An unknown gaseous substance has a density of 1.06 g/L at 31 °C and 371 torr. If the substance has the following percent composition: 88.8% C and 11.2% H, its molecular formula is ________.

Question

Quizplus · Accepted Answer

1. Calculate the molecular weight of the unknown substance:
- Divide the density by the molar volume at the given temperature and pressure to obtain the molar mass of the unknown gas: 
Molar volume = (RT) / P = [(0.08206 L.atm/mol.K)(31 °C + 273.15)] / (371/760 atm) = 0.02268 m³/mol
Molar mass = Density × Molar volume = 1.06 g/L × 0.02268 m³/mol × 1000 g/kg = 24.02 g/mol
- Calculate the number of moles of carbon and hydrogen in 100 g of the unknown substance:
88.8 g C / 12.01 g/mol C = 7.395 mol C
11.2 g H / 1.008 g/mol H = 11.11 mol H
2. Find the empirical formula of the unknown substance:
- Divide the number of moles of each element by the smallest number of moles to obtain the mole ratio:
C:H = 7.395 mol : 11.11 mol ≈ 2:3
- Multiply by a common factor to obtain whole-number subscripts:
C₂H₃
3. Determine the molecular formula of the unknown substance:
- Calculate the empirical formula mass of C₂H₃:
12.01 g/mol C × 2 + 1.008 g/mol H × 3 = 26.03 g/mol
- Divide the molar mass of the unknown substance by the empirical formula mass to obtain the molecular formula factor:
Molecular formula factor = 24.02 g/mol / 26.03 g/mol ≈ 0.923
- Multiply the subscripts of the empirical formula by the molecular formula factor and round to the nearest whole number:
C₄H₆
- Check that the molecular formula has the same percent composition of C and H as given in the problem:
(4 × 12.01 g/mol C) / 54.07 g/mol × 100% ≈ 88.8%
(6 × 1.008 g/mol H) / 54.07 g/mol × 100% ≈ 11.2%

An Unknown Gaseous Substance Has a Density of 1