3. Stoichiometry
- Syllabus
- 0620–2026–2027
- Section
- 3
- Level
- —
Exam analysis
3.1 Formulae
- Syllabus
- 0620–2026–2027
- Topic
- 3.1
- Level
- —
Recall and interpret chemical formulae
A chemical formula identifies each element by its symbol and uses subscripts to show how many atoms or ions occur in the stated unit.
| Name | Formula | Reading cue |
|---|---|---|
| oxygen | O₂ | elemental oxygen is diatomic |
| phosphorus | P₄ | one molecule contains four P atoms |
| water | H₂O | two H atoms and one O atom |
| ammonia | NH₃ | one N atom and three H atoms |
| nitric acid | HNO₃ | one H, one N and three O atoms |
| sulfuric acid | H₂SO₄ | two H, one S and four O atoms |
| calcium hydroxide | Ca(OH)₂ | one Ca ion and two hydroxide groups |
| ammonium sulfate | (NH₄)₂SO₄ | two ammonium ions for one sulfate ion |
Learn a named formula as an exact symbol-and-subscript pattern. Capital letters start element symbols; a following lower-case letter belongs to the same symbol, so Co and CO are different.
A subscript 1 is omitted. Never change capitalisation or add a charge unless the named species is an ion, such as NH₄⁺.
Define and read a molecular formula
A molecular formula gives the number and type of different atoms in one molecule of a compound.
| Formula feature | How to read it |
|---|---|
| no subscript | one atom of that element |
| subscript | number of atoms immediately before it |
| brackets with an outside subscript | multiply every atom inside the brackets |
| coefficient before a formula | number of whole molecules; it is not part of the molecular formula |
For C₄H₈O₂, one molecule contains 4 carbon atoms, 8 hydrogen atoms and 2 oxygen atoms. This actual count is the molecular formula even though 4:8:2 could be simplified.
A molecular formula describes one discrete molecule. Ionic compounds are represented by formula units, not molecular formulae.
Deduce a formula from a particle model
Treat the model boundary as one stated molecule or formula unit, then count each kind of particle inside that boundary.
| Step | Action |
|---|---|
| 1 | use the key or labels to identify each element |
| 2 | count every atom of each element in one displayed unit |
| 3 | write element symbols with those counts as subscripts |
| 4 | omit subscript 1 and check that every displayed atom was counted |
If one model contains 4 carbon atoms, 6 hydrogen atoms and 3 oxygen atoms, its molecular formula is C₄H₆O₃.
Do not simplify the counts for a molecular formula. Simplifying to the lowest whole-number ratio produces an empirical formula, which is a different description.
Construct word and balanced symbol equations
An equation records reactants becoming products. A word equation names the substances; a symbol equation uses their correct formulae and must conserve every kind of atom.
| Step | Check |
|---|---|
| 1 | place reactants on the left and products on the right |
| 2 | translate every name into its correct formula |
| 3 | count atoms on both sides |
| 4 | balance using whole-number coefficients before formulae |
| 5 | add (s), (l), (g) or (aq) when states are known |
magnesium + hydrochloric acid → magnesium chloride + hydrogen becomes Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). The coefficient 2 balances H and Cl without changing either formula.
State symbols describe physical state: (s) solid, (l) liquid, (g) gas and (aq) dissolved in water. They do not affect atom counts.
Balance with coefficients only. Changing MgCl₂ to Mg₂Cl or altering a subscript changes the substance rather than balancing the equation.
Define and obtain an empirical formula
An empirical formula gives the simplest whole-number ratio of the different atoms or ions in a compound.
| Starting information | Route to the empirical formula |
|---|---|
| molecular formula | divide every subscript by their greatest common factor |
| particle ratio | reduce all counts by the same factor |
| already simplest ratio | leave the subscripts unchanged |
C₄H₈O₂ has the ratio 4:8:2. Dividing every number by 2 gives the empirical formula C₂H₄O. No further common factor remains.
The empirical formula need not show the actual atoms in one molecule. For methanoic acid, CH₂O₂ is already simplest and must not be reduced unevenly.
Deduce an ionic formula from charges
An ionic compound is electrically neutral overall, so total positive charge must equal total negative charge in the smallest whole-number ratio.
| Step | Action |
|---|---|
| 1 | write the cation and anion with their charges |
| 2 | find the smallest numbers of each ion that make total charge zero |
| 3 | write the cation first and use those numbers as subscripts |
| 4 | put a polyatomic ion in brackets when more than one is required |
For Cr³⁺ and SO₄²⁻, the smallest common charge is 6: two Cr³⁺ give +6 and three SO₄²⁻ give −6. The formula is Cr₂(SO₄)₃.
If a model shows three Ca²⁺ ions for two PO₄³⁻ ions, the same neutral 3:2 ratio gives Ca₃(PO₄)₂.
Do not write ionic charges in the final neutral compound formula, and do not change the internal subscripts of a polyatomic ion such as SO₄²⁻.
