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This lesson develops the formal distinction between the empirical formula — the simplest whole-number ratio of atoms in a compound — and the molecular formula — the actual number of atoms present in one molecule. The empirical formula is the universal output of any compositional analysis: percentage-composition data, combustion analysis, and hydrated-salt heating experiments all return ratios first, identities second. To go from empirical to molecular, the relative molecular mass Mᵣ must be supplied independently — most commonly from mass spectrometry. This lesson links the mole concept developed in the previous lesson to the identification of organic compounds, the determination of stoichiometry for unknown inorganic salts, and the analytical foundation that NMR and IR later confirm.
Spec mapping (AQA 7405): This lesson maps directly to §3.1.2.3 (empirical and molecular formulae and the relationship between them). It builds on the mole concept of §3.1.2.2 covered in lesson 4 — every calculation here applies n = m/M at least once. The combustion-analysis material previews §3.3.6 (organic analysis, including the use of analytical data to deduce structure) and §3.3.15 (NMR), where empirical-formula determination is the first stage before any spectroscopic structure assignment is attempted. Hydrate-formula determination is an indicative practical activity referenced in the inorganic strand (§3.2). Refer to the official AQA specification document for the exact wording of each section.
Assessment objectives: Defining empirical and molecular formulae, and recognising that ionic compounds possess only an empirical formula, are AO1 recall items frequently appearing as the opening parts of Paper 1 and Paper 2 calculation questions. The percentage-composition method, combustion-analysis calculations, and water-of-crystallisation determinations are AO2 — apply knowledge and understanding to numerical data. Multi-step problems — for example, combining combustion-mass data with a mass-spectrometric Mᵣ to deduce a molecular formula, or evaluating an experimental procedure for a hydrate determination — test AO3 (analysis, evaluation, and synthesis). On Paper 3 (practical and synoptic), questions on empirical-formula determination almost always sit alongside an experimental procedure to evaluate.
For molecular substances — covalently bonded, discrete molecules — both formulae exist and the molecular formula is an integer multiple of the empirical formula:
Molecular formula = (empirical formula) × n
where n = Mᵣ(compound) / Mᵣ(empirical formula)
For ionic compounds — extended lattices with no discrete molecule — only the empirical formula is meaningful. The "formula" you write for NaCl, MgO, or CaF₂ is already the empirical formula: the simplest whole-number ratio of cations to anions in the lattice. There is no "molecular formula" of NaCl because there is no NaCl molecule. The lattice contains an essentially infinite number of Na⁺ and Cl⁻ ions in 1:1 ratio.
For macromolecular (giant covalent) substances — diamond, graphite, silicon dioxide — the situation is similar: the formula written (C for diamond, SiO₂ for silica) is empirical, and there is no discrete molecule with a finite molecular formula. A single crystal of diamond can be thought of as one "molecule" containing 10²² or more carbon atoms, but in practice we use the empirical formula C and characterise the substance by its lattice structure.
| Compound | Type | Molecular formula | Empirical formula | n |
|---|---|---|---|---|
| Water | molecular | H₂O | H₂O | 1 |
| Methane | molecular | CH₄ | CH₄ | 1 |
| Ethane | molecular | C₂H₆ | CH₃ | 2 |
| Ethene | molecular | C₂H₄ | CH₂ | 2 |
| Glucose | molecular | C₆H₁₂O₆ | CH₂O | 6 |
| Sucrose | molecular | C₁₂H₂₂O₁₁ | C₁₂H₂₂O₁₁ | 1 |
| Benzene | molecular | C₆H₆ | CH | 6 |
| Ethanoic acid | molecular | C₂H₄O₂ | CH₂O | 2 |
| Hydrogen peroxide | molecular | H₂O₂ | HO | 2 |
| Sodium chloride | ionic | — (none) | NaCl | n/a |
| Magnesium oxide | ionic | — (none) | MgO | n/a |
| Calcium fluoride | ionic | — (none) | CaF₂ | n/a |
| Silicon dioxide | macromolecular | — (none) | SiO₂ | n/a |
| Diamond | macromolecular | — (none) | C | n/a |
Key Point: Glucose (C₆H₁₂O₆), ethanal CH₃CHO (= C₂H₄O₂ if doubled — actually C₂H₄O empirical), methanal (HCHO = CH₂O), and ethanoic acid (C₂H₄O₂) all share the same empirical formula CH₂O. Empirical-formula identification alone cannot distinguish them — you need Mᵣ to identify a specific molecular formula, and you need NMR or IR to determine which structural isomer the molecule is. This is exactly why analytical workflows proceed in stages: combustion analysis → empirical formula → mass spectrometry → molecular formula → NMR/IR → structural identification.
