Define the mole and the Avogadro constant, interconvert mass, amount, gas volume and solution concentration, and apply stoichiometry including limiting reagent, percentage yield and atom economy and titration calculations

What this dot point is asking

SEAB wants you to define the mole and the Avogadro constant, move confidently between mass, amount of substance, gas volume and concentration, and apply stoichiometry to limiting-reagent problems, percentage yield, atom economy, and titrations. These calculations underpin almost every quantitative question on the paper, and the titration calculation appears in Paper 2 and the practical Paper 4.

The answer

The mole and the Avogadro constant

One mole is the amount of substance containing as many particles as there are atoms in $12$ g of carbon-12, namely the Avogadro constant $L = 6.02 \times 10^{23}$ per mole. The molar mass $M$ (g per mol) equals the relative formula mass.

The four key conversions

At room temperature and pressure (r.t.p.), one mole of any gas occupies about $24.0$ dm cubed. For non-standard conditions, use $pV = nRT$.

Stoichiometry and the limiting reagent

Balanced equations give mole ratios. When two reactants are mixed, the limiting reagent is the one that runs out first and fixes how much product forms. Find moles of each reactant, divide by its coefficient, and the smallest result is limiting.

Percentage yield and atom economy

Yield measures how much of the theoretical product is actually obtained; atom economy measures how much of the reactant mass ends up in the desired product (a green-chemistry metric).

The titration calculation, step by step

1. Calculate moles of the substance whose concentration and volume are known: $n = c \times V$.
2. Use the balanced equation to find moles of the other reactant by the mole ratio.
3. Divide by the volume (in dm cubed) to get the unknown concentration, or by $M$ to get a mass.

Always work in consistent units. A burette reading is recorded to $0.05$ cm cubed, and concordant titres (within $0.10$ cm cubed) are averaged before the calculation.

Examples in context

Example 1. Back-titration of an antacid. A common Paper 2 context dissolves an antacid tablet in a measured excess of HCl, then titrates the leftover acid with standard NaOH. Subtracting the moles of leftover acid from the total acid added gives the moles that reacted with the antacid, from which the mass of active ingredient is found. Back-titration is the standard route when the sample reacts too slowly or is insoluble.

Example 2. Atom economy in industry. Comparing two routes to an organic product, a route that produces only the desired molecule and water has higher atom economy than one that produces the same molecule plus a bulky salt by-product. SEAB uses this to test green-chemistry reasoning, expecting candidates to compute atom economy from molar masses and comment on waste.

Try this

Q1. Calculate the number of molecules in $0.25$ mol of carbon dioxide. [1 mark]

• Cue. $N = nL = 0.25 \times 6.02 \times 10^{23} = 1.51 \times 10^{23}$ molecules.

Q2. $4.0$ g of hydrogen reacts with $32.0$ g of oxygen to form water ($2\text{H}_2 + \text{O}_2 \rightarrow 2\text{H}_2\text{O}$). Identify the limiting reagent. [3 marks]

• Cue. $n(\text{H}_2) = 2.0$, $n(\text{O}_2) = 1.0$; ratio needs 2:1, available 2:1 exactly, so neither is in excess (both fully react).

Q3. $20.0$ cm cubed of $0.150$ mol per dm cubed NaOH is exactly neutralised by $18.0$ cm cubed of sulfuric acid. Calculate the acid concentration. ($2\text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + 2\text{H}_2\text{O}$.) [3 marks]

• Cue. $n(\text{NaOH}) = 3.0 \times 10^{-3}$; ratio 2:1 gives $n(\text{H}_2\text{SO}_4) = 1.5 \times 10^{-3}$; $c = 1.5\times10^{-3}/0.0180 = 0.0833$ mol per dm cubed.