Empirical & molecular formulae
Empirical & molecular formulae
Find empirical formulae from masses and percentages
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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.
Magnesium oxide
A compound contains 0.486 g Mg and 0.320 g O. Use Ar(Mg) = 24.3 and Ar(O) = 16.0.
Determine the empirical formula of the compound.
Worked answer
| Quantity | Mg | O |
|---|---|---|
| Mass / g | 0.486 | 0.320 |
| Ar | 24.3 | 16.0 |
| Amount / mol | n = m/Aᵣ n = 0.486/24.3 = 0.0200 | n = m/Aᵣ n = 0.320/16.0 = 0.0200 |
| ÷ smallest | Ratio = amount ÷ smallest amount 0.0200/0.0200 = 1 | Ratio = amount ÷ smallest amount 0.0200/0.0200 = 1 |
| Whole-number ratio | 1 | 1 |
Percentage composition
A compound contains 40.0% carbon, 6.67% hydrogen and 53.33% oxygen by mass. Determine its empirical formula. Use Aᵣ(C) = 12.0, Aᵣ(H) = 1.0 and Aᵣ(O) = 16.0.
Worked answer
| Quantity | C | H | O |
|---|---|---|---|
| Mass / g | 40.0 | 6.67 | 53.33 |
| Ar | 12.0 | 1.0 | 16.0 |
| Amount / mol | n = m/Aᵣ n = 40.0/12.0 = 3.333… | n = m/Aᵣ n = 6.67/1.0 = 6.67 | n = m/Aᵣ n = 53.33/16.0 = 3.333… |
| ÷ smallest | Ratio = amount ÷ smallest amount 3.333…/3.333… = 1 | Ratio = amount ÷ smallest amount 6.67/3.333… = 2 | Ratio = amount ÷ smallest amount 3.333…/3.333… = 1 |
| Whole-number ratio | 1 | 2 | 1 |
| Ratio containing | Multiply every part by | Example |
|---|---|---|
| 1.5 | 2 | 1 : 1.5 → 2 : 3 |
| 1.333… | 3 | 1 : 1.333… → 3 : 4 |
| 1.25 | 4 | 1 : 1.25 → 4 : 5 |
| 1.667… | 3 | 1 : 1.667… → 3 : 5 |
Recognise a half ratio
A phosphorus oxide is 43.6% P and 56.4% O. Use Ar(P) = 31.0 and Ar(O) = 16.0.
Determine the empirical formula of the phosphorus oxide from these percentages by mass.
Worked answer
| Quantity | P | O |
|---|---|---|
| Mass / g | 43.6 | 56.4 |
| Ar | 31.0 | 16.0 |
| Amount / mol | n = m/Aᵣ n = 43.6/31.0 = 1.40645… | n = m/Aᵣ n = 56.4/16.0 = 3.525 |
| ÷ smallest | Ratio = amount ÷ smallest amount 1.40645…/1.40645… = 1 | 2.506… ≈ 2.5 |
| Whole-number ratio | 2 | 5 |
Find an unknown metal
A metal chloride MCl2 contains 0.486 g M and 1.420 g Cl. Ar(Cl) = 35.5.
Calculate the relative atomic mass of M and identify the metal.
Worked answer
n(Cl atoms) = 1.420/35.5 = 0.0400 mol
n(M) = 0.0400/2 = 0.0200 mol
Ar(metal) = 0.486/0.0200 = 24.3
🧪 TEST IT
Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Use Ar: Mg 24.3; Cl 35.5; P 31.0; O 16.0.
