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The mole and stoichiometry

Chemistry · WAEC and JAMB · SS2 and SS3

The single most examined calculation in the paper, and the one most candidates lose marks on. Everything flows from one relationship: moles = mass / molar mass.

What you need to know

  • One mole contains 6.02 x 10^23 particles, the Avogadro constant.
  • One mole of any gas occupies 22.4 dm^3 at s.t.p., or 24 dm^3 at room temperature.
  • Molar mass is the relative molecular mass expressed in grams.
  • Always balance the equation before any mole calculation. The coefficients are the mole ratio.
  • Concentration in mol/dm^3 = moles / volume in dm^3. Divide cm^3 by 1000 first.
  • Percentage yield = (actual yield / theoretical yield) x 100.
  • The limiting reagent is the one that runs out first and therefore fixes the amount of product.

Key terms

Mole
The amount of substance containing as many particles as there are atoms in 12 g of carbon-12.
Molar volume
The volume occupied by one mole of a gas: 22.4 dm^3 at s.t.p.
Empirical formula
The simplest whole-number ratio of atoms in a compound.

Formulae

  • moles = mass / molar mass
  • moles = volume (dm^3) / 22.4 [gases at s.t.p.]
  • concentration (mol/dm^3) = moles / volume (dm^3)
  • CaVa / CbVb = na / nb [titration]

Worked example

What mass of CaCO3 is needed to produce 11.2 dm^3 of CO2 at s.t.p.? [Ca = 40, C = 12, O = 16]

  1. CaCO3 -> CaO + CO2, so the ratio is 1 : 1.
  2. Moles of CO2 = 11.2 / 22.4 = 0.5 mol
  3. So moles of CaCO3 = 0.5 mol
  4. Molar mass of CaCO3 = 40 + 12 + 48 = 100 g/mol
  5. Mass = 0.5 x 100

The mistake to avoid

Volumes given in cm^3 must be divided by 1000 before use in a concentration formula. This single slip turns a correct method into a wrong answer by a factor of a thousand.

In the exam

Write the balanced equation first, every time, even when the question seems not to need it. The mole ratio comes from nowhere else.