Empirical and Molecular Formulae

Spec Mapping — OCR H432 Module 2.1.3 — Amount of substance / empirical and molecular formulae, content statements covering the definition and calculation of empirical formulae from masses or percentage composition; the determination of molecular formulae when $M_r$Mr​ is also known; combustion-analysis interpretation; and the determination of the degree of hydration in a hydrated salt (refer to the official OCR H432 specification document for exact wording). This lesson is the experimental flip side of the mole concept — it works backwards from measured masses to formulae.

The mole equation $n = m/M$n=m/M is one direction of quantitative chemistry: given a known compound, calculate the amount. Empirical-formula determination is the inverse direction: given experimental masses or percentage composition, deduce the formula. This is the analytical-chemistry workflow that has characterised every unknown-compound investigation since Lavoisier's eighteenth-century elemental analyses of organic compounds. At A-Level you must be fluent in three problem types: (i) pure mass or percentage-composition data → empirical formula; (ii) empirical formula plus measured $M_r$Mr​ → molecular formula; and (iii) combustion analysis, where complete burning of a CHO-compound yields measured masses of CO₂ and H₂O from which the C and H content (and, by mass difference, the O content) are deduced. A fourth variant — hydrated-salt thermal decomposition — uses the same mole-ratio logic to determine the water of crystallisation $x$x in a formula like CuSO₄·xH₂O. The skill is procedural but unforgiving: a rounding slip in the divide-by-smallest step can change a 2:3 ratio into a 1:1 ratio and lose every mark.

Key Definitions:

• Empirical formula: the simplest whole-number ratio of atoms of each element in a compound.
• Molecular formula: the actual number of atoms of each element in a molecule. Always an integer multiple of the empirical formula.

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Definitions

Empirical formula

The simplest whole-number ratio of atoms of each element in a compound.

Molecular formula

The actual number of atoms of each element in a molecule.

The molecular formula is always an integer multiple of the empirical formula:

Compound	Molecular formula	Empirical formula	Multiplier $n$n
Ethane	C₂H₆	CH₃	2
Ethene	C₂H₄	CH₂	2
Butene	C₄H₈	CH₂	4
Cyclohexane	C₆H₁₂	CH₂	6
Benzene	C₆H₆	CH	6
Ethyne	C₂H₂	CH	2
Glucose	C₆H₁₂O₆	CH₂O	6
Methanoic acid	HCOOH	CH₂O	2
Ribose	C₅H₁₀O₅	CH₂O	5

Note that ethene and butene share the same empirical formula CH₂; glucose, methanoic acid and ribose all share CH₂O. The empirical formula alone does not uniquely identify a compound — you need $M_r$Mr​ as well to fix the molecular formula.

For ionic compounds (NaCl, MgO, Al₂O₃), the formula you write is already the empirical formula: there are no discrete molecules, so molecular formula does not apply. The "formula unit" simply expresses the simplest ratio of ions in the lattice.

Why empirical formulae matter

Empirical formulae are the experimental output of elemental analysis — historically combustion analysis (organic compounds), gravimetric methods (inorganic), and now also high-resolution mass spectrometry and combustion-MS hybrid instruments. Any new compound synthesised in a research laboratory is characterised first by elemental analysis to confirm its empirical formula; the molecular formula is then established separately via $M_r$Mr​ from vapour-density measurements, freezing-point depression, or — most commonly — mass spectrometry.

A note on percentage composition

A subtle convention: percentage composition is sometimes given by mass and sometimes by mole fraction in undergraduate work. At A-Level OCR exclusively uses percentage by mass — but it is worth noting the distinction. For a compound like CO₂, the percentages by mass are 27.3 % C and 72.7 % O, while the percentages by mole are 33.3 % C and 66.7 % O (because there are 1 C and 2 O per formula unit). The two diverge wherever the atomic masses of the constituents differ.

When the question says "percentage composition" without qualification, default to "by mass" in the A-Level context — the conversion to moles then proceeds via $n = m/A_r$n=m/Ar​ in the standard four-step procedure.

