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1.6.1Physical Chemistry

Empirical and Molecular Formulae — Edexcel AS Chemistry Revision Notes & Topical Questions

5 min read 6 parts Lesson 25 of 42

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Part 1 of 6Two Types of Formula

1.6.1 Empirical and Molecular Formulae
Molecular formula
  • Shows the actual number of atoms of each element in one molecule, for example C₂H₄O₂ for ethanoic acid.
Empirical formula
  • Gives the simplest whole-number ratio of atoms of each element, so ethanoic acid becomes CH₂O.
Ionic compounds
  • Formulae of ionic compounds are always empirical because they describe a ratio in a lattice, not a molecule.
Relationship
  • The molecular formula is a whole-number multiple of the empirical formula, found using the relative molecular mass.

Part 2 of 6Finding the Empirical Formula

1
Write down the mass or percentage of each element in the compound.
2
Divide each mass by the relative atomic mass of that element to find moles.
3
Divide every answer by the smallest number of moles to get a ratio.
4
If the ratio is not whole numbers, multiply by a small integer such as 2 or 3 until it is.
5
Write the formula using the whole-number ratio as subscripts.

Part 3 of 6From Empirical to Molecular Formula

From Empirical to Molecular Formula
Step one
  • Calculate the relative formula mass of the empirical formula by adding up its relative atomic masses.
Step two
  • Divide the given relative molecular mass by the empirical formula mass to get a whole number n.
Step three
  • Multiply every subscript in the empirical formula by n to obtain the molecular formula.
Check
  • The relative mass of the final molecular formula must match the value given in the question.
CompoundEmpirical formulaMolecular formula
EtheneCH₂C₂H₄
GlucoseCH₂OC₆H₁₂O₆
BenzeneCHC₆H₆
Hydrogen peroxideHOH₂O₂

Part 4 of 6Handling Awkward Ratios

Ratio of 1 : 1.5
  • Multiply both numbers by 2 to give 2 : 3.
Ratio of 1 : 1.33
  • Multiply by 3, because 1.33 is about one and a third, giving 3 : 4.
Ratio of 1 : 2.5
  • Multiply by 2 to obtain 2 : 5.
Rounding
  • Only round values that are very close to whole numbers, such as 1.98 or 3.02;
    • values like 1.5 need a multiplier.

Part 5 of 6Worked Example

Question

  • A compound contains 40.0% carbon, 6.7% hydrogen and 53.3% oxygen by mass.
  • Its relative molecular mass is 180.
  • Find the molecular formula.

  1. Moles of C = 40.0 ÷ 12.0 = 3.33; moles of H = 6.7 ÷ 1.0 = 6.7; moles of O = 53.3 ÷ 16.0 = 3.33.
  2. Divide by 3.33 to give a ratio C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O.
  3. Empirical formula mass = 12.0 + 2(1.0) + 16.0 = 30.0.
  4. n = 180 ÷ 30.0 = 6, so multiply every subscript by 6.

Answer Molecular formula = C₆H₁₂O₆

Part 6 of 6Exam Checklist & Tips

1.6.1 – Key Points for the Exam

  • Empirical formula Simplest whole-number ratio of atoms of each element in a compound.
  • Molecular formula Actual number of atoms of each element in one molecule.
  • Relative molecular mass Sum of the relative atomic masses in a molecule.
  • Percentage composition Mass of an element in a compound as a percentage of the total mass.

Common Mistakes

  • Dividing percentages by atomic number instead of relative atomic mass.
  • Rounding 1.5 to 2 instead of multiplying the ratio.
  • Giving an empirical formula for a molecule when the question asks for the molecular formula.
  • Forgetting to divide by the smallest value before reading off the ratio.

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