The most common A-Level task in this topic is to be given the percentage composition by mass of a compound and asked for its empirical formula. The method is a five-step recipe — once memorised, it works for any number of elements.
Step 1 — Percentages. Write down the percentage by mass of each element. If the percentages given do not add to 100%, the missing fraction is usually oxygen (or, less commonly, nitrogen). Compute the missing % by subtraction: 100% − (sum of given percentages).
Step 2 — Imagine 100 g. Treating the percentages as grams in a 100 g sample is the simplest mental device. The percentage of carbon then becomes the mass of carbon in the sample (in g), and so on. This step is implicit in the standard tabular layout.
Step 3 — Convert mass to moles. Divide each mass (= percentage) by the relative atomic mass Aᵣ of that element. This applies n = m/M (lesson 4) and gives the relative number of moles of each element present.
Step 4 — Divide by the smallest. Identify the smallest of the mole values and divide all the mole values by it. This produces a set of ratio values, at least one of which is 1.
Step 5 — Convert to whole-number ratio. If all the ratios are already integers (within rounding tolerance — within about ±0.05), write them as the empirical-formula subscripts. If a ratio is close to a simple fraction (1.5, 1.33, 1.25, 1.67), multiply every ratio by the appropriate integer to clear the fraction (×2 for 0.5, ×3 for 0.33/0.67, ×4 for 0.25/0.75). The final integer set is the empirical-formula subscript list.
Critical pedagogical point: Never round to whole numbers before Step 5. A ratio of 1.50 must be multiplied by 2 to give 3; rounding 1.50 to 2 directly gives the wrong empirical formula. The rounding tolerance is roughly ±0.05; outside that, treat the fraction as significant and apply the multiplier.
A compound contains 40.0% carbon, 6.7% hydrogen, and 53.3% oxygen by mass. Determine its empirical formula.
| Element | % by mass | ÷ Aᵣ | Moles | ÷ smallest | Ratio |
|---|---|---|---|---|---|
| C | 40.0 | ÷ 12.0 | 3.33 | ÷ 3.33 | 1 |
| H | 6.7 | ÷ 1.0 | 6.7 | ÷ 3.33 | 2 |
| O | 53.3 | ÷ 16.0 | 3.33 | ÷ 3.33 | 1 |
Empirical formula = CH₂O
This is the empirical formula shared by methanal, ethanoic acid, glycolaldehyde, glucose, fructose, ribose, and many other carbohydrate-class compounds — a useful reminder that the empirical formula on its own does not identify a substance. The percentages here (40.0/6.7/53.3) correspond to the textbook idealisation of "carbohydrate composition", which historically gave rise to the name "carbohydrate" = "hydrated carbon" — though that historical interpretation is structurally misleading and should not be taken literally.
A compound contains 52.2% carbon, 13.0% hydrogen, and 34.8% oxygen. Determine its empirical formula.
| Element | % by mass | ÷ Aᵣ | Moles | ÷ smallest | Ratio |
|---|---|---|---|---|---|
| C | 52.2 | ÷ 12.0 | 4.35 | ÷ 2.175 | 2.00 |
| H | 13.0 | ÷ 1.0 | 13.0 | ÷ 2.175 | 5.98 ≈ 6 |
| O | 34.8 | ÷ 16.0 | 2.175 | ÷ 2.175 | 1.00 |
Empirical formula = C₂H₆O
This is the empirical formula of ethanol (C₂H₆O = CH₃CH₂OH) and of its isomer methoxymethane (CH₃OCH₃). Note that C₂H₆O is already the molecular formula of ethanol — n = 1 here because the molecule cannot be simplified further. The ratio 2 : 6 : 1 contains no common factor greater than 1.
