3.1 Formulae
- Syllabus
- 0620–2026–2027
- Topic
- 3.1
- Level
- —
Learning objectives
3.1.1Formulae of the elements and compounds• State the formulae of the elements and compounds named in the subject content3.1.2Molecular formula of a compound as the• Define the molecular formula of a compound as the number and type of different atoms in one molecule3.1.3Deduce the formula of a simple• Deduce the formula of a simple compound from the relative numbers of atoms present in a model or a diagrammatic representation3.1.4Construct word equations and symbol• Construct word equations and symbol equations to show how reactants form products, including state symbols3.1.5Empirical formula of a compound as the• Define the empirical formula of a compound as the simplest whole number ratio of the different atoms or ions in a compound3.1.6Deduce the formula of an ionic• Deduce the formula of an ionic compound from the relative numbers of the ions present in a model or a diagrammatic representation or from the charges on the ions3.1.7Construct symbol equations with state• Construct symbol equations with state symbols, including ionic equations3.1.8Deduce the symbol equation with state• Deduce the symbol equation with state symbols for a chemical reaction, given relevant information
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.