IB DP Chemistry - S1.4.4 Empirical and molecular formulas - Study Notes - New Syllabus - 2026, 2027 & 2028
IB DP Chemistry – S1.4.4 Empirical and molecular formulas – Study Notes – New Syllabus
IITian Academy excellent Study Notes and effective strategies will help you prepare for your IB DP Chemistry exam.
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Chemistry 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Structure 1.4.4 — Empirical and Molecular Formulae
Structure 1.4.4 — Empirical and Molecular Formulae
- 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 of the compound.
Interconverting $\%$ Composition and Empirical Formula
Steps to determine empirical formula from % composition:
- Assume 100 g of the substance (so % becomes mass in g).
- Convert each mass to moles: \( \frac{\text{mass}}{\text{Ar}} \)
- Divide each mole value by the smallest number of moles.
- If necessary, multiply by a suitable factor to get whole numbers.
- Write the empirical formula using these whole-number ratios.
Determining Molecular Formula from Empirical Formula
- Step 1: Calculate molar mass of the empirical formula (Memp).
- Step 2: Use the actual molar mass (Mmol) from data to find the ratio:
\( \text{Ratio} = \frac{M_\text{mol}}{M_\text{emp}} \)
- Step 3: Multiply all subscripts in the empirical formula by this ratio to get the molecular formula.
Example
A compound is found to contain 40.0% carbon, 6.71% hydrogen and 53.29% oxygen by mass. Determine its empirical formula.
▶️ Answer/Explanation
Assume 100 g of compound:
- C: 40.0 g → \( \frac{40.0}{12.01} = 3.33 \, \text{mol} \)
- H: 6.71 g → \( \frac{6.71}{1.008} = 6.66 \, \text{mol} \)
- O: 53.29 g → \( \frac{53.29}{16.00} = 3.33 \, \text{mol} \)
Divide all by 3.33:
- C: 1, H: 2, O: 1
Empirical formula = CH2O
Example
A compound has an empirical formula of CH2O and a molar mass of 180 g mol⁻¹. Determine the molecular formula.
▶️ Answer/Explanation
Empirical formula mass:
\( M_\text{emp} = 12.01 + 2(1.008) + 16.00 = 30.03 \, \text{g mol}^{-1} \)
Ratio = \( \frac{180}{30.03} = 6 \)
Molecular formula = \( (CH_2O)_6 = C_6H_{12}O_6 \)
Example
When 0.25 g of a hydrocarbon was burned completely in oxygen, it produced 0.733 g of CO2 and 0.300 g of H2O. Find the empirical formula.
▶️ Answer/Explanation
From CO2: Moles of C = \( \frac{0.733}{44.01} \approx 0.0167 \) mol
From H2O: Moles of H = \( \frac{0.300}{18.02} \times 2 \approx 0.0333 \) mol
Mole ratio: C:H = 0.0167 : 0.0333 = 1:2
Empirical formula = CH2
Example
A compound is found to contain 29.1% Na, 40.5% S, and 30.4% O by mass. Determine its empirical formula.
▶️ Answer/Explanation
Assume 100 g → Na = 29.1 g, S = 40.5 g, O = 30.4 g
- Na: \( \frac{29.1}{22.99} \approx 1.27 \)
- S: \( \frac{40.5}{32.07} \approx 1.26 \)
- O: \( \frac{30.4}{16.00} = 1.90 \)
Divide by the smallest (≈1.26):
- Na ≈ 1.01 → 1
- S ≈ 1.00 → 1
- O ≈ 1.90 → ~2
Empirical formula = NaSO2