A compound contains 36.8% nitrogen and 63.2% oxygen. Determine its empirical formula.
| Element | % by mass | ÷ Aᵣ | Moles | ÷ smallest | Ratio |
|---|---|---|---|---|---|
| N | 36.8 | ÷ 14.0 | 2.629 | ÷ 2.629 | 1.000 |
| O | 63.2 | ÷ 16.0 | 3.950 | ÷ 2.629 | 1.502 |
The ratio 1 : 1.50 is not whole numbers. Multiply both by 2 to clear the half-integer:
Ratio = 2 : 3
Empirical formula = N₂O₃ (dinitrogen trioxide)
Exam Tip: Recognise the standard multipliers. A ratio of 1.5 means ×2 (gives 3); 1.33 or 1.67 means ×3 (gives 4 or 5); 1.25 or 1.75 means ×4 (gives 5 or 7); 1.2 or 1.4 etc. means ×5. Train your eye to spot these — agonising over whether 1.49 "should be rounded to 1" is wasted effort if you recognise it as 1.5.
1.35 g of aluminium reacts with 1.20 g of oxygen to form a single oxide. Determine the empirical formula of the oxide.
| Element | Mass (g) | ÷ Aᵣ | Moles | ÷ smallest | Ratio |
|---|---|---|---|---|---|
| Al | 1.35 | ÷ 27.0 | 0.0500 | ÷ 0.0500 | 1.00 |
| O | 1.20 | ÷ 16.0 | 0.0750 | ÷ 0.0500 | 1.50 |
Ratio = 1.00 : 1.50. Multiply by 2: ratio = 2 : 3.
Empirical formula = Al₂O₃
This is the empirical formula of aluminium oxide — and because Al₂O₃ is ionic (extended lattice), this is also the formula you write in any equation involving aluminium oxide. There is no "molecular formula" of Al₂O₃; the formula unit Al₂O₃ corresponds to two Al³⁺ ions and three O²⁻ ions, which balances charge: 2 × (+3) + 3 × (−2) = 0. Note the synoptic link: for ionic compounds, the empirical formula is determined by charge balance, which gives the same answer as the mole-ratio analysis.
When an organic compound containing carbon, hydrogen, and (optionally) oxygen is burned completely in excess oxygen, every carbon atom ends up as one CO₂ molecule and every two hydrogen atoms end up as one H₂O molecule:
CₓHᵧO_z + (x + y/4 − z/2) O₂ → x CO₂ + (y/2) H₂O
This gives a powerful analytical principle: weigh the CO₂ and H₂O produced, and you can back-calculate the moles of C and H in the original sample. Oxygen content is then found by mass balance — subtract the masses of C and H from the original sample mass; the remainder is oxygen. If the masses of C and H already sum to the original sample mass (within rounding tolerance), then there is no oxygen in the compound.
The general approach in five steps:
Step 1. Compute n(CO₂) = m(CO₂) / 44.0 and n(H₂O) = m(H₂O) / 18.0.
Step 2. Compute moles of C and H: n(C) = n(CO₂) and n(H) = 2 × n(H₂O). The factor of 2 is critical — each H₂O molecule contains 2 hydrogen atoms.
Step 3. Convert moles of C and H back to masses: m(C) = n(C) × 12.0 and m(H) = n(H) × 1.0.
Step 4. Determine m(O) by subtraction: m(O) = m(sample) − m(C) − m(H). If this value is essentially zero, the compound contains only C and H.
Step 5. If oxygen is present, compute n(O) = m(O) / 16.0. Then divide all mole values by the smallest to obtain the empirical-formula ratio.
