CLASS 11–12 CHEMISTRY · SOME BASIC CONCEPTS OF CHEMISTRY

Empirical and molecular formula from % composition

Assume 100 g, convert each % to moles (÷ atomic mass), divide by the smallest, and round to whole numbers. Molecular formula = (empirical formula) × n, where n = molar mass ÷ empirical formula mass.

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Worked examples

Example 1 · level 1

A compound contains C 80%, H 20% by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1)

CH₃
C₂H₅
CH₂
CH₄

Show solution

Take 100 g of compound: 80 g C, 20 g H.
Moles: C = 80 ÷ 12 = 6.667; H = 20 ÷ 1 = 20.
Divide by the smallest (6.667): C ≈ 1, H ≈ 3.
Empirical formula = CH₃.

Answer: CH₃

Example 2 · level 2

A compound contains C 20%, H 6.67%, N 46.67%, O 26.67% by mass. What is its empirical formula? (Atomic masses: C = 12, H = 1, N = 14, O = 16)

CH₄N₂O
CH₂NO
CH₄NO
C₂H₄N₂O

Show solution

Take 100 g of compound: 20 g C, 6.67 g H, 46.67 g N, 26.67 g O.
Moles: C = 20 ÷ 12 = 1.667; H = 6.67 ÷ 1 = 6.67; N = 46.67 ÷ 14 = 3.334; O = 26.67 ÷ 16 = 1.667.
Divide by the smallest (1.667): C ≈ 1, H ≈ 4, N ≈ 2, O ≈ 1.
Empirical formula = CH₄N₂O.

Answer: CH₄N₂O

Example 3 · level 3

A compound contains 40% C and 6.67% H and 53.33% O by mass. Its molar mass is 180 g/mol. What is its molecular formula? (Atomic masses: C = 12, H = 1, O = 16)

C₆H₁₂O₆
CH₂O
C₂H₄O₂
C₃H₆O₃

Show solution

Take 100 g of compound: 40 g C, 6.67 g H, 53.33 g O.
Moles: C = 40 ÷ 12 = 3.333; H = 6.67 ÷ 1 = 6.67; O = 53.33 ÷ 16 = 3.333.
Divide by the smallest (3.333): C ≈ 1, H ≈ 2, O ≈ 1.
Empirical formula = CH₂O.
Empirical formula mass = 30 g/mol.
n = 180 ÷ 30 = 6.
Molecular formula = (CH₂O)₆ = C₆H₁₂O₆.

Answer: C₆H₁₂O₆