Empirical ratios and ionic formulas
Distinguish an actual molecular formula from a simplest ratio and balance the charges of ions.
Section C3.1 · Formulae
Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.
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.
Sources for this lesson
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.
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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.