Determining Molecular Formulas from Empirical Data

The molecular formula of a compound is a whole-number multiple of its empirical formula, and it is found by dividing the molar mass of the compound by the molar mass of the empirical formula: n = M ÷ M_empirical. In brief: a compound with the empirical formula CH₂O (30.03 g/mol) and a molar mass of 180.2 g/mol has n = 180.2 ÷ 30.03 = 6, so its molecular formula is C₆H₁₂O₆.

The empirical formula gives the ratio of atoms but not their actual number. To get the molecular formula, one extra piece of information is needed: the molar mass. This article explains the method, shows the experimental ways of obtaining the molar mass (mass spectrometry, gas density, freezing point depression), compares empirical and molecular formulas in a table, works through six examples that begin with different kinds of data and gives some history.

How are empirical and molecular formulas related?

The molecular formula is the empirical formula multiplied by a whole number n: (empirical formula)_n. If n = 1, the two formulas are the same, as in water (H₂O) and methane (CH₄). Many compounds, especially organic compounds, have molecular formulas that are multiples of the empirical formula. The molar mass of the compound is n times the molar mass of the empirical formula.

What are the steps?

1. Find the empirical formula (from percent composition or masses).
2. Calculate the molar mass of the empirical formula (the empirical formula mass).
3. Divide the experimental molar mass of the compound by the empirical formula mass. The result should be close to a whole number n.
4. Multiply all subscripts in the empirical formula by n.
5. Check: calculate the molar mass of the molecular formula and compare it with the data.

Examples of empirical and molecular formulas

How is the molar mass measured?

• Mass spectrometry: the molecular ion peak with the highest mass-to-charge ratio gives the molar mass directly with high accuracy.
• Gas density: for volatile compounds, M = ρ × V_m at STP, or M = ρRT ÷ P in other conditions (the method of Dumas, 1826, and of Victor Meyer, 1878).
• Colligative properties: the freezing point depression or boiling point elevation of a solution of a known mass of the compound is proportional to the number of moles dissolved: ΔT_f = K_f × m.
• Osmotic pressure is used for large molecules such as polymers and proteins.
• Titration can give the equivalent mass of an acid or a base.

Worked examples

1. Simple multiple. The empirical formula is CH₂O and the molar mass is 180.2 g/mol. Empirical formula mass: 12.011 + 2(1.008) + 15.999 = 30.026. n = 180.2 ÷ 30.026 = 6.00. Molecular formula: C₆H₁₂O₆.
2. From percent composition and a mass. A hydrocarbon is 85.7% C and 14.3% H, and its molar mass is 56.1 g/mol. C: 85.7 ÷ 12.011 = 7.135 mol. H: 14.3 ÷ 1.008 = 14.19 mol. Ratio: C 1.000, H 1.988, which is 2. The empirical formula is CH₂ (14.027 g/mol). n = 56.1 ÷ 14.027 = 4.0. The molecular formula is C₄H₈.
3. From gas density. A gas consists of carbon and hydrogen with the empirical formula CH (13.019 g/mol). Its density at STP is 1.16 g/L. M = 1.16 × 22.414 = 26.0 g/mol. n = 26.0 ÷ 13.019 = 2.00. The molecular formula is C₂H₂ (acetylene).
4. From a freezing-point measurement. A sample of 2.50 g of a compound with the empirical formula CH₂O is dissolved in 50.0 g of water (K_f = 1.86 °C·kg/mol), and the freezing point is lowered by 0.516 °C. The compound does not ionize, so the molality is 0.516 ÷ 1.86 = 0.2774 mol/kg. The moles of compound in 0.0500 kg of water are 0.2774 × 0.0500 = 0.01387 mol. M = 2.50 ÷ 0.01387 = 180 g/mol. n = 180 ÷ 30.026 = 6.0, so the molecular formula is C₆H₁₂O₆.
5. Check the result. For C₄H₈: 4(12.011) + 8(1.008) = 48.044 + 8.064 = 56.108 g/mol, which agrees with the measured 56.1 g/mol.
6. Phosphorus oxide. The empirical formula is P₂O₅ (141.94 g/mol), and the molar mass is 283.9 g/mol. n = 283.9 ÷ 141.94 = 2.00. The molecular formula is P₄O₁₀, the formula of the molecules in the vapor.

A complete problem from percent composition

A compound is 40.00% carbon, 6.71% hydrogen and 53.29% oxygen by mass, and mass spectrometry shows that its molar mass is 180 g/mol.

1. In 100 g: C 40.00 ÷ 12.011 = 3.330 mol; H 6.71 ÷ 1.008 = 6.657 mol; O 53.29 ÷ 15.999 = 3.331 mol.
2. Divide by 3.330: C 1.000, H 1.999, O 1.000. Empirical formula CH₂O (30.026 g/mol).
3. n = 180 ÷ 30.026 = 5.99, which is 6.
4. The molecular formula is C₆H₁₂O₆. The identity of the compound (glucose, fructose or another sugar) needs more information about the structure.

How did chemists determine molecular formulas?

The determination of molar masses was crucial for developing the system of atomic masses. Jean-Baptiste Dumas used the vapor density method in 1826. In 1860 Stanislao Cannizzaro showed that a consistent set of atomic masses could be obtained by combining the vapor densities of volatile compounds with Avogadro’s hypothesis. François-Marie Raoult discovered the lowering of the freezing point of solutions in the 1880s, which provided a method for compounds that cannot be vaporized. In the 20th century, Francis Aston’s mass spectrograph (1919), and later modern mass spectrometers, allowed direct determination of molecular masses with high precision.

Which mistakes are common?

Mistake 1: Using the molar mass of the molecular formula to find n

n = M(compound) ÷ M(empirical formula). Divide by the empirical formula mass, not by the mass of the molecular formula.

Mistake 2: Multiplying only some of the subscripts

Multiply every subscript in the empirical formula by n.

Mistake 3: Accepting a non-integer n

If n is not close to a whole number, check the empirical formula and the molar mass for errors.

Mistake 4: Forgetting that isomers share a molecular formula

Ethanol and dimethyl ether are both C₂H₆O. The molecular formula does not tell the structure.