Construct complete and ionic equations
A complete symbol equation shows all substances. A net ionic equation keeps only the particles that undergo the chemical change.
| Step | Action |
|---|---|
| 1 | write and balance the complete equation with state symbols |
| 2 | split aqueous ionic substances into their ions |
| 3 | leave solids, liquids and gases undissociated |
| 4 | cancel identical spectator ions from both sides |
| 5 | check both atoms and total charge are balanced |
For lead(II) nitrate solution mixed with sodium sulfate solution, the net change is Pb²⁺(aq) + SO₄²⁻(aq) → PbSO₄(s). Sodium and nitrate ions are spectators.
A precipitate is written as a solid and is not split into ions. Never cancel an ion unless the identical species, including charge and state, appears on both sides.
Deduce a symbol equation from reaction information
When the equation is not supplied, convert every clue about reactants, products, charges and states into formulae before balancing.
| Evidence in the prompt | What it determines |
|---|---|
| named reactants and products | species on each side of the arrow |
| oxidation state or ion charge | correct ionic compound formula |
| ‘only other product’ or gas test | missing product identity |
| aqueous, precipitate, gas or heated solid | state symbols or decomposition pattern |
| atom counts | final whole-number coefficients |
Fe₃O₄ reacts with hydrochloric acid to form iron(II) chloride, iron(III) chloride and water. Writing the supplied species first and then balancing gives Fe₃O₄ + 8HCl → FeCl₂ + 2FeCl₃ + 4H₂O.
Read the finished equation back against the prompt: all and only the stated substances must appear, and every element and total charge must be conserved.
Do not begin by guessing coefficients. First deduce correct formulae; coefficients cannot repair a wrong species formula.
3.2 Relative masses of atoms and molecules
- Syllabus
- 0620–2026–2027
- Topic
- 3.2
- Level
- —
Describe relative atomic mass
Relative atomic mass, Aᵣ, is the weighted average mass of the isotopes of an element compared with 1/12 of the mass of one carbon-12 atom.
| Part of the definition | Meaning |
|---|---|
| average of the isotopes | isotope masses contribute according to their relative abundance |
| relative scale | masses are compared, not measured directly in grams |
| reference | one carbon-12 atom is exactly 12 units, so the comparison unit is 1/12 of its mass |
Aᵣ can be non-integer, such as chlorine at about 35.5, because it is an abundance-weighted average of different isotopes.
Aᵣ is not the mass number of one isotope and is not compared with the mass of a whole carbon-12 atom; the reference is one twelfth of that mass.
Calculate relative molecular and formula mass
Relative molecular mass, Mᵣ, is the sum of the relative atomic masses of all atoms in one molecule. For an ionic compound, the same calculation is called relative formula mass because the solid has formula units rather than molecules.
| Step | Action |
|---|---|
| 1 | read every element and subscript in the formula |
| 2 | multiply each Aᵣ by the number of that atom |
| 3 | apply an outside bracket subscript to every atom inside |
| 4 | add all contributions |
For Pb(NO₃)₂: Mᵣ = 207 + 2 × (14 + 3 × 16) = 331. The bracket means two N atoms and six O atoms in the formula unit.
The same sum can be rearranged to find an unknown Aᵣ when the total relative mass and formula are given.
Relative molecular or formula mass is a ratio and has no unit. Do not call an ionic formula unit a molecule.
Calculate reacting masses by simple proportion
A balanced equation fixes the particle ratio. Multiplying each formula mass by its coefficient converts that ratio into a reacting-mass ratio—without using the mole concept.
| Step | Action |
|---|---|
| 1 | write or use the balanced equation |
| 2 | calculate the relative mass of the required reactant and product |
| 3 | multiply each relative mass by its equation coefficient |
| 4 | form the required mass ratio and scale both sides by the same factor |
CaCO₃ → CaO + CO₂ gives relative masses 100 → 56 + 44. Therefore 100 g CaCO₃ produces 56 g CaO, so 10.0 g produces 10.0 × 56/100 = 5.6 g CaO.
The calculated product masses must respect conservation of mass. If a reactant is in excess, base the maximum product on the amount that actually reacts.
Use coefficient-weighted formula masses, not coefficients alone. This objective uses direct mass proportions; amount in moles belongs to the next Topic.
3.3 The mole and the Avogadro constant
- Syllabus
- 0620–2026–2027
- Topic
- 3.3
- Level
- —
State concentration units
Concentration tells how much solute is present per unit volume of solution.
| Unit | Meaning |
|---|---|
| g/dm³ | grams of solute in each dm³ of solution |
| mol/dm³ | moles of solute in each dm³ of solution |
Use the final solution volume, not merely the volume of solvent added. Since 1 dm³ = 1000 cm³, divide cm³ by 1000 before using mol/dm³.
A concentration value is incomplete without its unit; g/dm³ and mol/dm³ are not interchangeable without the solute's molar mass.
