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Empirical and Molecular Formula

Learn and apply Empirical and Molecular Formula in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Empirical and Molecular Formulae: Orientation

Empirical/molecular formula questions are always the same move: convert the data into moles of atoms, simplify to a ratio, then (if needed) scale to match Mᵣ. This lesson covers both common data types: composition by mass and combustion data.

Build this on Mole and Avogadro Constant and keep the Stoichiometry hub open so unit conversions and mole logic stay coherent.

Definitions (Must Know)

A. Empirical formula

The empirical formula is the simplest whole-number ratio of atoms in a compound.

B. Molecular formula

The molecular formula shows the actual number of each type of atom in one molecule.

C. Combustion data (combustion analysis)

Combustion data is the mass (or moles) of CO₂ and H₂O formed when a compound is completely combusted, used to deduce the compound’s formula.

Detailed Explanations

A. Empirical formula from composition (workflow)

Because the subscripts in a formula represent the ratio of atoms, you must convert mass data into moles (proportional to number of atoms).

  1. If given percentages, assume 100 g (so % becomes grams).
  2. Convert each element’s mass to moles: n = m/Aᵣ.
  3. Divide every mole value by the smallest to get the simplest ratio.
  4. Multiply the ratio to clear fractions, then write the empirical formula.

Mini example:

  • C = 24.0 g, H = 4.0 g → moles C = 2.00, H = 4.00 → ratio 1 : 2 → empirical formula CH₂.

B. Molecular formula from empirical formula (workflow)

Because the molecular formula is an integer multiple of the empirical formula:

  1. Calculate Mᵣ(empirical).
  2. Find the multiplier k = Mᵣ(molecule)/Mᵣ(empirical).
  3. Multiply every subscript in the empirical formula by k.

Mini example:

  • empirical CH₂O has Mᵣ = 30; if molecular mass is 180 then k = 180/30 = 6 → molecular formula C₆H₁₂O₆.

C. Empirical formula from combustion data (workflow)

For a compound containing C and H (and possibly O), combustion gives you CO₂ and H₂O, which lets you deduce moles of C and H directly.

  1. Find n(CO₂) and n(H₂O) from their masses.
  2. Convert to moles of atoms:
    • moles of C atoms = n(CO₂)
    • moles of H atoms = 2n(H₂O)
  3. If oxygen is present in the compound, find it by mass difference:
    • mass of C in the sample = n(C) × 12.0
    • mass of H in the sample = n(H) × 1.00
    • mass of O in the sample = mass of sample − mass of C − mass of H
    • moles of O atoms = m(O)/16.0
  4. Divide all mole values by the smallest to get the empirical ratio.

Worked Examples

Modelled example 1

Find an Empirical Formula from Percentages

Core

Problem

A compound contains 40.0% carbon, 6.67% hydrogen and 53.3% oxygen by mass. Find its empirical formula. Use Aᵣ(C) = 12.0, Aᵣ(H) = 1.00 and Aᵣ(O) = 16.0.

Study the worked solution

  1. Choose a percentage basis

    Method

    Assume a 100 g sample.

    Reason

    Each percentage then becomes the mass in grams without changing the composition ratio.

    Working

    m(C) = 40.0 g, m(H) = 6.67 g, m(O) = 53.3 g.

  2. Convert every mass to amount

    Reason

    Formula subscripts represent atom amounts, not mass proportions.

    Working

    n(C) = 40.0/12.0 = 3.33, n(H) = 6.67/1.00 = 6.67, n(O) = 53.3/16.0 = 3.33.

  3. Reduce to whole numbers

    Method

    Divide all amounts by the smallest value, 3.33.

    Reason

    An empirical formula is the simplest whole-number atom ratio.

    Working

    C:H:O = 1.00:2.00:1.00, so the empirical formula is CH₂O.

Guided practice 2

Scale an Empirical Formula

About 5 min

Problem

A compound has empirical formula CH₂O and molecular mass 180. Find its molecular formula.

Try this before viewing the solution

Hints

Hint 1: find the empirical-formula mass
Add the relative masses represented by one CH₂O unit.
Hint 2: find the integer multiplier
Divide the molecular mass by the empirical-formula mass.
View solution step by step

  1. Find empirical-formula mass

    Method

    Add the relative masses in CH₂O.

    Reason

    The molecular formula must contain a whole-number multiple of the empirical unit.

    Working

    Mᵣ(CH₂O) = 12.0 + 2(1.00) + 16.0 = 30.0

  2. Calculate the multiplier

    Reason

    The ratio of molecular mass to empirical-formula mass gives the number of empirical units.

    Working

    k = 180/30.0 = 6

  3. Scale all subscripts

    Method

    Multiply every empirical subscript by six.

    Reason

    Scaling only one element would change the empirical atom ratio.

    Working

    Molecular formula: C₆H₁₂O₆.

