Stoichiometry and the Mole Concept
This domain tests quantitative reasoning: converting between mass, moles, concentration and gas volume, deducing formulae, and identifying the limiting reagent in a reaction.
The mole and its conversions
n = m/M, where M is the sum of relative atomic masses in g mol−1. One mole contains 6.02 × 10²³ particles. For solutions, n = cV with V in dm³; for gases at 273 K and 101.3 kPa, one mole occupies 22.4 dm³. Always convert cm³ to dm³ by dividing by 1000 before substituting.
Formulae from data
Percentage composition divided by atomic mass gives a mole ratio; divide by the smallest value to get the empirical formula. The molecular formula is the empirical formula multiplied by an integer found from Mr/empirical formula mass. For hydrated salts, the water of crystallisation is found by heating to constant mass and comparing mass losses.
Limiting reagents, yield and atom economy
Convert every reactant to moles, divide by its stoichiometric coefficient, and the smallest quotient identifies the limiting reagent. Theoretical yield comes from the limiting reagent; percentage yield = actual/theoretical × 100. Atom economy = (Mr of desired product/Σ Mr of all reactants) × 100.
Worked example
1.20 g of magnesium is burned in excess oxygen and 1.85 g of magnesium oxide is obtained. n(Mg) = 1.20/24.3 = 0.0494 mol. From 2Mg + O2 → 2MgO, n(MgO) = 0.0494 mol, so theoretical mass = 0.0494 × 40.3 = 1.99 g. Percentage yield = 1.85/1.99 × 100 = 93%.
Traps that cost marks
- Choosing the limiting reagent by smallest mass rather than smallest mole quotient.
- Forgetting to balance the equation before using mole ratios.
- Leaving volumes in cm³ when using n = cV.
- Reporting yields above 100% caused by wet or impure product, without comment.
- Rounding intermediate answers too early; keep full precision and round at the end.