The molecular formula of the gas is (CH₂)₃, or C₃H₆
How can you solve the molecular formula of the gas?
Molar mass = (12.011 g/mol) + (1.008 g/mol) = 13.019 g/mol
The ideal gas law can be used to calculate the molar volume:
PV = nRT
734 mmHg × (1 atm/760 mmHg) = 0.965 atm
27°C + 273.15 = 300.15 K
(0.965 atm) × V = (1 mol) × (0.0821 L·atm/mol·K) × (300.15 K)
V = 24.51 L
This means that one mole of the gas occupies 24.51 liters at 27°C and 734 mmHg.
The density of the gas is given as 1.65 g/L. This means that one liter of the gas weighs 1.65 grams. Since one mole of the gas occupies 24.51 liters, then one mole of the gas weighs:
1.65 g/L × 24.51 L/mol = 40.05 g/mol
We know that the molar mass of the empirical formula is 13.019 g/mol. Therefore, the molecular formula of the gas is a multiple of the empirical formula. We can find the multiple by dividing the molar mass of the gas by the molar mass of the empirical formula:
40.05 g/mol ÷ 13.019 g/mol = 3.08
This means that the molecular formula of the gas is (CH₂)₃, or C₃H₆.