- Determine the empirical formula of a compound containing 1.215 g Mg and 3.550 g Cl. (3 marks)
- Determine the empirical formula of an oxide containing 43.7% phosphorus and 56.3% oxygen by mass. (3 marks)
- Explain why a mass ratio cannot be used directly as an empirical-formula ratio. (1 mark)
Total: 7 marks
Show answer & marking points
(a) n(Mg) = m/Aᵣ
n(Mg) = 1.215/24.3 = 0.0500 mol and n(Cl) = m/Aᵣ
n(Cl) = 3.550/35.5 = 0.100 mol. [1] Ratio 1:2. [1] MgCl₂. [1]
(b) For 100 g, n(P) = m/Aᵣ
n(P) = 43.7/31.0 = 1.40968… and n(O) = m/Aᵣ
n(O) = 56.3/16.0 = 3.51875 mol. [1] Ratio 1:2.496… is close to 1:2.5; multiply both by 2. [1] P₂O₅. [1]
(c) Formula subscripts give numbers of atoms; the elements have different atomic masses, so first convert masses to amounts. [1]
Find an oxide formula from heating data
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When a metal forms an oxide, the solid gains oxygen and its mass increases. First subtract the empty crucible mass so you are comparing the substances, not the apparatus.
The increase is the mass of oxygen gained, provided no solid is lost and the expected oxide is the only product. Convert the metal and oxygen masses to moles, then find their simplest ratio.
- Heat, cool and weigh again until consecutive masses agree within the balance’s resolution.
- Allow the apparatus to cool before weighing; hot apparatus gives unreliable readings.
- Keep the lid slightly open so oxygen can enter while loss of solid is reduced.
Incomplete oxidation makes the apparent oxygen gain too small, so the calculated oxygen-to-metal ratio is too low. Keep heating, cooling and weighing until the mass stops increasing. This is evidence that all the metal has reacted with oxygen.
Copper reacts with oxygen
Crucible mass 24.800 g; crucible + Cu 26.070 g; crucible + oxide 26.390 g. Use Ar(Cu) = 63.5 and Ar(O) = 16.0.
The copper is heated in oxygen to constant mass. Use the measurements to determine the empirical formula of the copper oxide.
Worked answer
| Quantity | Cu | O |
|---|---|---|
| Mass / g | 1.270 | 0.320 |
| Ar | 63.5 | 16.0 |
| Amount / mol | n = m/Aᵣ n = 1.270/63.5 = 0.0200 | n = m/Aᵣ n = 0.320/16.0 = 0.0200 |
| ÷ smallest | Ratio = amount ÷ smallest amount 0.0200/0.0200 = 1 | Ratio = amount ÷ smallest amount 0.0200/0.0200 = 1 |
| Whole-number ratio | 1 | 1 |
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Work on paper. Attempt each question before opening its answer. These are original CrackingChem questions, written in AQA style.
Heating 4.46 g iron in oxygen forms 6.38 g iron oxide. Use Aᵣ(Fe) = 55.8 and Aᵣ(O) = 16.0.
- Determine the empirical formula of the oxide. (4 marks)
- Explain the purpose of repeated heating, cooling and weighing until constant mass. (1 mark)
- Explain why the lid is kept slightly open during heating. (1 mark)
- Explain why the crucible must cool before weighing. (1 mark)
- State the effect of incomplete oxidation on the calculated oxygen-to-iron ratio. (1 mark)
Total: 8 marks
Show answer & marking points
(a) Oxygen gained = 6.38 − 4.46 = 1.92 g. [1] n(Fe) = m/Aᵣ
n(Fe) = 4.46/55.8 = 0.079928…; n(O) = m/Aᵣ
n(O) = 1.92/16.0 = 0.120 mol. [1] Fe:O = 1:1.501… ≈ 2:3. [1] Fe₂O₃. [1]
(b) Unchanged consecutive masses show no further measurable change under those conditions and support completion of oxidation. [1] They do not alone prove the solid’s identity or purity.
(c) It lets oxygen enter while reducing loss of solid. [1]
(d) Hot apparatus causes an unreliable balance reading, for example through convection. [1]
(e) The apparent oxygen-to-iron ratio is too low because oxygen gain is too small. [1]
Use relative mass to find a molecular formula
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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.
A carbohydrate
Empirical formula CH2O; Mr = 180.0. Use C 12.0, H 1.0 and O 16.0.
Determine the molecular formula of the carbohydrate.
Worked answer
Multiplier = Mᵣ/empirical-formula mass
Phosphorus oxide
Empirical formula P2O5; Mr = 284.0. Use P 31.0 and O 16.0.
