Stoichiometry and the mole concept
Stoichiometry and the mole concept tests mole calculations, equations, limiting reagents, yields and solution and gas volumes.
The mole
One mole contains 6.02 × 10^23 particles (Avogadro constant). Molar mass M is the mass of one mole in g mol−1. The key relationships are n = m ÷ M, n = V ÷ 24 for a gas at room temperature and pressure (V in dm3), and n = c × V for a solution (c in mol dm−3, V in dm3).
Equations and formulae
Balance equations before using mole ratios; the coefficients give the ratio of moles. Identify the limiting reagent by comparing the available moles with the ratio required; the other reagent is in excess. Percentage yield = (actual yield ÷ theoretical yield) × 100. Empirical formula is the simplest whole-number ratio; molecular formula is a multiple of it, found using Mr.
Worked example
Magnesium reacts with hydrochloric acid: Mg + 2HCl → MgCl2 + H2. Calculate the volume of hydrogen at r.t.p. from 4.8 g of magnesium. M(Mg) = 24 g mol−1, so n(Mg) = 4.8 ÷ 24 = 0.20 mol. The ratio Mg : H2 is 1 : 1, so n(H2) = 0.20 mol. V = n × 24 = 0.20 × 24 = 4.8 dm3. The acid is in excess, so magnesium is limiting.
Traps
Do not ignore coefficients: 2HCl means twice as many moles of acid as magnesium. Convert cm3 to dm3 by dividing by 1000 before using c × V. Do not use the excess reagent to find the product. Percentage yield cannot exceed 100% in a correct calculation; a higher value suggests an impure product or a weighing error. For hydrated salts, include water of crystallisation in the molar mass.