Question

When 0.72g of a liquid vaporized at 110.0oC and 0.967atm, the gas occupies a volume of 0.417L. The empirical formula of the gas is CH2. What is the molecular formula of the gas?

Answer 1

To find the molecular formula of the gas, its molecular weight must be calculated.

According to the ideal gas law:

pV = nRT or pV = mRT / Mr

where p - pressure, V - volume, n - number of moles, R - gas constant, T - temperature, m - mass, Mr - molecular weight.

From here:

Mr = mRT / pV

As R = 0.082057 L atm / K mol and T = 110.0 °C = 383.15 K:

Mr(gas) = (0.72 g × 0.082057 L atm / K mol × 383.15 K) / (0.967 atm × 0.417 L) = 56 g/mol

As a result, molecular weight of the gas is 56 g/mol and its empirical formula is CH₂. From here:

(12g/mol × x) + (2 × 1 g/mol × x) = 56 g/mol

where x - index coeeficient, 12g/mol and 1 g/mol - molecular weight of carbon and hydrogn respectively. From here:

12x + 2x = 56

14x = 56

x = 4

As a result, the molecular formula of the gas is C₄H₈

Answer: C₄H₈