If a relative molecular mass is supplied — typically from mass spectrometry, as developed in lesson 1 — the molecular formula follows from n = Mᵣ(compound) / Mᵣ(empirical formula). Mass spectrometry returns the M⁺ (molecular-ion) peak, whose m/z value equals Mᵣ to the nearest integer for the dominant isotopologue. This is the canonical empirical-to-molecular bridge in A-Level Chemistry: combustion gives composition, mass spectrometry gives Mᵣ, the two together yield the molecular formula.
0.900 g of an organic compound containing C, H, and O only was burned in excess oxygen. The products were 1.32 g of CO₂ and 0.540 g of H₂O. Determine the empirical formula.
Step 1 — Moles of products.
Step 2 — Moles of C and H.
Step 3 — Masses of C and H.
Step 4 — Mass of O by subtraction.
Step 5 — Ratio.
Empirical formula = CH₂O
This empirical formula is consistent with methanal (HCHO, Mᵣ = 30), ethanoic acid (CH₃COOH, Mᵣ = 60), glycolaldehyde (HOCH₂CHO, Mᵣ = 60), and glucose (C₆H₁₂O₆, Mᵣ = 180). Without an independent Mᵣ measurement the compound cannot be uniquely identified — this is precisely where mass spectrometry from lesson 1 enters the workflow.
0.560 g of a hydrocarbon produced 1.76 g of CO₂ and 0.720 g of H₂O on complete combustion. Determine the empirical formula and, given that mass spectrometry yields an M⁺ peak at m/z = 56, find the molecular formula.
Check for oxygen: m(C) = 0.0400 × 12.0 = 0.480 g; m(H) = 0.0800 × 1.0 = 0.0800 g; total = 0.560 g = original sample mass. No oxygen is present — confirming the compound is a hydrocarbon.
C : H = 0.0400 : 0.0800 = 1 : 2
Empirical formula = CH₂ (Mᵣ = 14.0)
n = Mᵣ(compound) / Mᵣ(empirical) = 56.0 / 14.0 = 4
Molecular formula = C₄H₈
Candidates: but-1-ene, but-2-ene (cis or trans), 2-methylpropene, methylcyclopropane — all share C₄H₈. Empirical-formula data alone cannot distinguish these structural isomers; ¹H NMR (lesson preview: §3.3.15) or IR spectroscopy would be needed to identify which specific isomer is present.
A compound has the empirical formula CH₂O and a relative molecular mass of 180 (obtained from the molecular-ion peak in its mass spectrum). Determine the molecular formula.
Mᵣ(CH₂O) = 12.0 + 2(1.0) + 16.0 = 30.0
n = 180 / 30.0 = 6
Molecular formula = C₆H₁₂O₆
This is the molecular formula of glucose, fructose, galactose, and a number of other hexose sugars. The empirical-to-molecular bridge has identified the molecular formula uniquely, but structural identification of which hexose requires NMR.
A compound contains 85.7% carbon and 14.3% hydrogen by mass. Its mass spectrum shows the M⁺ peak at m/z = 56. Determine the empirical and molecular formulae.
| Element | % | ÷ Aᵣ | Moles | ÷ smallest | Ratio |
|---|---|---|---|---|---|
| C | 85.7 | ÷ 12.0 | 7.14 | ÷ 7.14 | 1 |
| H | 14.3 | ÷ 1.0 | 14.3 | ÷ 7.14 | 2 |
Empirical formula = CH₂ (Mᵣ = 14.0)
n = 56 / 14.0 = 4
Molecular formula = C₄H₈
Same answer as Example 6 — note that two completely different experimental pipelines (percentage composition with Mᵣ, versus combustion analysis with Mᵣ) lead to the same molecular formula. Both methods report the same composition because both probe the same fundamental quantity: the mole ratio of elements in the compound.
0.730 g of a compound containing C, H, N, and O only was burned in excess oxygen, producing 1.32 g of CO₂ and 0.540 g of H₂O. A separate Dumas-method analysis on a different sample showed that the compound contains 19.2% nitrogen by mass. Determine the empirical formula.
