Empirical ratios and ionic formulas
Distinguish an actual molecular formula from a simplest ratio and balance the charges of ions.
Section 3.1 · Formulae
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Read molecular formulas and brackets.
- Find a smallest whole-number ratio and total charge.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is the simplest ratio 12:18?
What you will learn
- Find empirical formulas from atom ratios or molecular formulas.
- Deduce ionic formulas from stated charges or an ionic model.
Actual count versus simplest ratio
An empirical formula gives the simplest whole-number ratio of the different atoms or ions. A molecular formula gives the actual atom counts in one molecule. C₆H₁₂O₆ has empirical formula CH₂O after dividing all counts by six. A substance's empirical formula may equal its molecular formula, but this cannot be assumed from a ratio alone.
A model showing 12 magnesium ions and 24 chloride ions gives ratio 1:2 and formula MgCl₂. A small sample may contain many formula units; reduce the repeating ion counts to the smallest neutral ratio instead of writing Mg₁₂Cl₂₄ as the ionic formula.
Neutrality fixes the ion ratio
The charges in one ionic formula unit must sum to zero. Mg²⁺ needs two Cl⁻, giving MgCl₂. Al³⁺ and O²⁻ need total positive and negative charge of six, giving two aluminium ions and three oxide ions, Al₂O₃. Check the charge total rather than using a memorised crossing rule without understanding it.
A polyatomic ion behaves as a group in the ratio. Ca²⁺ needs two OH⁻ ions, so write Ca(OH)₂ with brackets. Do not change OH⁻ into O₂H⁻: the ion retains its identity. If charge magnitudes share a common factor, simplify the ratio; Mg²⁺ with O²⁻ gives MgO, not Mg₂O₂.
Pause and explain
What ionic formula balances Ca²⁺ and Cl⁻?
Worked example
Deduce the formula for aluminium sulfate from Al³⁺ and SO₄²⁻.
Show the worked solution
- The lowest common total charge magnitude is six.
- Two Al³⁺ supply +6 and three sulfate ions supply −6.
- Keep each sulfate group intact and use brackets for three groups.
Answer Al₂(SO₄)₃.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Give empirical formulas for C₄H₈ and C₆H₆, then say whether these give the actual molecule size.
Give me a hint
Divide each pair of atom counts by its greatest common factor.
Compare my reasoning
- C₄H₈ reduces by four to CH₂.
- C₆H₆ reduces by six to CH.
- The empirical ratios alone do not determine the actual molecular atom counts.
CH₂ and CH; neither ratio alone establishes molecule size.
Look for these in your work
- Every count is divided by the same factor.
- Ratio and actual molecular formula are distinguished.
Deduce formulas for Mg²⁺ with NO₃⁻ and for Al³⁺ with O²⁻.
Give me a hint
Find a neutral smallest ratio and keep nitrate intact.
Compare my reasoning
- One Mg²⁺ needs two nitrate ions, giving Mg(NO₃)₂.
- Two Al³⁺ and three O²⁻ balance +6 with −6.
- The aluminium oxide formula is Al₂O₃.
Mg(NO₃)₂ and Al₂O₃.
Look for these in your work
- Charges balance to zero.
- Brackets preserve the polyatomic nitrate ion.
A model has four calcium ions and eight hydroxide ions. A student writes Ca₄O₈H₈. Give the appropriate ionic formula and verify its charge using Ca²⁺ and OH⁻.
Give me a hint
Reduce the ion counts, then preserve hydroxide as a group.
Compare my reasoning
- The calcium-to-hydroxide ratio is 4:8, reducing to 1:2.
- Write Ca(OH)₂, rather than expanding an arbitrary sample count.
- One +2 charge and two −1 charges sum to zero.
Ca(OH)₂; +2 +2(−1)=0.
Look for these in your work
- The sample count is reduced to a formula-unit ratio.
- The ion group and total charge are preserved.
Common mistakes
- Treating an empirical formula as proof of a molecule's size.
- Changing a polyatomic ion's subscripts to balance charges.
What can you explain now?
Find the empirical formula of C₂H₄.
Compare with the explanation
CH₂
Divide both atom counts by their common factor two; the molecule still has formula C₂H₄.
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Make it stick
Explain why C₂H₄ and CH₂ express different information. Then derive an ionic formula from charges and show the charge sum explicitly.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.