Chemical formulae, equations and the mole concept / Formula kimia, persamaan kimia dan konsep mol
Chemical formulae, equations and the mole concept tests empirical and molecular formulae, balanced equations, molar mass, the Avogadro constant, concentration and stoichiometric calculations. The exam rewards clear unit handling and mole ratios.
Core
- Empirical formula = simplest ratio; molecular formula = actual number of atoms.
- Moles n = mass ÷ molar mass; n = number of particles ÷ 6.02 × 10²³.
- Gas volume: 22.4 dm³ mol⁻¹ at STP; 24 dm³ mol⁻¹ at room conditions. Use the value given in the question.
- Concentration c = n ÷ V, where V is in dm³.
- Balance the equation before using mole ratios; identify the limiting reactant.
Worked example
2.4 g of magnesium reacts completely with excess dilute hydrochloric acid. Find the volume of hydrogen at STP. Ar Mg = 24. n(Mg) = 2.4 ÷ 24 = 0.10 mol. Equation: Mg + 2HCl → MgCl₂ + H₂. The ratio is 1:1, so n(H₂) = 0.10 mol. Volume = 0.10 × 22.4 = 2.24 dm³, or 2240 cm³. Reasoning: convert mass to moles, use the balanced ratio, then use molar volume.
Traps
Never use mass ratios directly from an equation. If 0.10 mol Mg reacts with 0.05 mol HCl, HCl is limiting because it needs twice as many moles. Concentration in mol dm⁻³ is not the same as mass concentration in g dm⁻³. Hydrated salts include water of crystallisation in the molar mass. Round only at the end of a calculation.