Stoichiometry, Formulae and the Mole Concept
This domain tests the mole as the link between mass, particles, gas volume and solution concentration. STPM questions often combine balanced equations with limiting reagents and titration calculations.
Mole relationships
The mole is the amount containing 6.02×10^23 particles. n = m/M; n = N/NA; for gases n = V/Vm; for solutions n = cV. Empirical formula is the simplest whole-number ratio; molecular formula is (empirical formula)n. Percentage composition allows empirical formula determination.
Equations and limiting reagents
Balance the equation first, then use mole ratios. Identify the limiting reagent by comparing the actual mole ratio with the stoichiometric ratio. Theoretical yield comes from the limiting reagent; percentage yield = actual yield/theoretical yield ×100.
Solutions and titration
Concentration c = n/V, usually mol dm−3. Dilution: c1V1 = c2V2. In titration, moles of titrant = cV; use the equation to find moles of analyte.
Worked example
10.0 g of CaCO3 (M = 100 g mol−1) reacts with 100 cm3 of 2.0 mol dm−3 HCl. n(CaCO3) = 0.100 mol; n(HCl) = 0.200 mol. CaCO3 + 2HCl → CaCl2 + CO2 + H2O. 0.100 mol CaCO3 needs exactly 0.200 mol HCl, so neither is in excess. CO2 formed = 0.100 mol. At s.t.p., V = 0.100 × 22.4 = 2.24 dm3.
Traps that cost marks
Do not use mass ratios directly; convert to moles. Always check the limiting reagent. Molar volume depends on temperature and pressure: 22.4 dm3 mol−1 at s.t.p. but about 24 dm3 mol−1 at room conditions. Keep concentration units consistent and do not round too early.