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Difficulty: MediumEmpirical and Molecular Formula Calculations

An organic compound contains 40.0%40.0\% carbon, 6.7%6.7\% hydrogen, and 53.3%53.3\% oxygen by mass. If the vapour density of the compound is 3030, what is its molecular formula? [C=12,H=1,O=16][C = 12, H = 1, O = 16]

  1. A
    CH2OCH_2O
  2. C2H4O2C_2H_4O_2Answer
  3. C
    C3H6O3C_3H_6O_3
  4. D
    C4H8O4C_4H_8O_4

Answer

The molecular formula of the compound is C2H4O2C_2H_4O_2.
Dividing mass percentage by relative atomic mass yields a 1:2:11:2:1 mole ratio for C:H:O\text{C}:\text{H}:\text{O}, establishing an empirical formula of CH2OCH_2O (mass =30 g/mol= 30\text{ g/mol}). Multiplying the vapour density (3030) by 22 gives a molar mass of 60 g/mol60\text{ g/mol}. The molecular formula multiplier n=60/30=2n = 60 / 30 = 2, yielding C2H4O2C_2H_4O_2.

Step-by-Step Solution

1
Determine the mole ratio of each constituent element
Moles of C=40.012=3.33C = \frac{40.0}{12} = 3.33, Moles of H=6.71=6.70H = \frac{6.7}{1} = 6.70, Moles of O=53.316=3.33O = \frac{53.3}{16} = 3.33
Mass percentages are divided by their respective relative atomic masses to yield mole quantities.
2
Calculate the simplest whole-number ratio to obtain the empirical formula
Ratio C:H:O=3.333.33:6.703.33:3.333.33=1:2:1C : H : O = \frac{3.33}{3.33} : \frac{6.70}{3.33} : \frac{3.33}{3.33} = 1 : 2 : 1. Empirical formula is CH2OCH_2O.
Dividing all mole values by the smallest value (3.333.33) converts the mole ratio into simple integers.
3
Compute the molar mass from the given vapour density
Molar Mass=2×Vapour Density=2×30=60 g/mol\text{Molar Mass} = 2 \times \text{Vapour Density} = 2 \times 30 = 60\text{ g/mol}
The molar mass of a volatile substance is equal to twice its vapour density.
4
Find the molecular formula multiplier nn and determine the molecular formula
Empirical formula mass of CH2O=12+(2×1)+16=30 g/molCH_2O = 12 + (2 \times 1) + 16 = 30\text{ g/mol}. n=6030=2n = \frac{60}{30} = 2. Molecular formula = (CH2O)2=C2H4O2(CH_2O)_2 = C_2H_4O_2.
The molecular formula is obtained by multiplying the subscripts of the empirical formula by nn, where n=Molar MassEmpirical Massn = \frac{\text{Molar Mass}}{\text{Empirical Mass}}.

Key Concept

Deriving Empirical and Molecular Formulae using Vapour Density
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Topics

Empirical and Molecular Formula Calculations

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