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LearnYCH11WCH11 · Unit 1: Structure, Bonding and Introduction to Organic ChemistryFormulae, equations and amounts of substanceReacting quantities and titrations
Formulae, equations and amounts of substance · Subtopic 3 of 3

Reacting Quantities and Titrations

Every calculation runs through moles: mass (n=mMn=\frac{m}{M}), solution (n=cVn=cV, VV in dm³), gas (n=V24.0n=\frac{V}{24.0} dm³ at r.t.p.). Use the balanced equation's mole ratio to link substances. The limiting reagent is used up first and sets the yield. Titrations find unknown concentrations. In a back titration, an excess of reagent is added and the leftover is titrated; subtracting gives the amount that reacted. Percentage yield =actualtheoretical×100=\frac{\text{actual}}{\text{theoretical}}\times100.

Cambridge 9701 Paper 1Cambridge 9701 Paper 2Edexcel YCH11 WCH11AQA 7405 Paper 1≈ 50 minRevision 1 · updated 2026-10-30
Section 1 of 14

Learning objectives

  • Calculate reacting masses, gas volumes and concentrations.
  • Identify limiting reagents and calculate yields.
  • Carry out titration and back-titration calculations.
  • Deduce stoichiometry from experimental data.
Pause and think

Chalk (CaCO₃) is insoluble, so you can't titrate it directly. How could you find its purity?

Section 2 of 14

Core idea

Start hereMeasuring what you can't see

How much aspirin is in a tablet? How pure is a sample of limestone? Titrations answer these questions to within a fraction of a percent.

The big idea

Convert every measurement into moles, use the equation's ratio, then convert back. Back titrations handle insoluble or slow-reacting samples.

1. The mole hub

Converting to and from moles

Everything passes through moles.

2. Back titration

Worked exampleAmmonium salt analysis

2.00 g of an ammonium salt is warmed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH (excess). The leftover NaOH needs 20.0 cm³ of 1.00 mol dm⁻³ HCl. NH4++OH−→NH3+H2O\mathrm{NH_4^++OH^-\rightarrow NH_3+H_2O}.

Goal
% of NH₄⁺ by mass.
Strategy
Total − excess = reacted.
  1. n(OH−)total=0.0500n(\mathrm{OH^-})_{\text{total}}=0.0500
    1.00×0.05001.00\times0.0500.
  2. n(OH−)excess=0.0200n(\mathrm{OH^-})_{\text{excess}}=0.0200
    Equal to HCl used.
  3. n(NH4+)=0.0300⇒0.540 gn(\mathrm{NH_4^+})=0.0300\Rightarrow0.540\ \text{g}
    Mr(NH4+)=18M_r(\mathrm{NH_4^+})=18.
  4. 0.5402.00×100=27.0%\frac{0.540}{2.00}\times100=27.0\%
    Percentage.
Answer27.0%27.0\% ammonium by mass.
✓ CheckAmmonium sulfate is 27.3% NH₄⁺ ✓.
Exam tipSignificant figures Give answers to the same number of significant figures as the least precise data (usually 3).
Pause and think

Why is a pipette used for the aliquot but a burette for the titrant?

Summary

  • n=m/M=cV=V/24.0n=m/M=cV=V/24.0.
  • Limiting reagent.
  • Back titration: total − excess.
Key words
Titre
Volume of solution added from the burette.
Aliquot
A measured portion of a solution.
Limiting reagent
Reactant that is completely used up.
HistoryMohr's burette. Karl Friedrich Mohr designed the modern burette with a clamp in the 1850s and wrote the first textbook of titration analysis.
Fun factAntacid tests. Pharmaceutical labs check the acid-neutralising capacity of indigestion tablets by back titration, exactly the method in this lesson.
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