Stoichiometry & The Mole
Mole calculations, empirical formulae, and reacting masses
The mole is the chemist's counting unit. One mole = 6.02 x 10²³ particles (Avogadro's constant).
Molar mass = relative formula mass in grams. H₂O: Mr = 18, so 1 mol = 18g.
Key formulae:
Empirical formula = simplest whole number ratio of atoms. A compound is 40% C, 6.7% H, 53.3% O: divide by Ar → C:H:O = 3.33:6.7:3.33 = 1:2:1 → CH₂O.
Percentage yield = (actual yield / theoretical yield) × 100. Yields below 100% due to incomplete reactions, side reactions, or transfer losses.
Titration calculations: At equivalence point, moles of acid and base are related by the balanced equation. Use c₁V₁/n₁ = c₂V₂/n₂.
Key Points to Remember
- 11 mole = 6.02 x 10²³ particles (Avogadro's constant)
- 2n = m/Mr for mole calculations
- 3Empirical formula from percentage composition
- 4Percentage yield = (actual/theoretical) x 100
Pakistan Example
Fertiliser Production at Fauji Fertilizer — Mole Calculations at Scale
Fauji Fertilizer Company in Rahim Yar Khan produces millions of tonnes of urea (NH₂CONH₂, Mr = 60). Converting tonnes to moles: 1 million tonnes = 1.67 x 10¹⁰ mol. Industrial chemists use stoichiometry daily to optimise ammonia-to-urea conversion and minimise waste.