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The Mole Concept and Stoichiometry

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The Mole Concept and Stoichiometry

Turn mass, gas and solution evidence into amounts, formulae and balanced chemical relationships.

A-Level 8873, revised syllabus (2026-2027)

01

Keep relative mass, molar mass and amount distinct

Relative quantities have no units; molar mass converts grams into moles.

Relative masses compare a mass with one twelfth of the mass of a carbon-12 atom. They are ratios, so they have no unit. Relative isotopic mass refers to one isotope; relative atomic mass, Ar, is the weighted mean for the atoms in a specified sample of an element.

Four mass terms
TermWhat is compared with one twelfth of a carbon-12 atom
Relative isotopic massMass of one atom of the stated isotope
Relative atomic mass, ArWeighted mean mass of atoms of the element
Relative molecular mass, MrMass of one molecule; numerically the sum of its constituent Ar values
Relative formula mass, MrMass for one formula unit; the sum of Ar values in that formula, used for ionic compounds

One mole contains exactly 6.02214076 × 1023 specified entities. The Avogadro constant NA is that number per mole, with unit mol-1. Specify whether you mean atoms, molecules, ions, electrons or formula units. One mole of MgCl2 formula units contains one mole Mg2+ ions and two moles Cl- ions.

Use n = m/M for mass m in g and molar mass M in g mol-1; n is in mol. M has the same numerical value as the corresponding Ar or Mr, but different units. For particles, N = nNA.

Worked example

Weight the isotope data

A sample contains 75.0% isotope mass 35.0 and 25.0% isotope mass 37.0. Calculate its relative atomic mass using the supplied rounded isotopic masses.

  1. Convert percentages into fractions: 0.750 and 0.250.
  2. Weighted mean = (0.750 × 35.0) + (0.250 × 37.0).
  3. The result is 35.5; it lies between the two isotope masses and nearer the more abundant one.
Answer

Ar = 35.5, with no unit. An unweighted average of 36.0 would incorrectly treat the isotope populations as equal.

Worked example

From a weighed sample to atoms

How many atoms are in 0.486 g Mg, using Ar(Mg) = 24.3?

  1. n = 0.486 g / (24.3 g mol-1) = 0.0200 mol.
  2. N = 0.0200 mol × 6.02214076 × 1023 mol-1.
Answer

1.20 × 1022 Mg atoms, to three significant figures.

02

Turn mass evidence into an atom ratio

An empirical formula is the simplest ratio; a molecular formula counts atoms in one molecule.

An empirical formula gives the simplest whole-number ratio of elements. A molecular formula gives the actual number of each atom in one molecule. Ethene has molecular formula C2H4 and empirical formula CH2. Ionic substances such as MgO have formula units rather than discrete molecules.

Composition-by-mass method
  1. Choose a basis

    For percentages, imagine 100 g. Each percentage then becomes a mass in grams.

  2. Convert each element to moles

    Divide its mass by its atomic molar mass.

  3. Divide by the smallest

    Obtain the simplest mole ratio, not a mass ratio.

  4. Reach whole numbers carefully

    A reliable ratio such as 1:1.5 must be doubled, not rounded to 1:2.

Worked example

A percentage formula with a fractional ratio

An oxide contains 69.9% Fe and 30.1% O. Use Ar(Fe) = 55.8 and Ar(O) = 16.0.

  1. For 100 g, Fe amount = 69.9/55.8 = 1.253 mol; O amount = 30.1/16.0 = 1.881 mol.
  2. Divide both by 1.253: approximately 1 : 1.50.
  3. Multiply both values by two: 2 : 3.
Answer

The empirical formula is Fe2O3. The small departures from an exact ratio reflect the rounded input data.

Worked example

Use combustion and mass by difference

Complete combustion of 0.900 g of a compound containing only C, H and O gives 1.320 g CO2 and 0.540 g H2O. Its Mr is 90.0. Find both formulae; use C = 12.0, H = 1.00, O = 16.0.

  1. CO2 amount = 1.320/44.0 = 0.0300 mol, so C amount = 0.0300 mol and C mass = 0.360 g.
  2. H2O amount = 0.540/18.0 = 0.0300 mol. There are two H atoms per water molecule, so H amount = 0.0600 mol and H mass = 0.0600 g.
  3. Original O mass = 0.900 - 0.360 - 0.0600 = 0.480 g; O amount = 0.480/16.0 = 0.0300 mol.
  4. C:H:O = 0.0300:0.0600:0.0300 = 1:2:1, so the empirical formula is CH2O.
  5. Empirical formula mass = 30.0. The multiplier 90.0/30.0 = 3.
Answer

Empirical formula CH2O; molecular formula C3H6O3.

Check your understandingDoes an empirical formula of CH2 uniquely identify ethene?Think it through, then reveal the answer
No. C2H4, C3H6 and other molecules can share that simplest ratio. A relative molecular mass or other independent evidence is needed.
03

Balance atoms and charge separately

Oxidation numbers identify electron transfer; half-equations make the accounting visible.

