ZIMFundi
Sign inJoin free

Topic 3 of 8 · Form 3

Stoichiometry and the mole concept

Balanced equations with state symbols; ionic equations (e.g. Ag⁺(aq) + Cl⁻(aq) → AgCl(s)).

Notes for this topic 1 file, in order

Create a free account to download PDFs.

  1. PDFStoichiometry and the mole concept – Chemistry notes (Cambridge O Level)47.9 KB

Formulae and equations

Balanced equations with state symbols; ionic equations (e.g. Ag⁺(aq) + Cl⁻(aq) → AgCl(s)).

The mole

n = m/Mᵣ; gas volume = n × 24 dm³ at r.t.p.; concentration c = n/V (mol/dm³) or g/dm³.

Worked example: What mass of calcium oxide is produced by heating 50 g of calcium carbonate? (Ca = 40, C = 12, O = 16)

CaCO₃ → CaO + CO₂; n(CaCO₃) = 50 ÷ 100 = 0.50 mol → n(CaO) = 0.50 mol → mass = 0.50 × 56 = 28 g

Empirical and molecular formulae; percentage composition

Titration calculations

Worked example: 20.0 cm³ of 0.100 mol/dm³ HCl neutralises 25.0 cm³ of NaOH. Find the NaOH concentration.

n(HCl) = 0.00200 mol = n(NaOH) → c = 0.00200 ÷ 0.0250 = 0.0800 mol/dm³

Yield and purity

Percentage yield = actual ÷ theoretical × 100; percentage purity = mass of pure substance ÷ mass of sample × 100.

Limiting reactants

Identify which reactant runs out first using mole ratios.

Exam tip: Write the mole ratio from the equation before using it.

Check yourself

  1. Find the number of moles in 4.4 g of CO₂.
  2. What volume does 0.25 mol of gas occupy at r.t.p.?
  3. Find the percentage by mass of nitrogen in NH₄NO₃ (N = 14, H = 1, O = 16).
Show answers
  1. 0.10 mol
  2. 6.0 dm³
  3. 28 ÷ 80 × 100 = 35%
Next topicElectrochemistry

Questions about this topic (0)

No questions about this topic yet.