H2 Chemistry Mole Concept & Stoichiometry Notes

Study guideUpdated 17 Jul 2026

Mole calculations, limiting reagents, redox titrations, and stoichiometry problem-solving - step-by-step worked examples for A-Level 9476.

Q: What does H2 Chemistry Notes: Topic 6 - The Mole Concept and Stoichiometry cover?
A: Build systematic problem-solving routines for mole calculations, limiting reagents, redox titrations, and analytical stoichiometry in Core Idea 3 (Mole Concept and Stoichiometry).

Stoichiometry underpins quantitative chemistry-from gas calculations to titration analysis. This note structures the workflow and highlights the must-know techniques for Paper 2 and Paper 3.

Pair it with the broader revision plan available at https://eclatinstitute.sg/blog/h2-chemistry-notes.

Status: SEAB's current H2 Chemistry (9476) syllabus PDF is labelled for 2026, and the current Chemistry Data Booklet is labelled 8873/9476/9813 for use from 2026 in non-practical papers. Core Idea 3 Topic 6 is assessed across Papers 1-3.


The core idea is simple: Stoichiometry is a conversion workflow: given quantity to moles, mole ratio, then required quantity.

Use it as a working check: Write the balanced equation before calculating. The coefficients decide the ratio, not the numbers that appear in the question first.

Then go one layer deeper: Example: if 2 mol2\ \text{mol} of A\ce{A} reacts with 1 mol1\ \text{mol} of B\ce{B}, compare nA/2n_A/2

Route map: choose the stoichiometry pathway first

If the question gives you...Start with...Then connect to...Trap to avoid
Mass and molar massConvert mass to molesBalanced equation ratio, then required mass or amountDo not compare masses directly when coefficients differ.
Concentration and volumeConvert volume to litres, then use moles equals concentration times volumeTitration ratio, dilution, or concentration of the unknownDo not leave millilitres inside a concentration in mol per litre calculation.
Gas volume at r.t.p. or s.t.p.Use the stated molar volumeMole ratio, then gas volume, mass, or concentrationDo not use molar volume when the question gives non-standard temperature or pressure.
Two reactants with amountsDivide each mole amount by its coefficientThe smaller normalised value gives the limiting reagentDo not choose the reactant with fewer raw moles by default.
Actual yield or impure sample data

Use this map before calculating. Most errors in Topic 6 happen before the arithmetic, when the wrong pathway is chosen.

Quick revision box

  • What this topic tests: Mole workflows, limiting reagents, stoichiometric ratios, and titration calculations.
  • Top mistakes to avoid: Premature rounding; wrong limiting reagent choice; missing units/significant figures.
  • 20-minute sprint plan: 5 min stoichiometry workflow; 10 min limiting/titration practice; 5 min unit + s.f. checks.

1 Fundamental Relationships

FormulaDescription
n=mMn = \dfrac{m}{M}Moles from mass and molar mass.
n=CVn = CV

Always state units. If conditions differ from the data booklet definitions of s.t.p./r.t.p., revert to the ideal gas equation PV=nRT PV = nRT or use any alternative value provided in the question stem.


2 Stoichiometric Method

  1. Write a balanced equation.
  2. Convert all given quantities to moles.
  3. Use mole ratios from the equation to relate substances.
  4. Convert back to required quantity (mass, volume, concentration).

When numerical work is required, take relative atomic masses and constants directly from the SEAB Chemistry Data Booklet (exams from 2026) rather than rounded memory values.

2.1 Limiting Reagent Logic

Calculate moles of each reactant and compare the ratio with the balanced equation. The reactant yielding the smallest amount of product is limiting. Show working to secure method marks.

A fast check is to compare nstoichiometric coefficient\dfrac{n}{\text{stoichiometric coefficient}} for each reactant: the smaller value identifies the limiting reagent, and all theoretical-yield/purity calculations should then be based on that reagent.