Balance an equation by changing coefficients, never subscripts inside a correct formula. A coefficient multiplies every atom in the formula. In an ionic equation, conserve total signed charge as well as every element. Include state symbols when they matter, especially for precipitation, gases and aqueous ions.

Oxidation is electron loss and an increase in oxidation number; reduction is electron gain and a decrease. A reducing agent loses electrons and is oxidised; an oxidising agent gains electrons and is reduced. An element in its elemental form has oxidation number zero; a monatomic ion has oxidation number equal to its charge. Oxygen is usually -2 and hydrogen usually +1 in the compounds used here; the sum equals the species' charge.

Worked example

Locate the changing atom

What happens to iron and manganese when Fe2+ reacts with acidified MnO4-?

  1. Fe2+ becomes Fe3+: oxidation number rises from +2 to +3, so each Fe loses one electron.
  2. In MnO4-, four oxygens total -8 and the ion totals -1, so Mn is +7.
  3. Mn becomes Mn2+: +7 to +2 means gain of five electrons.
Answer

Fe2+ is the reducing agent and is oxidised. MnO4- is the oxidising agent and is reduced.

Construct a half-equation in acidic solution
  1. Balance the changing element

    Start MnO4- → Mn2+.

  2. Balance oxygen with water

    Add 4H2O on the right.

  3. Balance hydrogen with H+

    Add 8H+ on the left.

  4. Balance charge with electrons

    Left is +7 before electrons, right +2: add five electrons on the left.

Reduction: MnO4- + 8H+ + 5e- → Mn2+ + 4H2O.
Oxidation: Fe2+ → Fe3+ + e-.
Multiply the iron half-equation by five, add, then cancel electrons.

Overall: MnO4-(aq) + 8H+(aq) + 5Fe2+(aq) → Mn2+(aq) + 4H2O(l) + 5Fe3+(aq). The charge totals +17 on each side. The 1:5 mole ratio follows from the electrons transferred, not from comparing ion charges.

04

Find the limiting reactant before the product

Compare amounts divided by stoichiometric coefficients.

One reliable calculation route
  1. Mass, solution or gas data

    Convert each relevant quantity into moles.

  2. Balanced equation

    Use coefficient ratios, not mass ratios.

  3. Limiting reactant

    Compare available moles divided by the coefficient for each reactant.

  4. Requested quantity

    Convert the product or excess amount into the requested units.

Worked example

The smaller mass is not automatically limiting

2.50 g CaCO3 is added to 50.0 cm3 of 0.600 mol dm-3 HCl. Find CO2 mass and unreacted CaCO3, using M(CaCO3) = 100.0 g mol-1.

  1. CaCO3 + 2HCl → CaCl2 + CO2 + H2O.
  2. CaCO3 amount = 2.50/100.0 = 0.0250 mol. HCl amount = 0.600 × 0.0500 = 0.0300 mol.
  3. HCl can react with only 0.0300/2 = 0.0150 mol CaCO3, so HCl is limiting.
  4. CO2 amount = 0.0150 mol; mass = 0.0150 × 44.0 = 0.660 g.
  5. CaCO3 remaining = 0.0250 - 0.0150 = 0.0100 mol, equivalent to 1.00 g.
Answer

0.660 g CO2 forms and 1.00 g CaCO3 remains. At an explicitly supplied molar gas volume of 24.0 dm3 mol-1, the gas volume would be 0.360 dm3.

If the sample is impure, only the reacting component contributes its amount to the equation. If an isolated product yield is below the theoretical amount, percentage yield = actual amount/theoretical amount × 100%, using consistent units. Neither purity nor yield changes the stoichiometric coefficients.

Check your understandingIn the worked example, would doubling the carbonate mass double the carbon dioxide produced?Think it through, then reveal the answer
No. HCl already limits the reaction. More carbonate would remain unreacted while the maximum CO2 amount stays 0.0150 mol, unless more acid is supplied.
05

A gas volume can reveal an unknown formula

Compare volumes only at the same temperature and pressure.

For gases measured at the same temperature and pressure, volume is proportional to amount. Use a supplied molar gas volume Vm through n = V/Vm; keep cm3 and dm3 consistent. Do not silently assume a molar gas volume for unspecified conditions.

Complete combustion follows CxHy + (x + y/4)O2 → xCO2 + (y/2)H2O. If the products are cooled enough to condense water, that water no longer contributes to the measured gas volume. Carbon dioxide and unreacted oxygen still do.

Worked example

Infer the hydrocarbon from gas data

20.0 cm3 of a gaseous hydrocarbon burns completely in 100.0 cm3 O2. After cooling to the original conditions and condensing water, 80.0 cm3 gas remains. Removing CO2 leaves 40.0 cm3 O2. Find the hydrocarbon formula.

