The complete lecture — same content as the written version, but every concept comes alive in the live panel on the right as you read. Scroll down; the animation keeps pace.
1 — What is stoichiometry?
The word comes from the Greek stoicheion (element) and metron (measurement).
- Stoichiometry — the study of the quantitative relationship between reactants and products in a chemical reaction, using a balanced chemical equation.
Assumptions
- Reactants are completely converted into products.
- The reaction is irreversible.
- There is no side reaction.
Exam point: stoichiometric calculations cannot be applied to reversible reactions.
Qualitative vs quantitative
- Qualitative analysis — which substances are present.
- Quantitative analysis — how much of each. Stoichiometry is the quantitative tool.
2 — Laws of chemical combination
- Conservation of Mass (Lavoisier) — total mass of products = total mass of reactants.
- Definite Proportions (Proust) — a pure compound always has the same elements in the same fixed ratio by mass.
- Multiple Proportions (Dalton) — masses of one element combining with a fixed mass of another are in simple whole-number ratios.
Conservation check
2H₂ + O₂ → 2H₂O
reactants 2(2)+32 = 36 g · products 2(18) = 36 g ✓
3 — Atomic, molecular & formula mass
- Atomic mass — mass of one atom vs carbon-12. H = 1, O = 16 a.m.u.
- Molecular mass — sum of atomic masses in a molecule. H₂O = 18 a.m.u.
- Formula mass — sum for one formula unit of an ionic compound. NaCl = 58.5 a.m.u.
a.m.u = dalton (Da) = 1/12 the mass of a C-12 atom = 1.66 × 10⁻²⁴ g.
4 — The mole & Avogadro's number
- Mole — the amount of a substance containing 6.02 × 10²³ particles.
- Avogadro's number (Nₐ) — 6.02 × 10²³ particles per mole, for any substance.
Key formulasmoles = mass / molar mass
particles = moles × 6.02 × 10²³
5 — Molar mass & molar volume
- Molar mass — mass in grams of 1 mole (g/mol).
- Molar volume — volume of 1 mole of gas at STP = 22.4 dm³.
Avogadro's law: equal volumes of gases (same T, P) hold equal numbers of molecules — not equal mass.
6 — The master relationship
| 1 mole of… | Mass | Particles | Volume (STP) |
|---|
| H₂ | 2 g | 6.02 × 10²³ | 22.4 dm³ |
| CH₄ | 16 g | 6.02 × 10²³ | 22.4 dm³ |
| NaCl | 58.5 g | 6.02 × 10²³ | — |
Three conversionsmoles = mass / molar mass
moles = particles / 6.02 × 10²³
moles = volume / 22.4 dm³
7 — Mole / mass / particle problems
moles from mass
CO₂: 2400 / 44 = 54.5 mol
atoms from mass
0.3 g C → (0.3 × 6.02×10²³)/12 = 1.505 × 10²² atoms
mass from molecules
3.01×10²³ H₂O → 0.5 mol → 9 g
8 — Percentage composition
Percentage of an element% = (mass of element in 1 mole / molar mass) × 100
water
%H = (2/18)×100 = 11.1 %
%O = (16/18)×100 = 88.9 %
carbon dioxide
%C = (12/44)×100 = 27.3 % · %O = 72.7 %
9 — Empirical & molecular formula
- Empirical formula — simplest whole-number ratio of atoms. CH₂O.
- Molecular formula — actual atom counts. C₆H₁₂O₆.
Linked by nMolecular = n × Empirical, n = molecular mass / empirical mass
% → empirical
40% C, 6.7% H, 53.3% O
÷ atomic mass → 3.33 : 6.7 : 3.33 → ÷3.33 → CH₂O
empirical → molecular
CH₂O (mass 30), molecular mass 180 → n = 6 → C₆H₁₂O₆
10 — Exponential notation
Write numbers as N × 10ⁿ with 1 ≤ N < 10. Greater than one → positive exponent; less than one → negative.
| Standard | | Exponential |
|---|
| 400000000 | = | 4.0 × 10⁸ |
| 0.000045 | = | 4.5 × 10⁻⁵ |
11 — Significant figures
| Rule | Example |
|---|
| Non-zero digits count | 183 → 3 s.f. |
| Zeros between digits count | 1008 → 4 s.f. |
| Leading zeros (<1) don't | 0.00122 → 3 s.f. |
| Trailing zeros after a decimal count | 3.0000 → 5 s.f. |
12 — Rounding off data
| Digit to drop | Do | Example |
|---|
| > 5 | add 1 | 3.768 → 3.77 |
| < 5 | drop | 6.823 → 6.82 |
| = 5, even before | drop | 4.865 → 4.86 |
| = 5, odd before | round up | 4.835 → 4.84 |
13 — Stoichiometric calculations
- Mole ratio — ratio of coefficients in the balanced equation. 2H₂ + O₂ → 2H₂O ⇒ 2 : 1 : 2.
Mass–Mass (mole method)
49 g KClO₃ → 49/122.5 = 0.4 mol
2 KClO₃ → 3 O₂ ⇒ 0.6 mol O₂ → 19.2 g O₂
14 — Theoretical, practical & % yield
Percentage yield% yield = (practical / theoretical) × 100
worked
theoretical 15.648 g O₂, collected 14.9 g
% yield = (14.9 / 15.648) × 100 = 95.2 %
15 — Limiting reactant
- Limiting reactant — consumed first; controls how much product forms.
- Excess reactant — whatever is left over.
Find it by amount ÷ coefficient — the smaller value limits. Change the sliders in the live panel and press React to watch one reactant run out.
16 — Exam recap
- Define stoichiometry + its 3 assumptions.
- Laws: conservation of mass, definite proportions.
- Mole, Avogadro's number, molar volume (values).
- Atomic vs molecular vs formula mass.
- Percentage composition; empirical vs molecular formula.
- Limiting reactant; theoretical vs % yield.