry (8462) specification.

Sample question

(Higher tier) In the reaction Fe + S → FeS, 14 g of iron (Ar: Fe = 56) is heated with 12 g of sulfur (Ar: S = 32). What is the maximum mass of iron(II) sulfide, FeS (Mr = 88), that can form?

See the answer

22 g

Moles of Fe = 14 ÷ 56 = 0.25 mol. Moles of S = 12 ÷ 32 = 0.375 mol. The ratio is 1 : 1, so iron is the limiting reactant and sulfur is in excess. Moles of FeS = 0.25 mol, mass = 0.25 × 88 = 22 g. (33 g comes from using the sulfur, which is in excess; 26 g simply adds the two masses, but the excess sulfur does not react.)

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