Stoichiometry and the Mole Concept
Stoichiometry and the mole concept tests quantitative reasoning: converting between mass, moles, concentration and gas volume, deducing formulae, identifying limiting reagents and calculating yields.
Core rules
- n = m/M, where M is molar mass in g mol−1.
- For solutions, n = cV with V in dm3 and c in mol dm−3. Convert cm3 to dm3 by dividing by 1000.
- At room temperature and pressure, one mole of any gas occupies about 24 dm3, so n = V/24 with V in dm3.
- Empirical formula is the simplest whole-number ratio; molecular formula = empirical formula × n, where n = Mr/mass of empirical formula.
- Use the balanced equation for mole ratios. The limiting reagent is used up first and determines the theoretical yield.
- Percentage yield = actual yield/theoretical yield × 100. Atom economy = Mr of desired product/total Mr of reactants × 100.
Worked example
4.8 g of magnesium (Ar = 24) is 0.20 mol. For Mg + 2HCl → MgCl2 + H2, 0.20 mol Mg needs 0.40 mol HCl. If only 0.25 mol HCl is present, HCl is limiting. HCl : H2 is 2 : 1, so H2 = 0.125 mol. Volume at r.t.p. = 0.125 × 24 = 3.0 dm3. Checking the limiting reagent before using the metal ratio prevents an overestimate.
Traps
Using an unbalanced equation; substituting cm3 directly into cV; rounding mole ratios too early; assuming the larger mass is in excess; confusing percentage yield with atom economy; forgetting that atom economy uses molar masses from the equation, not actual masses.