Combustion of a C/H/N/O compound returns the CO₂ and H₂O quantities expected from the C and H content, but nitrogen is not quantified — nitrogen-containing compounds yield N₂ on complete combustion (after passing through a catalytic bed to convert any NOₓ formed). The Dumas method measures the volume of N₂ produced and back-calculates % N. (Strictly, Dumas-method analysis lies beyond AQA spec coverage, but understanding the principle clarifies why combustion alone cannot determine N content.)
Step 1 — From combustion data:
Step 2 — Nitrogen content from Dumas analysis:
Step 3 — Oxygen by mass balance:
Step 4 — Ratio. Divide by the smallest (n(N) = 0.0100):
Within rounding (1.06 ≈ 1), empirical formula = C₃H₆NO.
Spec note: Quantitative nitrogen-by-Dumas determination is not on the AQA A-Level Chemistry specification. The example is included to illustrate the mass-balance principle when a fourth element is present — it shows that combustion analysis on its own is insufficient for nitrogen-containing compounds, and that an additional analytical technique is required.
Many salts crystallise with water molecules occupying defined positions in the crystal lattice — these are hydrates or hydrated salts. The water molecules are not bonded covalently to the cation in the same way a ligand is, but they occupy stoichiometric, well-defined sites in the unit cell. The formula is written with a dot, for example CuSO₄·5H₂O for copper(II) sulfate pentahydrate or Na₂CO₃·10H₂O for sodium carbonate decahydrate.
The water of crystallisation is determined experimentally by heating the hydrate to constant mass — that is, continuing to heat and reweigh until further heating produces no further loss of mass. This ensures all the water has been driven off and only the anhydrous salt remains. The phrase "to constant mass" is examinable: heating once for a fixed time is not sufficient because partial dehydration may give an intermediate hydrate (e.g. CuSO₄·5H₂O → CuSO₄·H₂O → CuSO₄).
3.21 g of hydrated magnesium sulfate (MgSO₄·xH₂O) was heated until constant mass. The anhydrous salt weighed 1.57 g. Determine the value of x.
Step 1 — Mass of water lost.
Step 2 — Moles of anhydrous salt and water.
Step 3 — Ratio of water to salt.
x = 7, so the formula is MgSO₄·7H₂O (Epsom salts — magnesium sulfate heptahydrate).
5.00 g of hydrated copper(II) sulfate was heated to constant mass. The anhydrous residue weighed 3.20 g. Determine the water of crystallisation and verify against the standard formula.
Step 1 — Mass of water lost.
Step 2 — Moles.
Step 3 — Ratio.
x = 5, confirming the formula CuSO₄·5H₂O.
Note also the multiple-of-the-empirical-formula interpretation: the empirical formula of the hydrate is CuSO₄·5H₂O itself (or, written out atom-by-atom, CuSO₉H₁₀), and there is no integer multiple n > 1 that produces a "molecular formula" — hydrates are ionic and have no molecular formula. The hydrate formula CuSO₄·5H₂O simply specifies the composition of the unit cell.
Common Misconception: When calculating the Mᵣ of a hydrated salt, the water must be included. Mᵣ(CuSO₄·5H₂O) = 63.5 + 32.1 + 4 × 16.0 + 5 × 18.0 = 249.6, not 159.6. The anhydrous salt has Mᵣ = 159.6; the hydrate adds 5 × 18.0 = 90.0 to that. Forgetting the water term is one of the most common errors in hydrated-salt mole calculations and will derail an entire titration calculation if not noticed.
Why heating to constant mass matters: Partial dehydration can produce intermediate hydrates that look thermally stable over short time periods. CuSO₄·5H₂O, for example, loses water in stages: CuSO₄·5H₂O → CuSO₄·3H₂O → CuSO₄·H₂O → CuSO₄ (anhydrous). Each stage requires progressively higher temperatures and longer heating. Reweighing after every heating cycle and continuing until consecutive weighings agree to within experimental tolerance is the only way to guarantee full dehydration.
Experimental determination of an empirical formula from combustion or hydrate-heating data is prone to a small number of characteristic errors that examiners frequently target on Paper 3.
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