3.30620 · Section 3: Stoichiometry

The mole and the Avogadro constant

Key learnings

The mole, Avogadro constant and n = m/Mr
The mole (mol) is the unit of amount of substance; one mole contains the Avogadro constant $L = 6.02 \times 10^{23}$ particles. Mass, amount and relative formula mass are linked by $n = \dfrac{m}{M_r}$ ( $n$ in mol, $m$ in g). Rearrange it: $m = n \times M_r$ to find a mass, $M_r = m/n$ to identify an unknown. $M_r$ is the sum of the $A_r$ of every atom in the formula — e.g. $M_r(\text{H}_2\text{SO}_4) = 2(1) + 32 + 4(16) = 98$ .
Reacting masses, yield and empirical formula
Reacting masses follow four steps: find $M_r$ values, convert the given mass to moles, apply the equation's mole ratio, convert back to mass. Percentage yield $= \dfrac{\text{actual mass}}{\text{theoretical mass}} \times 100\%$ . Percentage by mass of an element $= \dfrac{\text{count} \times A_r}{M_r} \times 100\%$ . Empirical formula: divide each percentage by its $A_r$ , then by the smallest result; a ratio like 1.5 is multiplied by 2, never rounded.
(Extended) Concentration, titration, gas volume
Concentration $c = \dfrac{n}{V}$ in mol/dm³, with $V$ in dm³ ( $1\ \text{dm}^3 = 1000\ \text{cm}^3$ , so divide cm³ by 1000). Mass concentration in g/dm³ $= c \times M_r$ . Titration: use the balanced-equation mole ratio to link the known solution to the unknown, then $c = n/V$ for the unknown. Gas volume at RTP: the molar gas volume is $24\ \text{dm}^3/\text{mol}$ , so $V = 24n$ . Keep volumes in consistent units.

Common mistakes

Drawn from real examiner reports.

Mr left as a sum, or unit omitted

Examiners flagged $M_r$ left as $23 + 16 + 1$ without evaluating to 40, and mass answers like "28" written without the unit "g". The evaluated number earns the accuracy mark; a missing unit costs marks in many schemes. Always write the sum, evaluate to one number, and add the unit on any measured quantity — the same for concentrations (mol/dm³) and volumes (cm³ or dm³).

W22 Paper 42 Q2(c)(iv) — units omitted on final answers; answers left as sums without a final stated value.

Wrong (or skipped) mole ratio

A common stoichiometry error is going from moles of one species to mass of another with no equation ratio. In $2\text{Mg} + \text{O}_2 \rightarrow 2\text{MgO}$ the Mg:MgO ratio is 1:1 but Mg: $\text{O}_2$ is 2:1 — the wrong ratio leaves the answer a factor of 2 out. Write the equation, mark both species and coefficients, then apply the ratio before converting to mass.

W22 Paper 21 Q9 — moles of one species found correctly but the wrong stoichiometric ratio applied.

Rounding a non-integer mole ratio

After dividing by the smallest mole value the ratio is not always whole. A ratio of 1 : 1.5 must be multiplied by 2 to give 2 : 3 — never rounded to 1 : 1 or 1 : 2. Common multipliers: ×2 for a 0.5 remainder, ×3 for 0.33, ×4 for 0.25. Rounding 1.5 is named in the mark scheme as a wrong method and scores nothing.

S22 Paper 41 Q5(h) — non-integer mole ratios rounded (1.5 to 1 or 2) instead of multiplying all values by 2.

Not stating the number is the Mr

Examiners record candidates computing a value but never saying it is the $M_r$ , so it reads as a random number, and giving fractions (e.g. 8/40) as final answers — both lose credit. Label what you have found (" $M_r = 40$ ") and evaluate every fraction to a decimal or whole number. A quantity left unlabelled or as a fraction is treated as incomplete.

W22 Paper 41; W22 Paper 42 — number not stated to be the Mr; fractions given as final answers.

Dividing mass by Ar, not Mr

In $n = m/M_r$ the denominator is the $M_r$ of the whole formula, not the $A_r$ of one element. For MgO, divide by $M_r = 24 + 16 = 40$ , not by $A_r(\text{Mg}) = 24$ ; $8.0\ \text{g}$ then gives $8.0/40 = 0.20\ \text{mol}$ , not $0.33$ . Always build the full $M_r$ before dividing.

Forgetting the limiting reagent

When two reactant masses are given, one runs out first (the limiting reagent) and it alone fixes how much product forms. Convert both to moles, divide each by its coefficient, and the smaller result is limiting — base the product calculation on that species. Using the reactant in excess overstates the yield.

W22 Paper 21 Q9 — forgetting to identify the limiting reagent.

Exam tips

Five-step stoichiometry scaffold
Show every step — marks are for method, not just the answer: (1) balanced equation; (2) $M_r$ of each species; (3) convert to moles ( $n = m/M_r$ , $n = cV$ , or $n = V/24$ for gases at RTP); (4) apply the mole ratio; (5) convert back and add the unit.
(Extended) Titration order of steps
For a titration work in one direction: moles of the known solution ( $n = cV$ ), multiply by the equation ratio to get moles of the unknown, then divide by the unknown's own volume. Always divide by the volume of the substance you are finding, not the one you started with.
Sanity-check the final answer
Quick checks catch silly errors: a percentage yield can never exceed 100% (if it does, the theoretical mass is wrong); empirical ratios are multiplied up to whole numbers, never rounded from 1.5; and any mass, volume or concentration answer needs its unit.

Revision notes