Stoichiometry: the mole, the limiting reactant and five worked problems

Stoichiometry is the bookkeeping of chemistry: it tells you how much reactant you need and how much product you get. It rests on a single concept, the mole, and on a balanced equation. This lesson explains the method and applies it to worked problems of the kind found in exams.

The mole

A mole contains 6.022·10²³ particles (Avogadro’s number, N_A). The molar mass M of a substance, in g/mol, is numerically equal to its molecular mass in atomic mass units: H₂O has M = 2·1.008 + 16.00 = 18.02 g/mol. The three relations to know are:

• n = m / M (moles from mass);
• N = n · N_A (number of particles);
• n = P·V / (R·T) for gases, or V = n · 22.4 L at 0 °C and 1 atm.

The method in four steps

1. Write and balance the equation.
2. Convert the data into moles.
3. Use the coefficients of the equation as the mole ratio.
4. Convert the moles you found into the requested quantity (mass, volume, number of particles).

Problem 1: how much product

How many grams of water form when 10.0 g of methane burn?

CH₄ + 2 O₂ → CO₂ + 2 H₂O. The molar mass of methane is 16.04 g/mol, so n = 10.0/16.04 = 0.623 mol. One mole of CH₄ gives two moles of H₂O: 1.247 mol, that is 1.247 · 18.02 = 22.5 g of water.

Problem 2: the limiting reactant

5.00 g of hydrogen react with 32.0 g of oxygen. How much water forms and what is left over?

2 H₂ + O₂ → 2 H₂O. Moles: H₂ = 5.00/2.016 = 2.48 mol; O₂ = 32.0/32.00 = 1.00 mol. 1.00 mol of O₂ needs 2.00 mol of H₂, and there are 2.48: oxygen runs out first, it is the limiting reactant. 2.00 mol of water form = 36.0 g, and 0.48 mol of H₂ is left, that is 0.97 g.

The trick to spot the limiting reactant: divide the moles of each reactant by its coefficient. The smallest value marks the limiting one (here 2.48/2 = 1.24 against 1.00/1 = 1.00).

Problem 3: percent yield

Heating 50.0 g of CaCO₃ gives 25.2 g of CaO. What is the yield?

CaCO₃ → CaO + CO₂. Moles of carbonate: 50.0/100.09 = 0.4996 mol; theoretical yield of CaO (M = 56.08): 0.4996 · 56.08 = 28.0 g. Percent yield = 25.2 / 28.0 · 100 = 90.0%.

Problem 4: empirical formula

A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass. What is its empirical formula?

Per 100 g: C 40.0/12.01 = 3.33 mol; H 6.7/1.008 = 6.65 mol; O 53.3/16.00 = 3.33 mol. Dividing by the smallest value: C 1, H 2, O 1. Empirical formula CH₂O. If the measured molar mass is 180 g/mol, the molecular formula is six times the empirical one: C₆H₁₂O₆, glucose.

Problem 5: a gas produced

How many litres of hydrogen, at 25 °C and 1 atm, are released when 6.54 g of zinc react with excess hydrochloric acid?

Zn + 2 HCl → ZnCl₂ + H₂. Moles of zinc: 6.54/65.38 = 0.100 mol, hence 0.100 mol of H₂. V = nRT/P = 0.100 · 0.0821 · 298 = 2.45 L.

Common mistakes

• Using grams instead of moles in the stoichiometric ratio: the coefficients only hold between moles.
• Forgetting to balance the equation before starting.
• Using the temperature in degrees Celsius in the gas equation.
• Rounding too early: keep at least one extra digit in intermediate steps.