Stoichiometry: study of the relative amounts of substances in a chemical reaction, from a balanced equation. Mole: atomic, molecular or formula mass in grams. Molar volume: 1 mole of any gas occupies 24 dm³ at RTP (25 °C, 1 atm).
Limiting reactant: the reactant completely used up, which gives the least product. Reactant in excess: the one left over.
Theoretical yield: product calculated from the equation. Actual yield: product really obtained. Percentage yield is below 100% because of side reactions, reversible reactions and mechanical loss in filtering or crystallising.
Empirical formula: simplest whole-number ratio of atoms. Molecular formula: actual number of atoms.
Molarity: moles of solute per dm³ of solution. Strength: grams of solute per dm³ (or cm³) of solution.
Formulas
Moles = mass / molar mass. Gas at RTP: moles = volume (dm³) / 24. Volume in dm³ = cm³ / 1000.
Dilution: M₁V₁ = M₂V₂. Titration (acid and base, 1:1): M₁V₁ = M₂V₂ (use the mole ratio from the equation if it is not 1:1).
% composition of an element = (mass of the element in 1 mole / molar mass) x 100. % yield = (actual / theoretical) x 100. % purity = (mass of pure compound / total mass of sample) x 100.
Empirical formula steps: (1) moles of each element from mass or percentage (treat percentages as grams in 100 g); (2) divide by the smallest; (3) whole-number ratio gives the subscripts (if ratio has .5, multiply all by 2). Molecular formula: n = molar mass / empirical formula mass, then multiply all subscripts by n.
Past-paper questions with model answers
Concentration in mol/dm³ of 20 g NaOH in 1 dm³ (2026 B): M = 40, moles = 20/40 = 0.5, so 0.5 mol/dm³.
10 g NaCl in 1 dm³ (B): M = 58.5, moles = 10/58.5 = 0.171, so 0.171 mol/dm³.
10 g Na₂CO₃ in 1 dm³ (M): M = 106, moles = 10/106 = 0.094, so 0.094 mol/dm³.
9.8 g H₂SO₄ in 0.5 dm³ (B): M = 98, moles = 9.8/98 = 0.1, molarity = 0.1/0.5 = 0.2 mol/dm³.
Concentration in g/dm³ of 0.25 mol KNO₃ in 1 dm³ (B): M = 39 + 14 + 48 = 101, mass = 0.25 x 101 = 25.25 g/dm³.
Empirical CH₂O, molar mass 180 (B and D): empirical mass = 30, n = 180/30 = 6, so C₆H₁₂O₆. For molar mass 60 (M): n = 2, so C₂H₄O₂.
C 38.71%, H 9.68%, O 51.61%, molar mass 62 (B): moles C 3.226, H 9.68, O 3.226. Divide by 3.226 giving 1 : 3 : 1, so empirical formula CH₃O (mass 31). n = 62/31 = 2, so molecular formula C₂H₆O₂.
C 40%, H 6.67%, O 53.33%, molar mass 90 (D): moles 3.33 : 6.67 : 3.33 = 1 : 2 : 1, so CH₂O (30). n = 3, so C₃H₆O₃.
C 54.55%, H 9.09%, O 36.36%, molar mass 88 (M): moles 4.546 : 9.09 : 2.273 = 2 : 4 : 1, so C₂H₄O (44). n = 2, so C₄H₈O₂.
150 g Zn with 200 g dil. H₂SO₄ (B), Zn + H₂SO₄ → ZnSO₄ + H₂ (the printed equation has an unbalanced 2ZnSO₄, use the 1:1 balanced one): moles Zn = 150/63.5 = 2.36, moles H₂SO₄ = 200/98 = 2.04. H₂SO₄ gives fewer moles of H₂, so it is the limiting reactant. H₂ = 2.04 mol x 2 = 4.08 g.
68.1 g CaCO₃ with 51.6 g HCl (D), CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O: moles CaCO₃ = 68.1/100 = 0.681; moles HCl = 51.6/36.5 = 1.414. HCl needed for 0.681 mol CaCO₃ is 1.362, which is less than 1.414, so CaCO₃ is the limiting reactant. CaCl₂ = 0.681 x 111 = 75.6 g.
159 g Na₂CO₃ with 73 g HCl (M), Na₂CO₃ + 2HCl → 2NaCl + CO₂ + H₂O: moles Na₂CO₃ = 159/106 = 1.5; moles HCl = 73/36.5 = 2. 1.5 mol Na₂CO₃ would need 3 mol HCl, so HCl is the limiting reactant. NaCl = 2 mol x 58.5 = 117 g.
1.2 g Mg with excess HCl (M), Mg + 2HCl → MgCl₂ + H₂: moles Mg = 1.2/24 = 0.05, so H₂ = 0.05 mol, mass = 0.05 x 2 = 0.1 g.
Other numericals from the book exercise (likely in future papers)
10.0 g NaCl in 500 cm³: strength = 10/0.5 = 20 g/dm³; molarity = (10/58.5)/0.5 = 0.34 mol/dm³.
25.0 cm³ of 0.10 M NaOH neutralises 20.0 cm³ HCl: M(HCl) = (0.10 x 25.0)/20.0 = 0.125 mol/dm³.
Mass of NaCl for 250 cm³ of 0.5 M: moles = 0.5 x 0.25 = 0.125; mass = 0.125 x 58.5 = 7.31 g. In g/dm³ it is 0.5 x 58.5 = 29.25.
Percentage yield example: theoretical 2.5 g, actual 1.85 g, yield = 1.85/2.5 x 100 = 74%.
Dilution: to make 250 cm³ of 0.1 M from 12 M HCl: V₁ = (0.1 x 250)/12 = 2.08 cm³.
Percentage composition of MgO: Mg = 24/40 = 60%, O = 16/40 = 40%.