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1.4.3Additional lessons

Empirical Formula — IB Chemistry SL Revision Notes & Topical Questions

5 min read 5 parts Lesson 13 of 98

📚 IB Chemistry SL Hub / Topical Revision
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Part 1 of 5Definitions

1.4.3 Empirical Formula

The formula can show the simplest ratio or the actual number of atoms.

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 a molecule, a whole-number multiple of the empirical formula.
Example
  • Glucose has the molecular formula C₆H₁₂O₆ and the empirical formula CH₂O.
Ionic compounds
  • Their formulas are always empirical, because they show the ratio of ions in the lattice.

Part 2 of 5Finding the Empirical Formula

Convert percentages or masses into mole ratios.

1
Write the mass of each element, treating percentages as masses in a 100 g sample.
2
Divide each mass by the molar mass of the element to find moles of atoms.
3
Divide all the mole values by the smallest one to obtain a ratio.
4
If a ratio is close to a decimal like 1.5, multiply all values by 2 to obtain whole numbers.

Part 3 of 5To the Molecular Formula

To the Molecular Formula
Needed data
  • The molar mass of the compound is needed in addition to the empirical formula.
Multiplier
  • Divide the molar mass by the empirical formula mass to find the whole number n.
Final step
  • Multiply every subscript in the empirical formula by n.
Combustion data
  • Masses of CO₂ and H₂O from burning a sample give the carbon and hydrogen present in the sample.

Part 4 of 5Worked Example

Question

  • A compound contains 40.0% C, 6.7% H and 53.3% O by mass and has a molar mass of 180 g mol⁻¹.
  • Find its empirical and molecular formulas.

  1. In 100 g: n(C) = 40.0 ÷ 12.01 = 3.33; n(H) = 6.7 ÷ 1.01 = 6.63; n(O) = 53.3 ÷ 16.00 = 3.33.
  2. Dividing by 3.33 gives C : H : O = 1 : 2 : 1, so the empirical formula is CH₂O.
  3. Empirical formula mass = 12.01 + 2.02 + 16.00 = 30.03; 180 ÷ 30.03 = 6.
  4. Multiply each subscript by 6.

Answer

  • Empirical formula CH₂O;
    • molecular formula C₆H₁₂O₆

Part 5 of 5Exam Checklist & Tips

1.4.3 – Key Points for the Exam

  • Empirical formula The simplest whole-number ratio of the atoms of each element in a compound.
  • Molecular formula The actual numbers of atoms of each element in one molecule.
  • Mole ratio The ratio of the amounts in moles of the elements in a compound.
  • Percentage composition The mass of each element as a percentage of the total mass of the compound.

Common Mistakes

  • Using ratios of masses instead of ratios of moles.
  • Rounding 1.5 to 2 instead of doubling all the ratios.
  • Giving the empirical formula when the question asks for the molecular formula.
  • Dividing by the largest value rather than the smallest in the ratio step.

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