Determine the molecular formula of the phosphorus oxide.
Worked answer
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 AQA style.
A compound has empirical formula C₂H₄O and Mᵣ = 88.0. Use Aᵣ: C 12.0, H 1.0, O 16.0.
- Determine the molecular formula. (3 marks)
- Distinguish an empirical formula from a molecular formula. (2 marks)
Total: 5 marks
Show answer & marking points
Multiplier = Mᵣ/empirical-formula mass
(a) Empirical-formula mass = 2(12.0) + 4(1.0) + 16.0 = 44.0. [1] Multiplier = 88.0/44.0 = 2. [1] C₄H₈O₂. [1]
(b) The empirical formula gives the simplest whole-number atom ratio. [1] The molecular formula gives the actual numbers of atoms of each element in one molecule. [1]
Find a formula from combustion products
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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.
n amount of the specified substance or particles, in mol.
m mass of the specified sample or substance, in g.
A CHO compound
A 0.460 g sample gives 0.880 g CO2 and 0.540 g H2O. Use Mr(CO2) = 44.0 and Mr(H2O) = 18.0; Ar(C) = 12.0, H = 1.0, 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) = n × Aᵣ
m(C) = 0.0200 × 12.0 = 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/Aᵣ n = 0.240/12.0 = 0.0200 | n = m/Aᵣ n = 0.0600/1.0 = 0.0600 | n = m/Aᵣ 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 AQA style.
Complete combustion of 0.880 g of a compound containing only C, H and O gives 1.760 g CO₂ and 0.720 g H₂O. Mᵣ(compound) = 88.0. Use Aᵣ: C 12.0, H 1.0, O 16.0.
- Determine the empirical formula, showing how the oxygen in the original sample is found. (6 marks)
- Determine the molecular formula. (2 marks)
- Explain why not all the oxygen in the combustion products came from the original compound. (1 mark)
- Explain why the oxygen-by-mass-difference method would fail if the original compound also contained nitrogen. (1 mark)
Total: 10 marks
Show answer & marking points
(a) n(C) = n(CO₂) = m/Mᵣ
n(CO₂) = 1.760/44.0 = 0.0400 mol. [1] n(H) = 2m(H₂O)/Mᵣ(H₂O)
n(H) = 2(0.720/18.0) = 0.0800 mol. [1] m(C) = 0.480 g and m(H) = 0.0800 g. [1] Oxygen in sample = 0.880 − 0.480 − 0.0800 = 0.320 g, so n(O) = 0.0200 mol. [1] Ratio C:H:O = 2:4:1 [1]; C₂H₄O. [1]
(b) Empirical-formula mass = 44.0; multiplier = 88.0/44.0 = 2. [1] Molecular formula C₄H₈O₂. [1]
(c) Oxygen gas supplied for combustion also contributes oxygen atoms to the products. [1]
(d) The remaining sample mass would include nitrogen as well as oxygen, so it could not all be assigned to oxygen. [1]
Complete combustion of 1.480 g of a compound containing only C, H and O gives 2.640 g CO₂ and 1.080 g H₂O. Mᵣ(compound) = 148.0. Use Aᵣ: C 12.0, H 1.0, O 16.0. Deduce its molecular formula. (7 marks)
Total: 7 marks
Show answer & marking points
n(C) = m(CO₂)/Mᵣ(CO₂)
n(C) = 2.640/44.0 = 0.0600 mol. [1] n(H) = 2m(H₂O)/Mᵣ(H₂O)
n(H) = 2(1.080/18.0) = 0.120 mol. [1] m(C) = 0.720 g, m(H) = 0.120 g, so m(O in sample) = 1.480 − 0.720 − 0.120 = 0.640 g. [1] n(O) = m/Aᵣ
n(O) = 0.640/16.0 = 0.0400 mol. [1] Ratio 1.5:3:1 becomes 3:6:2, empirical formula C₃H₆O₂. [1] Empirical-formula mass = 74.0; multiplier = 148.0/74.0 = 2. [1] C₆H₁₂O₄. [1]