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Finding the Empirical Formula from Mass Data

The four-step method

1. Divide the mass of each element by its $A_r$Ar​ to get moles.
2. Divide every mole value by the smallest mole value.
3. If the resulting ratio is not whole numbers, multiply through by 2, 3 or 4 to clear half-integers, thirds or quarters.
4. Read off the empirical formula from the integer ratio.

A neat tabular layout (elements across the top, working steps in rows: mass → $\div A_r$÷Ar​ → $\div$÷ smallest → ratio) earns method marks even if arithmetic slips.

Worked Example 1 — Iron oxide

A compound contains $1.12\text{ g}$1.12 g Fe and $0.48\text{ g}$0.48 g O.

	Fe	O
Mass / g	1.12	0.48
$A_r$Ar​	55.8	16.0
$n = m/A_r$n=m/Ar​ / mol	0.02007	0.03000
$\div\ \text{smallest}$÷ smallest	1.000	1.495
$\times 2$×2	2	3

Empirical formula = Fe₂O₃

Worked Example 2 — Percentage composition

A compound: 40.0 % C, 6.7 % H, 53.3 % O. Treat percentages as grams in 100 g of sample.

	C	H	O
Mass / g	40.0	6.7	53.3
$A_r$Ar​	12.0	1.0	16.0
$n$n / mol	3.333	6.700	3.331
$\div\ \text{smallest}$÷ smallest	1.001	2.011	1.000
Ratio	1	2	1

Empirical formula = CH₂O.

Worked Example 3 — Hydrocarbon

85.7 % C, 14.3 % H.

	C	H
$n$n / mol	$85.7/12.0 = 7.142$85.7/12.0=7.142	$14.3/1.0 = 14.300$14.3/1.0=14.300
$\div\ \text{smallest}$÷ smallest	1.000	2.002

Empirical formula = CH₂.

Worked Example 4 — Needing a factor of 3 (Na₂SO₄)

32.4 % Na, 22.6 % S, 45.0 % O.

	Na	S	O
$n$n / mol	1.4087	0.7040	2.8125
$\div\ \text{smallest}$÷ smallest	2.001	1.000	3.994
Ratio	2	1	4

Empirical formula = Na₂SO₄.

Worked Example 5 — A 3:2 ratio (Mg₃N₂)

72.2 % Mg, 27.8 % N.

	Mg	N
$n$n / mol	$72.2/24.3 = 2.971$72.2/24.3=2.971	$27.8/14.0 = 1.986$27.8/14.0=1.986
$\div\ \text{smallest}$÷ smallest	1.496	1.000

Recognise $1.496 \approx 1.5$1.496≈1.5, multiply by 2: ratio becomes 3:2. Empirical formula = Mg₃N₂.

Worked Example 5b — Recognising thirds (Fe₃O₄ magnetite)

A magnetite sample contains 72.4 % Fe and 27.6 % O.

	Fe	O
$n$n / mol	$72.4/55.8 = 1.297$72.4/55.8=1.297	$27.6/16.0 = 1.725$27.6/16.0=1.725
$\div\ \text{smallest}$÷ smallest	1.000	1.330

The ratio 1.330 is close to 4/3 (1.333), so multiply by 3 to give a 3:4 ratio. Empirical formula = Fe₃O₄.

Magnetite is a mixed-valence oxide: $\text{Fe}^{2+}\text{Fe}_2^{3+}\text{O}_4^{2-}$Fe2+Fe23+​O42−​ — one Fe²⁺ and two Fe³⁺ per formula unit, balancing four O²⁻. The empirical-formula calculation does not reveal the oxidation-state mixing — that requires further characterisation (e.g. Mössbauer spectroscopy in undergraduate inorganic).

Worked Example 5c — Quarters (TiCl₄ titanium(IV) chloride)

A compound contains 25.2 % Ti and 74.8 % Cl by mass.

	Ti	Cl
$n$n / mol	$25.2/47.9 = 0.5261$25.2/47.9=0.5261	$74.8/35.5 = 2.108$74.8/35.5=2.108
$\div\ \text{smallest}$÷ smallest	1.000	4.007

Empirical formula = TiCl₄ — volatile liquid used in industrial Ti metal extraction.

