Empirical & molecular formulae
Empirical & molecular formulae
Empirical formulae from composition
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An empirical formula gives the simplest whole-number ratio of atoms. A mass ratio is not an atom ratio because different atoms have different masses.
- Find the mass of each element. For percentages, assume a 100 g sample so each percentage becomes grams.
- Divide each mass by its Aᵣ to find moles.
- Divide every mole amount by the smallest one.
- If a simple fraction remains, multiply the entire ratio to obtain whole numbers.
- Use those whole numbers as the formula subscripts.
Do not round 1 : 1.5 to 1 : 2. Multiply both values by 2 to get 2 : 3. For a ratio of 1 : 1.33…, multiply both by 3 to get 3 : 4. This may look like 1 : 1.3 if rounded too early, so keep the full calculator values when choosing the multiplier.
| Ratio after division | Next step |
|---|---|
| 1 : 1.5 | Multiply both by 2 → 2 : 3 |
| 1 : 1.33… | Multiply both by 3 → 3 : 4 |
| 1 : 1.25 | Multiply both by 4 → 4 : 5 |
An iron oxide
1.116 g Fe forms 1.596 g of an oxide. Use Ar(Fe) = 55.8 and Ar(O) = 16.0.
Determine the empirical formula of the iron oxide.
Worked answer
| Quantity | Fe | O |
|---|---|---|
| Mass / g | 1.116 | 0.480 |
| Ar | 55.8 | 16.0 |
| Amount / mol | n = m/M n = 1.116/55.8 = 0.0200 | n = m/M n = 0.480/16.0 = 0.0300 |
| ÷ smallest | Ratio = amount ÷ smallest amount 0.0200/0.0200 = 1 | Ratio = amount ÷ smallest amount 0.0300/0.0200 = 1.5 |
| Whole-number ratio | 2 | 3 |
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A compound contains 0.486 g magnesium and 1.420 g chlorine. Determine its empirical formula. Use Ar(Mg) = 24.3 and Ar(Cl) = 35.5. Show your working. (4 marks)
Total: 4 marks
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| Quantity | Mg | Cl |
|---|---|---|
| Mass / g | 0.486 | 1.420 |
| M / g mol−1 | 24.3 | 35.5 |
| Amount / mol | n = m/M n(Mg) = 0.486/24.3 = 0.0200 [1] | n = m/M n(Cl atoms) = 1.420/35.5 = 0.0400 [1] |
| ÷ smallest | 0.0200/0.0200 = 1 | 0.0400/0.0200 = 2 |
The simplest whole-number ratio Mg : Cl is 1 : 2. [1] Empirical formula: MgCl2. [1] No further scaling is needed.
A 1.674 g sample of iron reacts completely with oxygen to form 2.314 g of an oxide. Determine the empirical formula of the oxide. Use Ar(Fe) = 55.8 and Ar(O) = 16.0. Show how you obtain the mass of oxygen. (6 marks)
Total: 6 marks
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Mass of oxygen = mass of oxide − mass of iron
m(O) = 2.314 − 1.674 = 0.640 g. [1]
| Quantity | Fe | O |
|---|---|---|
| Mass / g | 1.674 | 0.640 |
| M / g mol−1 | 55.8 | 16.0 |
| Amount / mol | n = m/M n(Fe) = 1.674/55.8 = 0.0300 [1] | n = m/M n(O atoms) = 0.640/16.0 = 0.0400 [1] |
| ÷ smallest | 0.0300/0.0300 = 1 | 0.0400/0.0300 = 1.333… |
| Multiply both by 3 | 1 × 3 = 3 | 1.333… × 3 = 4 |
Divide both amounts in moles by 0.0300 to obtain 1 : 1.333… [1] Multiply the complete ratio by 3 to obtain 3 : 4. [1] Empirical formula: Fe3O4. [1] Keep the unrounded ratio when choosing the multiplier.
Molecular formulae and percentage composition
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The empirical formula gives the simplest atom ratio. The molecular formula gives the actual numbers of atoms in one molecule, so it is a whole-number multiple of the empirical formula.
- Find the empirical formula first. If composition is given as percentages, begin with a 100 g sample.
- Add the Aᵣ contributions to find the empirical-formula mass.
- Divide the supplied Mᵣ by the empirical-formula mass.
- Multiply every empirical-formula subscript by that whole-number multiplier.
Check that the final molecular formula adds up to the supplied Mᵣ. The multiplier applies to every element, not just the last one.
Mr relative molecular or formula mass; no unit.
From percentages to a molecular formula
A compound contains 54.5% C, 9.1% H and 36.4% O and has Mr = 88.0. Use C 12.0, H 1.0 and O 16.0.
These are percentages by mass. Determine the empirical formula and then the molecular formula.