  1. CO2 formed = 80.0 - 40.0 = 40.0 cm3. The hydrocarbon:CO2 ratio is 20.0:40.0 = 1:2, so x = 2.
  2. O2 consumed = 100.0 - 40.0 = 60.0 cm3. The hydrocarbon:O2 ratio is 1:3.
  3. Thus x + y/4 = 3. With x = 2, y = 4.
  4. Check: C2H4 + 3O2 → 2CO2 + 2H2O. The final dry gas is 40.0 CO2 plus 40.0 excess O2.
Answer

C2H4. All volume comparisons assume the same temperature and pressure and complete combustion.

06

Make dilution and titration steps explicit

A pipette transfers an amount; a burette delivers the amount needed to react with it.

For a solution, n = cV with concentration in mol dm-3 and volume in dm3. Divide a volume in cm3 by 1000 first. Dilution changes concentration and volume but not the solute amount: c1V1 = c2V2 when no reaction or loss occurs.

Worked example

A redox titration binds volume to a 1:5 ratio

25.0 cm3 Fe2+ solution requires 14.60 cm3 of 0.0200 mol dm-3 acidified MnO4-. Find the Fe2+ concentration.

  1. Permanganate amount = 0.0200 × 0.01460 = 2.92 × 10-4 mol.
  2. The balanced acidic equation requires five Fe2+ per MnO4-, so Fe2+ amount = 1.46 × 10-3 mol.
  3. That amount is in 0.0250 dm3. Concentration = 1.46 × 10-3 / 0.0250.
Answer

[Fe2+] = 0.0584 mol dm-3. The answer has three significant figures, matching the limiting input precision.

Worked example

Deduce a stoichiometric ratio from experiment

0.0100 mol of an unknown acid reacts exactly with 0.0200 mol NaOH. What does the evidence suggest?

  1. NaOH supplies one OH- per formula unit.
  2. Two moles of OH- react per mole of the acid.
  3. Under the stated complete neutralisation conditions, this is consistent with two replaceable protons per acid molecule.
Answer

The acid is dibasic in this neutralisation. The evidence gives a 1:2 ratio, not the acid's full molecular identity.

Use raw burette readings to obtain the delivered volume, and use concordant titres rather than an isolated overshoot. Record measured precision, retain guard digits during calculation, then round the final result appropriately. Units such as mol, mol dm-3, g and cm3 communicate different quantities; a correct number with the wrong unit is not a correct result.

Quick revision

Revisit the essentials, then return to an explanation when you need it.

Quantity calculation route
  1. Data → moles

    Use m/M, cV or V/Vm, with units stated.

  2. Moles → moles

    Apply the balanced coefficient ratio; check limiting reactants.

  3. Moles → answer

    Convert to requested mass, volume, concentration or particles.

Essential checks
CalculationCheck
Formula from percentagesDivide by atomic masses before comparing.
Combustion formulaOne C per CO2; two H per H2O; original oxygen by difference only when justified.
RedoxBoth atoms and total charge balance.
Gas analysisSame temperature/pressure; state whether water is condensed.
TitrationUse delivered volume, not a burette reading, and include the coefficient ratio.

Scope and references

Learning outcomes and sources

5. The Mole Concept and Stoichiometry (8873, 2026 revision; examinations 2026 and 2027). Use the outcome map to find the explanation for a particular syllabus requirement.

See the learning outcome map

  1. 5(a) Define the four relative masses.

    • Relative atomic, isotopic, molecular and formula masses; formula mass for ionic substances.

    Keep relative mass, molar mass and amount distinct

  2. 5(b) Define the mole through the Avogadro constant.

    • Specified entities and N = nNA.

    Keep relative mass, molar mass and amount distinct

  3. 5(c) Calculate isotope-weighted relative atomic mass.

    • Use relative abundances and isotope masses.

    Keep relative mass, molar mass and amount distinct

  4. 5(d) Distinguish empirical and molecular formulae.

    • Simplest ratio versus actual atom counts in a molecule.

    Turn mass evidence into an atom ratio

  5. 5(e) Calculate formulae from analytical evidence.

    • Combustion data and composition by mass; molecular-mass multiplier.

    Turn mass evidence into an atom ratio

  6. 5(f) Construct balanced equations.

    • Conserve atoms and charge; use correct formulae.

    Balance atoms and charge separately

  7. 5(g) Explain redox through electrons and oxidation numbers.

    • Fe3+/Fe2+ and MnO4-/Mn2+ examples.

    Balance atoms and charge separately

  8. 5(h) Combine relevant redox half-equations.

    • Balance electrons and cancel to obtain overall equation.

    Balance atoms and charge separately

  9. 5(i) Calculate reacting quantities.

    • (i) Reacting masses from formulae/equations.
    • (ii) Gas volumes including hydrocarbon combustion.
    • (iii) Solution volumes/concentrations; appropriate significant figures.

    Find the limiting reactant before the productA gas volume can reveal an unknown formulaMake dilution and titration steps explicit

  10. 5(j) Infer stoichiometric relationships.

    • Use quantity data rather than assuming the unknown ratio.

    Find the limiting reactant before the productA gas volume can reveal an unknown formulaMake dilution and titration steps explicit