Worked Example 5d — Element X by elimination

A sample contains 28.0 % Mg, 14.3 % C and 57.6 % O. Verify the compound is magnesium carbonate.

	Mg	C	O
$n$n / mol	$28.0/24.3 = 1.152$28.0/24.3=1.152	$14.3/12.0 = 1.192$14.3/12.0=1.192	$57.6/16.0 = 3.600$57.6/16.0=3.600
$\div\ \text{smallest}$÷ smallest	1.000	1.035	3.125

The ratio 1 : 1 : 3 is essentially what the data shows (1.035 rounds to 1; 3.125 to 3 within experimental tolerance). Empirical formula = MgCO₃, confirming magnesium carbonate. The slight deviations (1.035 and 3.125 rather than exact 1 and 3) reflect rounding in the percentage data — examiners accept reasonable rounding tolerance at A-Level.

Decision tree — ratio handling

flowchart TD
    A[Moles of each element] --> B[Divide all by smallest mole value]
    B --> C{Is ratio whole-number to within ±0.05?}
    C -->|Yes| D[Empirical formula complete]
    C -->|No| E{Closest fraction?}
    E -->|0.5| F[Multiply by 2]
    E -->|0.33 or 0.67| G[Multiply by 3]
    E -->|0.25 or 0.75| H[Multiply by 4]
    F --> D
    G --> D
    H --> D

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Finding the Molecular Formula from Empirical Formula and $M_r$Mr​

1. Calculate the empirical-formula mass.
2. Divide $M_r$Mr​ by the empirical-formula mass to give an integer $n$n.
3. Multiply every subscript in the empirical formula by $n$n.

Worked Example 6 — Glucose

Empirical formula CH₂O; $M_r = 180.0$Mr​=180.0.

• Empirical mass $= 12.0 + 2(1.0) + 16.0 = 30.0$=12.0+2(1.0)+16.0=30.0
• $n = 180.0 / 30.0 = 6$n=180.0/30.0=6
• Molecular formula = $(\text{CH}_2\text{O})_6 =$(CH2​O)6​= C₆H₁₂O₆

Worked Example 7 — Hydrocarbon with $M_r = 56$Mr​=56

Empirical formula CH₂; $M_r = 56.0$Mr​=56.0.

• Empirical mass $= 14.0$=14.0; $n = 4$n=4
• Molecular formula = C₄H₈

Worked Example 8 — Benzene from elemental + $M_r$Mr​ data

92.3 % C, 7.7 % H, $M_r = 78.0$Mr​=78.0.

	C	H
$n$n / mol	$92.3/12.0 = 7.692$92.3/12.0=7.692	$7.7/1.0 = 7.700$7.7/1.0=7.700
$\div\ \text{smallest}$÷ smallest	1.000	1.001

Empirical formula = CH (empirical mass 13.0); $n = 78.0/13.0 = 6$n=78.0/13.0=6; molecular formula = C₆H₆ (benzene).

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Combustion Analysis

When a hydrocarbon or CHO-compound is burned in excess oxygen, all carbon becomes CO₂ and all hydrogen becomes H₂O. The masses of CO₂ and H₂O produced give the moles of C and H in the original compound; oxygen, if present, is calculated by mass difference.

Key relationships

• $n(\text{C in compound}) = n(\text{CO}_2)$n(C in compound)=n(CO2​) — every CO₂ contains exactly one C atom.
• $n(\text{H in compound}) = 2 \times n(\text{H}_2\text{O})$n(H in compound)=2×n(H2​O) — every H₂O contains two H atoms.
• $m(\text{O in compound}) = m(\text{compound}) - m(\text{C}) - m(\text{H})$m(O in compound)=m(compound)−m(C)−m(H).

Worked Example 9 — Combustion analysis of ethanoic acid

Burning $2.40\text{ g}$2.40 g of CHO-compound X produces $3.52\text{ g}$3.52 g CO₂ and $1.44\text{ g}$1.44 g H₂O. $M_r(\text{X}) = 60.0$Mr​(X)=60.0.

