Stoichiometry and the Mole
What the exam tests
Stoichiometry and the Mole covers the mole concept, molar mass, percentage composition, empirical and molecular formulae, limiting reagents, solution concentration and titration calculations.
Core rules
- One mole contains 6.022 × 10²³ particles (Avogadro's number).
- n = m/M, where M is molar mass in g/mol.
- Percentage composition = (mass of element in one mole / molar mass) × 100.
- Empirical formula comes from mole ratios; molecular formula = empirical formula × a whole-number factor from the molar mass.
- Concentration c = n/V, with V in litres or dm³.
- At STP one mole of gas occupies 22.4 L.
- Coefficients in a balanced equation give mole ratios; find the limiting reagent by dividing moles by the coefficient.
Worked example
4.0 g of calcium reacts with excess hydrochloric acid: Ca + 2HCl → CaCl₂ + H₂. Find the volume of hydrogen at STP. M(Ca) = 40 g/mol, so n = 4.0/40 = 0.10 mol. Ratio Ca : H₂ = 1 : 1, so n(H₂) = 0.10 mol. V = 0.10 × 22.4 = 2.24 L.
Traps
- Using the unbalanced equation for mole ratios.
- Forgetting to convert cm³ to dm³ (divide by 1000) in titrations.
- Assuming the reactant with the greater mass is in excess.
- Rounding moles too early in a multi-step calculation.
- Mixing up empirical and molecular formulae when the molar mass is given.