Define the mole and Avogadro constant
The mole, mol, is the unit of amount of substance. One mole contains 6.02 × 10²³ specified particles; this number is the Avogadro constant, Nₐ.
| Substance description | Particle counted |
|---|---|
| element such as He | atoms |
| molecular substance such as CO₂ | molecules |
| ionic compound such as NaCl | formula units; ions can then be counted from the formula |
One mole of MgCl₂ contains 1 mol of Mg²⁺ ions and 2 mol of Cl⁻ ions, so it contains 2 × 6.02 × 10²³ chloride ions.
Always name the particle. One mole of molecules does not necessarily contain one mole of atoms.
Link mass, amount and particle number
Molar mass connects a measured mass to amount of substance; the Avogadro constant then connects amount to particle count.
| Find | Relation |
|---|---|
| amount, n | n = mass ÷ molar mass |
| mass | mass = n × molar mass |
| molar mass | molar mass = mass ÷ n |
| particles, N | N = n × 6.02 × 10²³ |
| amount from particles | n = N ÷ (6.02 × 10²³) |
For 22.0 g CO₂, n = 22.0/44.0 = 0.500 mol, so molecules = 0.500 × 6.02 × 10²³ = 3.01 × 10²³.
Use molar mass in g/mol with mass in g. Then apply any atoms or ions per formula unit after finding the number of formula units.
Use molar gas volume at r.t.p.
At room temperature and pressure, r.t.p., one mole of any gas occupies 24 dm³, which is 24 000 cm³.
| Given | Amount of gas |
|---|---|
| volume in dm³ | n = V ÷ 24 |
| volume in cm³ | n = V ÷ 24 000 |
| amount n | V = n × 24 dm³, or n × 24 000 cm³ |
0.0200 mol CO₂ occupies 0.0200 × 24 = 0.480 dm³, or 480 cm³, at r.t.p.
The 24 dm³ value applies to gases at r.t.p.; do not use it for liquids or solids, and keep cm³ and dm³ consistent.
Solve multi-step stoichiometry
Convert every available quantity to moles, use the balanced-equation ratio, then convert the required moles into the requested mass, gas volume or solution quantity.
| Quantity | Convert to or from moles |
|---|---|
| mass | n = m/M |
| gas at r.t.p. | n = V/24 dm³ |
| solution | n = cV, with V in dm³ |
| g/dm³ concentration | mass concentration = molar concentration × molar mass |
For two reactants, calculate n ÷ equation coefficient for each. The smaller value identifies the limiting reactant; use it to calculate the maximum product. The other reactant is in excess.
Convert cm³ to dm³ by dividing by 1000. After using the mole ratio, convert back to the unit requested.
Never compare reactant masses or moles directly when their equation coefficients differ; compare coefficient-adjusted amounts.
Calculate from titration data
At the titration end-point, the measured portions have reacted in the exact mole ratio shown by the balanced equation.
| Step | Action |
|---|---|
| 1 | choose the concordant titre and convert cm³ to dm³ |
| 2 | calculate known moles with n = cV |
| 3 | use the balanced-equation coefficient ratio |
| 4 | calculate unknown concentration with c = n/V, or unknown volume with V = n/c |
25.0 cm³ of 0.0500 mol/dm³ KOH contains 0.00125 mol. For H₂SO₄ + 2KOH, acid moles are 0.000625 mol; if its titre is 20.0 cm³, c = 0.000625/0.0200 = 0.03125 mol/dm³.
Use the equation ratio between solutes, not the ratio of burette and pipette volumes. Volumes may differ even at exact neutralisation.
Calculate empirical and molecular formulae
An empirical formula is the simplest whole-number mole ratio of elements. A molecular formula is a whole-number multiple of the empirical formula.
| Step | Action |
|---|---|
| 1 | use masses directly, or assume 100 g for percentage data |
| 2 | divide each element's mass by its Aᵣ |
| 3 | divide every mole value by the smallest |
| 4 | multiply all ratios if needed to obtain whole numbers |
Find the empirical-formula mass, then multiplier = Mᵣ ÷ empirical-formula mass. Multiply every empirical subscript by this same integer.
For a hydrate, calculate moles of anhydrous salt and water separately; divide both by the salt moles to obtain salt : water = 1 : x.
Do not round a ratio such as 1:1.5 directly; multiply all ratios by 2. A molecular multiplier must be a whole number.
Calculate yield, composition and purity
Each percentage compares a selected part with its correct whole, then multiplies by 100.
| Quantity | Calculation |
|---|---|
| percentage yield | actual product ÷ theoretical product × 100 |
| percentage composition by mass | mass contribution of element in formula ÷ Mᵣ × 100 |
| percentage purity | mass of pure substance ÷ total mass of impure sample × 100 |
Find theoretical yield from the limiting reactant and balanced equation before applying the yield formula. Actual and theoretical quantities must use the same unit.
In NH₄NO₃, nitrogen contributes 2 × 14 = 28 to Mᵣ 80, so nitrogen composition = 28/80 × 100 = 35%.
Yield measures process success, composition measures a formula's mass fraction, and purity measures a sample. Their numerators and denominators are not interchangeable.