3 Percentage Yield and Purity

  • Percentage yield: %yield=actual amounttheoretical amount×100\%\text{yield} = \dfrac{\text{actual amount}}{\text{theoretical amount}} \times 100
  • Percentage purity: %purity=mass of pure substancetotal mass of sample×100\%\text{purity} = \dfrac{\text{mass of pure substance}}{\text{total mass of sample}} \times 100

Use mass or moles consistently throughout. For purity problems, the impure mass is often the quantity measured experimentally; set up stoichiometric equations using only the pure component.


4 Redox and Acid-Base Titrations

4.1 Typical Workflow

  1. Write ionic equations (especially for redox).
  2. Convert primary standard volume x concentration into moles.
  3. Apply mole ratio to find moles of analyte.
  4. Convert to requested quantity (concentration, mass, % purity).

4.2 Aliquot and dilution checkpoint

Before using a titre, identify which solution volume the titre actually reacts with. This keeps stock-solution concentration, diluted-solution concentration, aliquot volume, and average titre from being swapped.

Quantity in the questionWhat it representsFirst calculation move
Pipetted aliquotFixed volume transferred into the conical flaskUse this as the analyte volume in the mole ratio, not the full volumetric-flask volume.
Average titreVolume delivered from the burette to react with the aliquotConvert to litres, then use n=CVn = CV for the titrant.
Volumetric-flask volumeFinal volume after dilutionUse it only when scaling from aliquot concentration back to the diluted solution or stock solution.
Dilution statementHow the stock solution was made less concentratedApply the dilution factor after finding the concentration of the diluted solution.

Misconception check: the titre does not usually react with the whole volumetric flask. It reacts with the aliquot in the flask, so scale back to the original solution only after the mole ratio step is complete.

4.3 Common Redox Equations

  • MnOX4X\ce{MnO4-} in acidic medium:
    MnOX4X+8HX++5eXMnX2++4HX2O \ce{MnO4- + 8H+ + 5e- -> Mn^{2+} + 4H2O}

State oxidation numbers to justify electron counts if required.


5 Empirical and Molecular Formulae

  1. Divide percentage or mass data by relative atomic mass to get mole ratio.
  2. Divide all moles by the smallest value to obtain simplest whole-number ratio.
  3. Determine molecular formula using molar mass:
    Result: MmolecularMempirical=integer multiplier \dfrac{M_{\text{molecular}}}{M_{\text{empirical}}} = \text{integer multiplier}

Be ready for combustion analysis questions: convert mass of COX2\ce{CO2} and HX2O\ce{H2O}

2D structural formula of carbon dioxide
Carbon dioxide: one-carbon product species used in combustion-analysis mole bookkeeping.

Each COX2\ce{CO2} molecule contains one carbon atom, so moles of COX2\ce{CO2}

Combustion formula bookkeeping checkpoint

Combustion product or dataMole linkCommon trap
Mass of COX2\ce{CO2}n(C)=n(COX2)n(C)=n(\ce{CO2})

Worked check: an organic compound contains only C, H, and O. A 0.300 g\pu{0.300 g} sample gives 0.440 g\pu{0.440 g} of COX2\ce{CO2} and

n(C)=n(COX2)=0.44044.0=0.0100 mol n(C)=n(\ce{CO2})=\frac{0.440}{44.0}=\pu{0.0100 mol}

n(H)=2n(HX2O)=2(0.18018.0)=0.0200 mol n(H)=2n(\ce{H2O})=2\left(\frac{0.180}{18.0}\right)=\pu{0.0200 mol}

Masses: m(C)=0.0100×12.0=0.120 gm(C)=0.0100\times12.0=\pu{0.120 g}, m(H)=0.0200×1.0=0.0200 gm(H)=0.0200\times1.0=\pu{0.0200 g}

Misconception check: oxygen by difference is a mass step before it is a mole-ratio step.


