Quantitative chemistry — section overview

Section C3 is the maths of chemistry — using relative atomic masses, the mole concept and stoichiometry to calculate amounts in reactions.

Relative masses

• Relative atomic mass (Ar): mass of an atom relative to 1/12 of a carbon-12 atom
• Relative formula mass (Mr): sum of Ar for all atoms in a formula

Example: Mr of H₂O = 2(1) + 16 = 18

The mole

The mole is the unit of amount in chemistry.

$$n = \frac{m}{M_r}$$

• n = moles, m = mass (g), Mr = relative formula mass

Avogadro's number: 1 mole = 6.02 × 10²³ particles.

Balanced equations and mole ratios

In the equation: 2H₂ + O₂ → 2H₂O

Ratio: 2 moles H₂ : 1 mole O₂ : 2 moles H₂O

Use ratios to find masses: if 4 g H₂ reacts (n = 4/2 = 2 mol), it produces 2 mol H₂O = 2 × 18 = 36 g.

Percentage yield

$$\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100$$

Yield < 100% because: reaction doesn't go to completion; product lost in purification; side reactions.

Atom economy (HT)

$$\text{Atom economy} = \frac{M_r \text{ of desired product}}{M_r \text{ of all reactants}} \times 100$$

High atom economy = less waste = more sustainable chemistry.

Concentration (HT)

$$c = \frac{n}{V}$$

c = concentration (mol/dm³), n = moles, V = volume (dm³)

Remember: 1 dm³ = 1 litre = 1000 cm³.

Limiting reactant (HT)

The reactant that is completely used up first, determining the maximum yield. The other reactant is in excess.

Common exam mistakes in C3

1. Mr = gram, not mole — Mr is dimensionless; n = m/Mr gives moles
2. Percentage yield > atom economy — they measure different things; yield is actual vs theoretical; atom economy is structural efficiency
3. Forgetting to balance the equation — always balance before calculating mole ratios