Worked answer
| Quantity | C | H | O |
|---|---|---|---|
| Mass / g | 54.5 | 9.1 | 36.4 |
| Ar | 12.0 | 1.0 | 16.0 |
| Amount / mol | n = m/M n = 54.5/12.0 = 4.54167… | n = m/M n = 9.1/1.0 = 9.10 | n = m/M n = 36.4/16.0 = 2.275 |
| ÷ smallest | 1.996… ≈ 2 | Ratio = amount ÷ smallest amount 9.10/2.275 = 4 | Ratio = amount ÷ smallest amount 2.275/2.275 = 1 |
| Whole-number ratio | 2 | 4 | 1 |
Multiplier = Mᵣ/empirical-formula mass
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
A molecular compound is 55.8% C, 7.0% H and 37.2% O by mass. Its Mr is 86.0. Use Ar: C 12.0, H 1.0, O 16.0.
(a) Determine its empirical and molecular formulae. (5 marks)
(b) Another compound has empirical formula C2H4O and Mr = 44.0. State its molecular formula. (1 mark)
Total: 6 marks
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(a) For 100 g, amounts are C: 55.8/12.0 = 4.65; H: 7.0/1.0 = 7.0; O: 37.2/16.0 = 2.325 mol. [1] Dividing by 2.325 gives approximately 2 : 3 : 1, so the empirical formula is C2H3O. [1] Its formula mass is 43.0 [1]; 86.0/43.0 = 2 [1]; molecular formula C4H6O2. [1]
(b) C2H4O, because the multiplier is 44.0/44.0 = 1. [1]
Formulae from combustion analysis
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Complete combustion lets you count carbon and hydrogen through the products they form. For a compound containing only C, H and O, all its carbon enters CO₂ and all its hydrogen enters H₂O.
- Find moles of CO₂. Each mole contains one mole of C atoms, so n(C) = n(CO₂).
- Find moles of H₂O. Each mole contains two moles of H atoms, so n(H) = 2n(H₂O).
- Convert the C and H amounts into masses. Subtract them from the original sample mass to find the oxygen originally in the compound.
- Convert that oxygen mass to moles, then find the empirical atom ratio.
Oxygen in the combustion products also comes from the oxygen gas used in the reaction. That is why you cannot simply count it all as oxygen from the sample. If Mᵣ is supplied, use it to find the molecular formula.
A compound containing C, H and O
0.460 g of compound forms 0.880 g CO2 and 0.540 g H2O. Use C 12.0, H 1.0 and O 16.0.
The compound contains carbon, hydrogen and oxygen only and is burned completely. Determine its empirical formula.
Worked answer
n(C) = 0.880/44.0 = 0.0200 mol; m(C) = 0.240 g
n(H) = 2(0.540/18.0) = 0.0600 mol; m(H) = 0.0600 g
| Quantity | C | H | O |
|---|---|---|---|
| Mass / g | 0.240 | 0.0600 | 0.160 |
| Ar | 12.0 | 1.0 | 16.0 |
| Amount / mol | n = m/M n = 0.240/12.0 = 0.0200 | n = m/M n = 0.0600/1.0 = 0.0600 | n = m/M n = 0.160/16.0 = 0.0100 |
| ÷ smallest | Ratio = amount ÷ smallest amount 0.0200/0.0100 = 2 | Ratio = amount ÷ smallest amount 0.0600/0.0100 = 6 | Ratio = amount ÷ smallest amount 0.0100/0.0100 = 1 |
| Whole-number ratio | 2 | 6 | 1 |
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in OCR A style.
Complete combustion of 0.440 g of a compound containing only C, H and O produces 0.880 g CO2 and 0.360 g H2O. Its Mr is 88.0. Use M(CO2) = 44.0 and M(H2O) = 18.0 g mol−1; Ar: C 12.0, H 1.0, O 16.0.
(a) Determine its empirical and molecular formulae. (7 marks)
(b) Explain why the oxygen atoms in the combustion products cannot all be assigned to the original compound. (1 mark)
Total: 8 marks
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(a) n(C) = m(CO₂)/M(CO₂)
n(C) = 0.880/44.0 = 0.0200 mol. [1] n(H) = 2m(H₂O)/M(H₂O)
n(H) = 2 × 0.360/18.0 = 0.0400 mol. [1] Masses of C and H are 0.240 and 0.0400 g, giving sample oxygen mass 0.440 − 0.240 − 0.0400 = 0.160 g. [1] n(O) = m/M
n(O) = 0.160/16.0 = 0.0100 mol. [1] Ratio 2 : 4 : 1 gives C2H4O. [1] Empirical-formula mass = 44.0, so the multiplier is 88.0/44.0 = 2. [1] Molecular formula C4H8O2. [1]
(b) Some of that oxygen came from the oxygen supply used to burn the sample. [1]