Empirical and molecular formulas from data
Turn masses into an atom ratio, then use molecular mass to find the actual molecular formula.
Section 3.3 · The mole and the Avogadro constant
Exam scope and assessment notes
Extended only (Supplement) · Chemistry 0971, 2026–2028. Core concentration units are introduced in a separate lesson. 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
- Distinguish empirical and molecular formulas.
- Convert mass into amount and calculate relative formula mass.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which starting quantities give an atom ratio from measured element masses?
What you will learn
- Deduce an empirical formula from element masses or percentage composition.
- Use relative molecular mass to deduce the molecular formula from an empirical formula.
Convert each element's mass before comparing
An empirical formula gives the simplest whole-number ratio of atoms. A gram ratio is not generally an atom ratio because atoms of different elements have different masses. Divide each element's mass by its relative atomic mass to obtain proportional amounts in moles of atoms, then divide every amount by the smallest. If percentage composition is given, assume a 100 g sample so each percentage becomes a convenient mass in grams.
Inspect the reduced ratio rather than rounding everything to the nearest integer. A ratio 1:1.5 becomes 2:3 by multiplying every part by two; 1:2.5 becomes 2:5. Values very close to integers because of stated experimental precision can be treated as those integers, but a genuine half or third should be resolved by multiplying the whole ratio. If an element's percentage is found by difference, the sample must contain only the specified elements.
Use molecular mass to choose the multiplier
Calculate the relative mass of one empirical-formula unit. Divide the supplied relative molecular mass by that value to get a whole-number multiplier, then multiply every empirical subscript by it. For empirical CH₂O with relative empirical mass 30, a relative molecular mass of 180 gives multiplier six and molecular formula C₆H₁₂O₆. Empirical composition alone cannot identify that multiplier.
Combustion data can provide the starting atom amounts when the question gives a complete combustion model: one mole of CO₂ contains one mole of carbon atoms, and one mole of H₂O contains two moles of hydrogen atoms. For a compound known to contain only carbon, hydrogen and oxygen, calculate the carbon and hydrogen masses from the products, then use the original sample mass to find its oxygen mass by difference. Oxygen in combustion products also comes from the supplied oxygen gas, so it cannot all be assigned to the original compound.
Pause and explain
A reduced atom ratio is 1:1.5. What is its simplest whole-number ratio?
Worked example
A compound contains 40.0% C, 6.67% H and 53.33% O. Its relative molecular mass is 180. Find empirical and molecular formulas using C=12, H=1 and O=16.
Show the worked solution
- For 100 g, proportional atom amounts are 40/12=3.333… mol C, 6.67/1=6.67 mol H and 53.33/16=3.333125 mol O.
- Divide by the smallest to obtain approximately 1:2:1, so empirical formula is CH₂O with relative mass 30.
- 180/30=6; multiply every subscript by six to obtain C₆H₁₂O₆.
Answer Empirical CH₂O; molecular C₆H₁₂O₆.
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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.
A compound contains 2.7 g Al and 2.4 g O. Find its empirical formula using Al=27 and O=16.
Give me a hint
Calculate both amounts, then resolve the resulting half-integer ratio.
Compare my reasoning
- Al amount is 2.7/27=0.100 mol; O amount is 2.4/16=0.150 mol.
- Divide by 0.100 to give Al:O=1:1.5.
- Multiply both by two, giving 2:3 and Al₂O₃.
Al₂O₃
Look for these in your work
- Masses are converted before forming the ratio.
- The half-integer is handled without arbitrary rounding.
A compound contains 80.0% carbon and 20.0% hydrogen. Its relative molecular mass is 30. Find both formulas using C=12 and H=1.
Give me a hint
Assume 100 g and find the smallest atom ratio first.
Compare my reasoning
- Proportional amounts are 80/12=6.666… mol C and 20/1=20 mol H.
- Divide by 6.666… to get 1:3, so empirical formula is CH₃ with relative mass 15.
- 30/15=2, so molecular formula is C₂H₆.
Empirical CH₃; molecular C₂H₆.
Look for these in your work
- The percentage data are treated as masses, then converted.
- The multiplier applies to both element subscripts.
A 2.30 g compound containing only C, H and O gives 4.40 g CO₂ and 2.70 g H₂O in complete combustion. Find its empirical formula using C=12, H=1 and O=16.
Give me a hint
Obtain C and H amounts from product molecules; find original oxygen mass by subtracting C and H masses from 2.30 g.
Compare my reasoning
- CO₂ amount is 4.40/44=0.100 mol, giving 0.100 mol C atoms and carbon mass 1.20 g. H₂O amount is 2.70/18=0.150 mol, giving 0.300 mol H atoms and hydrogen mass 0.300 g.
- Original oxygen mass is 2.30−1.20−0.300=0.800 g, giving 0.800/16=0.0500 mol O atoms.
- C:H:O amounts 0.100:0.300:0.0500 reduce to 2:6:1, giving C₂H₆O.
C₂H₆O
Look for these in your work
- Hydrogen atom count includes two atoms per water molecule.
- Only the original sample's oxygen is found by mass difference.
Common mistakes
- Using percentage or gram ratios directly as atom ratios.
- Rounding a ratio such as 1:1.5 into 1:2.
- Changing only one subscript when applying the molecular multiplier.
- Assigning all oxygen in combustion products to the original sample.
What can you explain now?
A compound has empirical formula CH₂ and relative molecular mass 56. Use C=12 and H=1 to find its molecular formula.
Compare with the explanation
C₄H₈
Empirical relative mass is 12+2=14. The multiplier is 56/14=4, giving C₄H₈.
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Make it stick
Deduce a formula from a new percentage table, explaining each division. Show what extra information is needed to move from the empirical formula to a molecular formula.
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.