Quantitative Chemistry

Moles — recap

n = m ÷ M (moles = mass ÷ molar mass). Use coefficients from balanced equations for molar ratios.

Percentage yield

In practice, we rarely obtain the theoretical (maximum possible) amount of product. Reasons include:

• Reaction does not go to completion (reversible reactions reach equilibrium).
• Side reactions producing unwanted products.
• Loss during transfer, filtration, evaporation.

% yield = (actual yield ÷ theoretical yield) × 100

Example: CaCO₃(s) → CaO(s) + CO₂(g) Theoretical: 100 g CaCO₃ → 56 g CaO (from molar masses Mᵣ 100 and 56). Actual yield in experiment: 45 g CaO. % yield = (45 ÷ 56) × 100 = 80.4%.

Atom economy

Atom economy measures how efficiently atoms in the reactants are incorporated into the desired product. High atom economy = more sustainable chemistry (less waste).

% atom economy = (Mᵣ of desired product ÷ sum of Mᵣ of all products) × 100

Equivalently (since mass is conserved): % atom economy = (Mᵣ of desired product ÷ Mᵣ of all reactants) × 100

Example: Production of hydrogen by steam reforming: CH₄ + H₂O → CO + 3H₂

Mᵣ of all reactants = 16 + 18 = 34. Desired product = H₂ (3 mol × Mᵣ 2 = 6). % atom economy = (6 ÷ 34) × 100 = 17.6% — quite low; CO is "wasted."

Example — addition vs substitution:

• Addition reactions have 100% atom economy because all atoms in reactants end up in the single product.
• Substitution reactions produce a by-product, so atom economy < 100%.

Gas volumes (Avogadro's law — Higher)

At the same temperature and pressure, equal volumes of gases contain equal numbers of moles. At room temperature and pressure (rtp: 25 °C, 1 atm), 1 mole of any gas occupies 24 dm³ (24,000 cm³).

Volume (dm³) = n × 24

Example: 0.5 mol of O₂ at rtp occupies 0.5 × 24 = 12 dm³.

Concentration

c = n ÷ V (mol/dm³). Convert cm³ to dm³ by dividing by 1000.

Limiting reagent

The limiting reagent is the reactant that is completely consumed first; it determines how much product forms. The other reactant is in excess.

To identify: calculate moles of each reactant, divide by the stoichiometric coefficient, and compare. The smaller value gives the limiting reagent.

Example: 3 mol H₂ and 2 mol N₂ react: N₂ + 3H₂ → 2NH₃. H₂ ratio: 3 ÷ 3 = 1. N₂ ratio: 2 ÷ 1 = 2. H₂ gives the smaller value → H₂ is limiting.