Empirical and Molecular Formulae

Learning Objectives (OCR Spec 2.1.3)

• Define empirical formula and molecular formula
• Calculate empirical formulae from masses or percentage composition
• Determine molecular formulae when Mr is also known
• Interpret combustion analysis data to determine formulae
• Find the degree of hydration in a hydrated salt
• Deduce empirical formulae from reaction mass data

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 of a compound.

The molecular formula is always a whole-number multiple of the empirical formula. For example:

Compound	Molecular Formula	Empirical Formula	Multiplier
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 — the empirical formula alone does not uniquely identify a compound. Similarly, glucose, methanoic acid and ribose share CH₂O despite being very different molecules. This illustrates why you need both empirical formula and Mr to identify a specific compound.

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 come directly from experimental measurements of composition by mass. Historically, elemental analysis (combustion analysis, gravimetric methods) gave the empirical formula of an unknown compound, and molecular formulae had to be deduced separately from measured molar mass (via vapour density, freezing point depression, mass spectrometry, etc.). Even today, any new compound synthesised in the lab is characterised first by elemental analysis.

Finding Empirical Formula from Mass Data

The Four-Step Method

1. Divide the mass of each element by its Ar to get moles
2. Divide all mole values by the smallest mole value
3. If the ratio is not whole numbers, multiply by a factor to get whole numbers (×2, ×3, ×4)
4. Write the formula in the ratio you obtained

A neat layout: write the elements across the top and work down in rows (mass, ÷Ar, ÷smallest, ratio).

Worked Example 1: Iron Oxide

Analysis shows a compound contains 1.12 g of iron and 0.48 g of oxygen. Find the empirical formula.

	Fe	O
Mass / g	1.12	0.48
Ar	55.8	16.0
n = m/Ar	0.02007	0.03000
÷ smallest (0.02007)	1.00	1.495
× 2	2	3

Empirical formula = Fe₂O₃

Worked Example 2: Percentage Composition

A compound contains 40.0% C, 6.7% H and 53.3% O by mass. Find the empirical formula.

Trick: assume you have exactly 100 g of the compound. Then percentages become grams directly.

	C	H	O
Mass / g	40.0	6.7	53.3
Ar	12.0	1.0	16.0
n / mol	3.333	6.700	3.331
÷ smallest (3.331)	1.001	2.011	1.000
Ratio	1	2	1

Empirical formula = CH₂O

Worked Example 3: When Ratios Are Not Whole

A hydrocarbon contains 85.7% C and 14.3% H.

	C	H
Mass / g	85.7	14.3
Ar	12.0	1.0
n / mol	7.142	14.300
÷ smallest	1.000	2.002

Empirical formula = CH₂

Worked Example 4: Needing a Factor of 3

A compound contains 32.4% Na, 22.6% S and 45.0% O.

	Na	S	O
Mass / g	32.4	22.6	45.0
Ar	23.0	32.1	16.0
n / mol	1.4087	0.7040	2.8125
÷ smallest (0.7040)	2.001	1.000	3.994
Ratio	2	1	4

Empirical formula = Na₂SO₄

Worked Example 5: A Tricky Non-Integer Ratio

A compound contains 72.2% Mg and 27.8% N. Find its empirical formula.

	Mg	N
Mass / g	72.2	27.8
Ar	24.3	14.0
n / mol	2.971	1.986
÷ smallest (1.986)	1.496	1.000

1.496 ≈ 1.5, so multiply by 2:

| Ratio | 3 | 2 |

Empirical formula = Mg₃N₂ (magnesium nitride)

Finding Molecular Formula from Empirical Formula and Mr

Once you have the empirical formula and the relative molecular mass Mr of the compound:

1. Calculate the empirical formula mass
2. Divide Mr by the empirical formula mass to get an integer n
3. Multiply every subscript in the empirical formula by n

Worked Example 6: Glucose

Empirical formula is CH₂O (from Example 2). Mr = 180.0. Find the molecular formula.

• Empirical mass = 12.0 + 2(1.0) + 16.0 = 30.0
• n = 180.0 / 30.0 = 6
• Molecular formula = (CH₂O)₆ = C₆H₁₂O₆

Worked Example 7: A Hydrocarbon

A hydrocarbon has empirical formula CH₂ and Mr = 56.0.

• Empirical mass = 12.0 + 2(1.0) = 14.0
• n = 56.0 / 14.0 = 4
• Molecular formula = C₄H₈

Worked Example 8: Benzene Derivative

An aromatic compound has 92.3% C and 7.7% H with Mr = 78.0.

	C	H
Mass / g	92.3	7.7
Ar	12.0	1.0
n / mol	7.692	7.700
÷ smallest	1.000	1.001

Empirical formula = CH (empirical mass 13.0) n = 78.0 / 13.0 = 6 Molecular formula = C₆H₆ (benzene)

Combustion Analysis

When a hydrocarbon or CHO-compound is burned in excess oxygen, all of the carbon becomes CO₂ and all of the hydrogen becomes H₂O. The masses of CO₂ and H₂O produced allow calculation of the moles of C and H, from which the formula follows.

Key Relationships

• n(C in compound) = n(CO₂) — each mole of CO₂ came from one C atom
• n(H in compound) = 2 × n(H₂O) — each mole of H₂O came from two H atoms
• Mass of O in compound = total mass − mass of C − mass of H

Worked Example 9: Combustion Analysis