Step 1 — moles of C: $n(\text{CO}_2) = 3.52/44.0 = 0.0800\text{ mol}$n(CO2​)=3.52/44.0=0.0800 mol, so $n(\text{C}) = 0.0800\text{ mol}$n(C)=0.0800 mol; $m(\text{C}) = 0.0800 \times 12.0 = 0.960\text{ g}$m(C)=0.0800×12.0=0.960 g.

Step 2 — moles of H: $n(\text{H}_2\text{O}) = 1.44/18.0 = 0.0800\text{ mol}$n(H2​O)=1.44/18.0=0.0800 mol; $n(\text{H}) = 0.160\text{ mol}$n(H)=0.160 mol; $m(\text{H}) = 0.160\text{ g}$m(H)=0.160 g.

Step 3 — mass of O by difference: $m(\text{O}) = 2.40 - 0.960 - 0.160 = 1.28\text{ g}$m(O)=2.40−0.960−0.160=1.28 g; $n(\text{O}) = 1.28/16.0 = 0.0800\text{ mol}$n(O)=1.28/16.0=0.0800 mol.

Step 4 — empirical formula:

	C	H	O
$n$n / mol	0.0800	0.160	0.0800
$\div\ \text{smallest}$÷ smallest	1	2	1

Empirical formula = CH₂O (empirical mass 30.0). $n = 60.0/30.0 = 2$n=60.0/30.0=2, so molecular formula = C₂H₄O₂ (ethanoic acid, CH₃COOH).

Worked Example 10 — Clean combustion (ethanol)

Burning $4.60\text{ g}$4.60 g of an alcohol gives $8.80\text{ g}$8.80 g CO₂ and $5.40\text{ g}$5.40 g H₂O. $M_r = 46.0$Mr​=46.0.

• $n(\text{CO}_2) = 0.200 \Rightarrow n(\text{C}) = 0.200\text{ mol}; m(\text{C}) = 2.40\text{ g}$n(CO2​)=0.200⇒n(C)=0.200 mol;m(C)=2.40 g
• $n(\text{H}_2\text{O}) = 0.300 \Rightarrow n(\text{H}) = 0.600\text{ mol}; m(\text{H}) = 0.600\text{ g}$n(H2​O)=0.300⇒n(H)=0.600 mol;m(H)=0.600 g
• $m(\text{O}) = 4.60 - 2.40 - 0.600 = 1.60\text{ g} \Rightarrow n(\text{O}) = 0.100\text{ mol}$m(O)=4.60−2.40−0.600=1.60 g⇒n(O)=0.100 mol

	C	H	O
$n$n / mol	0.200	0.600	0.100
$\div\ \text{smallest}$÷ smallest	2	6	1

Empirical formula = C₂H₆O (empirical mass 46.0); $n = 1$n=1; molecular formula = C₂H₆O (ethanol).

Worked Example 10b — A trickier combustion (with rounding tolerance)

Burning $3.20\text{ g}$3.20 g of an unknown CHO compound gives $4.40\text{ g}$4.40 g CO₂ and $1.80\text{ g}$1.80 g H₂O. $M_r = 32.0$Mr​=32.0.

• $n(\text{CO}_2) = 0.100\text{ mol} \Rightarrow n(\text{C}) = 0.100\text{ mol}; m(\text{C}) = 1.20\text{ g}$n(CO2​)=0.100 mol⇒n(C)=0.100 mol;m(C)=1.20 g
• $n(\text{H}_2\text{O}) = 0.100\text{ mol} \Rightarrow n(\text{H}) = 0.200\text{ mol}; m(\text{H}) = 0.200\text{ g}$n(H2​O)=0.100 mol⇒n(H)=0.200 mol;m(H)=0.200 g
• $m(\text{O}) = 3.20 - 1.20 - 0.200 = 1.80\text{ g}; n(\text{O}) = 1.80/16.0 = 0.1125\text{ mol}$m(O)=3.20−1.20−0.200=1.80 g;n(O)=1.80/16.0=0.1125 mol

	C	H	O
$n$n / mol	0.100	0.200	0.1125
$\div\ \text{smallest}$÷ smallest	0.889	1.778	1.000
$\times 9$×9	8	16	9