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1.2.4Physical Chemistry

Empirical and Molecular Formula

5 min read 6 parts Lesson 10 of 63

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Section 1 of 6Two Types of Formula

1.2.4 Empirical and Molecular Formula
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 one molecule.
Relationship
  • The molecular formula is a whole-number multiple of the empirical formula.
Examples
  • Ethene has molecular formula C₂H₄ and empirical formula CH₂.

Section 2 of 6Finding the Empirical Formula

This works because the ratio of moles of atoms equals the ratio of atoms in the formula.

1
List each element with its mass or percentage by mass in the sample.
2
Divide each mass by the Ar of that element to find the number of moles.
3
Divide every mole value by the smallest one to get a ratio.
4
If the ratio is not whole, multiply all values by the smallest number that makes them whole, such as 2 or 3.
5
Write the formula using these whole numbers as subscripts.

Section 3 of 6From Empirical to Molecular

From Empirical to Molecular
Empirical mass
  • Add up the Ar values of the atoms in the empirical formula.
Divide
  • Divide the given Mr by the empirical formula mass to find the multiple, n.
Multiply
  • Multiply every subscript in the empirical formula by n.
Check
  • The final molecular formula must have an Mr equal to the value given.

Section 4 of 6Combustion Data

Carbon
  • All carbon in a compound ends up in carbon dioxide, so moles of C = moles of CO₂.
Hydrogen
  • All hydrogen ends up in water, so moles of H = 2 × moles of H₂O.
Oxygen
  • Find the mass of oxygen by subtracting the masses of C and H from the sample mass.
Ratio
  • Then proceed as usual to the lowest whole-number ratio of moles.

Section 5 of 6Worked Example

Question

  • A hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass, and has Mr = 56.
  • Find the empirical and molecular formulae.

  1. Moles C = 85.7 / 12.0 = 7.14; moles H = 14.3 / 1.0 = 14.3.
  2. Divide by the smaller, 7.14: C = 1.00 and H = 2.00, so the empirical formula is CH₂.
  3. Empirical formula mass = 12.0 + 2(1.0) = 14.0.
  4. n = 56 / 14.0 = 4, so molecular formula = C₄H₈.

Answer

  • Empirical formula CH₂;
    • molecular formula C₄H₈.

Section 6 of 6Exam Checklist & Tips

1.2.4 – Key Points for the Exam

  • 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 present in one molecule.
  • Percentage composition The percentage by mass of each element in a compound.
  • Hydrated salt A salt containing water of crystallisation, shown as a formula such as CuSO₄·5H₂O.

Common Mistakes

  • Using the ratio of masses instead of the ratio of moles.
  • Rounding 1.33 or 1.5 to a whole number instead of multiplying through.
  • Giving an empirical formula when the question asks for the molecular formula.
  • Forgetting to find the mass of oxygen by difference in combustion analysis.
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