Skip to notes
wise otter.Chemistry
Contents
The Mole Concept and Stoichiometry

Topic 2 of 5

Find an empirical or molecular formula

Convert composition or combustion products into atom ratios.

A-Level 9476 (2026-2027)

A formula records an atom ratio, not a mass ratio

Divide each mass by its atomic molar mass before comparing the numbers.

An empirical formula gives the simplest whole-number ratio of atoms of each element. A molecular formula gives the actual numbers in a molecule. Benzene has empirical formula CH and molecular formula C6H6. An ionic formula such as MgCl2 describes a formula-unit ratio, not a molecule.

From mass composition to molecular formula
  1. Choose a convenient sample

    For percentages, assume 100 g. Convert each elemental mass to moles using its atomic molar mass.

  2. Find the simplest ratio

    Divide all amounts by the smallest. If a value is near 1.5 or 1.33, multiply the whole ratio by 2 or 3; do not round these straight to integers.

  3. Use the independent molar mass

    Molecular multiplier = molecular molar mass / empirical-formula molar mass. Multiply every subscript by the same integer.

Worked example

Distinguish the two formulae

A compound contains 40.0% C, 6.67% H and 53.3% O by mass. Its molar mass is 180 g mol-1. Use C = 12.0, H = 1.00 and O = 16.0.

  1. In 100 g: n(C) = 40.0/12.0 = 3.333; n(H) = 6.67/1.00 = 6.67; n(O) = 53.3/16.0 = 3.331 mol.
  2. Divide by the smallest to obtain approximately 1 : 2 : 1, so the empirical formula is CH2O.
  3. Empirical-formula molar mass = 30.0 g mol-1; 180/30.0 = 6.
Answer

Empirical formula CH2O; molecular formula C6H12O6. Composition alone cannot choose the multiplier.

Combustion transfers carbon into carbon dioxide and hydrogen into water

Count atoms from products, then use mass difference only when the remaining element is known.

For complete combustion of a compound containing C and H, every mole of CO2 contains one mole of original carbon atoms, while every mole of H2O contains two moles of original hydrogen atoms. Thus n(C) = n(CO2) and n(H) = 2n(H2O). Oxygen in the products comes from both the compound and the oxygen supply, so product oxygen cannot directly give the original oxygen content.

Worked example

Find oxygen by a justified mass difference

Complete combustion of 0.460 g of a compound containing only C, H and O produces 0.880 g CO2 and 0.540 g H2O. Its molar mass is 46.0 g mol-1.

  1. n(C) = 0.880/44.0 = 0.0200 mol; carbon mass = 0.240 g.
  2. n(H) = 2(0.540/18.0) = 0.0600 mol; hydrogen mass = 0.0600 g.
  3. Original oxygen mass = 0.460 - 0.240 - 0.0600 = 0.160 g; n(O) = 0.0100 mol.
  4. The ratio C : H : O is 2 : 6 : 1. Its formula mass is 46.0, so the molecular multiplier is 1.
Answer

C2H6O. Its complete-combustion equation is C2H6O + 3O2 → 2CO2 + 3H2O. The formula does not distinguish ethanol from methoxymethane.

This deduction assumes complete combustion and quantitative product collection. A water absorber can also collect moisture from the incoming gas unless the gas is dried. Incomplete combustion, escaping CO2 or a compound containing an unaccounted element makes the simple mass balance unreliable. State the assumptions when evaluating experimental data.