Common misconception 3

Correct a Partly Scaled Formula

Find and correct the mistake

Learner attempt

A compound has empirical formula NO₂ and molecular mass 92. A learner calculates Mᵣ(NO₂) = 46 and k = 2, then writes NO₄. Identify the first error and give the molecular formula.

Find the first error

First error

View solution step by step

  1. Confirm the multiplier

    Method

    Retain k = 92/46 = 2.

    Reason

    The learner’s mass calculation is correct; the first error occurs when applying the multiplier.

    Working

    k = 2

  2. Scale the whole empirical unit

    Method

    Multiply the implicit N subscript and the O subscript by two.

    Reason

    A molecular formula preserves the empirical atom ratio.

    Working

    (NO₂)₂ = N₂O₄

Examiner practice 4

Use Combustion Data to Find a Formula

5 marks

Problem

A 1.50 g compound containing only carbon, hydrogen and oxygen produces 2.20 g CO₂ and 0.900 g H₂O on complete combustion. Find its empirical formula. Use M(CO₂) = 44.0 g mol⁻¹ and M(H₂O) = 18.0 g mol⁻¹. [5 marks]

Try this before viewing the solution

View solution step by step

  1. Find carbon amount

    1 mark

    Method

    Convert carbon dioxide mass to amount and use its one-carbon formula.

    Reason

    Each mole of CO₂ contains one mole of carbon atoms.

    Working

    n(C) = n(CO₂) = 2.20/44.0 = 0.0500 mol

  2. Find hydrogen amount

    1 mark

    Method

    Convert water mass to amount, then double it.

    Reason

    Each mole of water contains two moles of hydrogen atoms.

    Working

    n(H) = 2(0.900/18.0) = 0.100 mol

  3. Find oxygen by mass difference

    1 mark

    Reason

    The original compound contains only C, H and O, so its remaining mass belongs to oxygen.

    Working

    m(C) = 0.600 g, m(H) = 0.100 g; m(O) = 1.50-0.600-0.100 = 0.800 g.

  4. Convert oxygen and reduce the ratio

    1 mark

    Reason

    All three elements must be compared as amounts.

    Working

    n(O) = 0.800/16.0 = 0.0500 mol; C:H:O = 1:2:1.

  5. State the empirical formula

    1 mark

    Method

    Translate the smallest whole-number ratio into subscripts.

    Reason

    The 1:2:1 ratio is already in its simplest form.

    Working

    Empirical formula: CH₂O.

Challenge 5

Infer an Oxide Formula after Reduction

Minimal support

Problem

A 1.60 g sample of a copper oxide is completely reduced to 1.28 g copper. Determine the empirical formula of the oxide. Use Aᵣ(Cu) = 63.5 and Aᵣ(O) = 16.0.

Try this before viewing the solution

Hints

Hint 1: recover oxygen mass
The mass lost during complete reduction is the oxygen formerly combined with copper.
Hint 2: compare amounts
Convert both copper and oxygen masses to moles before finding their ratio.
View solution step by step

  1. Find each element mass

    Method

    Subtract the copper mass from the oxide mass to obtain oxygen mass.

    Reason

    Complete reduction removes oxygen and leaves all the copper from the original sample.

    Working

    m(O) = 1.60-1.28 = 0.320 g

  2. Convert to amounts

    Reason

    Empirical subscripts compare numbers of atoms rather than their masses.

    Working

    n(Cu) = 1.28/63.5 = 0.0202 mol; n(O) = 0.320/16.0 = 0.0200 mol.

  3. Interpret the ratio

    Method

    Treat the small difference as measurement and rounding variation around 1:1.

    Reason

    The amounts agree within the precision of the supplied masses and should form small whole-number subscripts.

    Working

    Cu:O ≈ 1:1, so the empirical formula is CuO.

Mind Stretchers

Mind stretcher 1Extension

1.50 g of a compound containing C, H and O produces 2.20 g of CO₂ and 0.900 g of H₂O on complete combustion. Its molecular mass is 60.0. Find its molecular formula.

Show Hint

Convert every mass or percentage to moles, divide by the smallest amount, then use the molar mass only after finding the empirical formula.

Show Answer

Mark scheme:

  • From combustion (as in Example 3), empirical formula is CH₂O with Mᵣ = 30.0.
  • Multiplier k = 60.0/30.0 = 2.
  • Molecular formula: C₂H₄O₂.

Mind stretcher 2: Finding water of crystallisationExtension

Question. Heating 2.50 g of CuSO₄.xH₂O leaves 1.60 g of anhydrous CuSO₄. Given M(CuSO₄) = 160 g mol⁻¹, find x.

Show Hint

The lost mass is water. Compare moles of water with moles of anhydrous salt.

Show Answer

Water lost = 0.90 g, so n(H₂O) = 0.90/18.0 = 0.0500 mol. Also n(CuSO₄) = 1.60/160 = 0.0100 mol. The ratio is 1:5, so x = 5.