6 Worked Example

Question: An impure sample of potassium iodide (KI\ce{KI}) weighing 0.700 g\pu{0.700 g} is titrated with 0.0200 molL1\pu{0.0200 mol L^{-1}} KX2CrX2OX7\ce{K2Cr2O7}

Solution:

  1. Ionic equation (acidic medium): CrX2OX7X2+14HX++6IX2CrX3++3IX2+7HX2O\ce{Cr2O7^{2-} + 14H+ + 6I- -> 2Cr^{3+} + 3I2 + 7H2O}

Remember to report with appropriate significant figures based on experimental data.


7 Practical Tips

  • Use consistent decimal places in titration tables (e.g. two decimal places for burette readings).
  • In Paper 4 planning sections, specify standard solutions (e.g. primary standard NaX2COX3\ce{Na2CO3}

8 Common Mistakes

  • Forgetting dilution effect after mixing solutions.
  • Applying molar ratios incorrectly when coefficients differ.
  • Ignoring spectator ions in ionic equations, leading to unbalanced charge.
  • Using molar volume 24.0 dm3mol1\pu{24.0 dm3.mol-1} at non-RTP conditions.

9 Quick Drills

  1. A hydrate CuSOX4xHX2O\ce{CuSO4.xH2O}

Check answers with method sheets to ensure your working lines follow the balanced-equation → mole ratio → final quantity structure.


Common exam mistakes

  • Identifying the wrong limiting reagent: Dividing each reactant's moles by its stoichiometric coefficient gives the correct comparison; the smaller value identifies the limiting reagent. Students who compare raw moles without using coefficients consistently pick the wrong reagent.
  • Premature rounding of intermediate values: Rounding moles to 2 s.f. mid-calculation introduces cumulative error; carry at least one extra significant figure through each step and round only in the final answer.
  • Forgetting the dilution effect after mixing solutions: When two solutions are mixed, total volume increases; recalculating concentrations using the new total volume before applying the equilibrium or buffer equation is mandatory.
  • Using the wrong molar volume for the conditions: 24.0 dm3mol1\pu{24.0 dm^3.mol-1} applies at r.t.p. and 22.7 dm3mol1\pu{22.7 dm^3.mol-1}

Frequently asked questions

Do I need to memorise atomic masses for the exam?
No. Relative atomic masses are provided in the SEAB Chemistry Data Booklet. However, knowing common values e.g.H=1,C=12,O=16,Na=23,Cl=35.5 e.g. H = 1, C = 12, O = 16, Na = 23, Cl = 35.5

When should I use the molar volume shortcut versus PV = nRT?
Use n=V/Vmn = V / V_m only when the question states r.t.p. or s.t.p. explicitly and the gas is treated as ideal. In all other cases - non-standard temperatures, pressures, or when ideal-gas assumptions are being tested - use PV=nRTPV = nRT with appropriate unit conversions.

How do I handle a limiting reagent question where one reagent is in excess?
Calculate moles of each reactant from the data given, apply the stoichiometric ratio, identify the limiting reagent, and base all subsequent calculations (theoretical yield, percentage yield, remaining excess) on the limiting reagent only.

What is the difference between an empirical formula and a molecular formula?
The empirical formula gives the simplest whole-number ratio of atoms; the molecular formula gives the actual number of atoms per molecule. To find the molecular formula, divide the given molar mass by the empirical formula mass to find the integer multiplier.


Struggling with The Mole Concept and Stoichiometry? Our H2 Chemistry tuition programme covers this topic with structured practice, Paper 4 practical drills, and worked exam solutions.


Fluent stoichiometry keeps later topics (equilibria, kinetics, redox) manageable. Keep rehearsing with mixed-problem sets, consult https://eclatinstitute.sg/blog/h2-chemistry-notes for integrated practice, and use the official data booklet guide for RTP/STP values, constants, and quick unit checks you’ll apply repeatedly: https://eclatinstitute.sg/blog/h2-chemistry-notes/H2-Chemistry-Data-Booklet-2026.


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Reviewed by
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Sources

  1. SEAB H2 Chemistry (9476) Syllabus 2026
  2. SEAB Chemistry Data Booklet (8873/9476/9813)
  3. NIST Reference on Constants, Units, and Uncertainty