Quantitative reference
The calculation chain from the mole outward — with canonical worked examples, HL-only calculations marked, and the maths you do not need.
Editorial framing Every calculation in IB Chemistry hangs off a single backbone: the mole. Master the chain below and every multi-step HL question becomes a sequence of sub-steps you already know. The chain is read top-to-bottom; each step links to the sub-topic primer that owns it, and each step exposes its parent so the dependency order is explicit, not just visual.
The central chain
Worked example: Mass → moles via n = m/M
Given: 2.45 g of H₂SO₄. M(H₂SO₄) = 98.08 g mol⁻¹.
Substitution: n = m/M = 2.45 g / 98.08 g mol⁻¹
Result: n = 0.02498 mol ≈ 0.0250 mol (3 s.f.)
- R2.1.2MOLE RATIO from a balanced equation → reacting masses / volumes / concentrationsSLR2.1 primer →
Worked example: Limiting reactant and percentage yield
Given: 5.00 g Fe₂O₃ reacts with 3.00 g CO. Actual Fe produced = 2.80 g.
Step 1 — moles: n(Fe₂O₃) = 5.00/159.7 = 0.0313 mol; n(CO) = 3.00/28.0 = 0.107 mol
Step 2 — limiting reactant: Fe₂O₃ + 3CO → 2Fe + 3CO₂. 0.0313 mol Fe₂O₃ needs 0.0939 mol CO. We have 0.107 mol CO, so Fe₂O₃ is limiting.
Step 3 — theoretical yield: n(Fe) = 2 × 0.0313 = 0.0626 mol → m(Fe) = 0.0626 × 55.85 = 3.50 g
Step 4 — percentage yield: % yield = (2.80/3.50) × 100 = 80.0% (3 s.f.)
Branch: Calorimetry and enthalpy
Descends from the mole ratio step (R2.1.2).
Branch: Rates
Descends from the mole ratio step (R2.1.2).
Branch: Equilibrium
Descends from the mole ratio step (R2.1.2).
Branch: Acid–base and titration
Descends from the mole ratio step (R2.1.2).
Worked example: Acid–base titration
Given: 25.0 cm³ of HCl of unknown concentration is titrated with 0.100 mol dm⁻³ NaOH. Mean titre = 23.50 cm³.
Step 1: n(NaOH) = cV = 0.100 × (23.50/1000) = 0.00235 mol
Step 2 — mole ratio: HCl + NaOH → NaCl + H₂O, ratio 1:1. n(HCl) = 0.00235 mol
Step 3: c(HCl) = n/V = 0.00235 / (25.0/1000) = 0.0940 mol dm⁻³ (3 s.f.)
Branch: Redox and electrochemistry
Descends from the mole ratio step (R2.1.2).
Branch: HL entropy / Gibbs
Descends from the mole ratio step (R2.1.2).
Worked example: HL: ΔG⦵ = ΔH⦵ − TΔS⦵ (watch the units)
Given: ΔH⦵ = +131 kJ mol⁻¹, ΔS⦵ = +134 J K⁻¹ mol⁻¹, T = 298 K.
Step 1 — unit reconciliation: Convert ΔS⦵ to kJ: 134 J K⁻¹ mol⁻¹ ÷ 1000 = 0.134 kJ K⁻¹ mol⁻¹
Step 2 — substitution: ΔG⦵ = 131 kJ mol⁻¹ − (298 K × 0.134 kJ K⁻¹ mol⁻¹)
Step 3: ΔG⦵ = 131 − 39.9 = +91.1 kJ mol⁻¹ (3 s.f.)
Interpretation: ΔG⦵ > 0, so the reaction is non-spontaneous at 298 K under standard conditions.
Editorial The J/kJ unit slip is the most frequent single error in ΔG calculations. Always convert ΔS⦵ from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ before subtracting from ΔH⦵ in kJ mol⁻¹.
Other required calculations
These calculations belong to sub-topics that carry quantitative content but do not sit on the central mole chain. They are linked to their parent step where relevant.
- S2.4.2Position in bonding triangle from electronegativity data (no % ionic character)SLS2.4 primer →
- S3.2.8–11MS fragment m/z; IR wavenumber assignment; ¹H NMR integration ratios and n+1 splittingHLS3.2 primer →
You do NOT need to
The guide explicitly excludes the following mathematics. Do not spend time learning these:
- Quadratic equations in equilibrium or K_a/K_b calculations.
- Non-integer reaction orders.
- Mathematical treatment of real-gas deviation (no van der Waals equation).
- Percentage ionic character calculations.
- Construction of a complete Born–Haber cycle (interpretation only).
- Quantitative treatment of heterogeneous equilibria (homogeneous only).
- Acidity of hydrated transition-element ions / Al³⁺(aq).
- Hybridization beyond sp, sp² and sp³.
Maths prerequisites
The course silently assumes a set of maths skills — rearranging equations, logarithms, standard form, unit conversion, significant figures, uncertainty propagation and more. The full list with where each skill bites and where students get stuck lives in one place: