The combustion of the hydrocarbon released 0.22 g of carbon dioxide and 0.09 g of water vapor.

The combustion of the hydrocarbon released 0.22 g of carbon dioxide and 0.09 g of water vapor. The density of this substance in the air is 1.45. Determine the molecular formula of the hydrocarbon.

m (CO2) = 0.22 g
m (H2O) = 0.09 g
D = 1.45
Decision:
M (a.v.) = 29 * 1.45 = 42 grams / mol
n (CO2) = m / M = 0.22 / 44 = 0.005 mol
n (CO2) = n (C) = 0.05 mol
n (H2O) = m / M = 0.09 / 18 = 0.005 mol
n (H) = n (H2O) * 2 = 0.005 * 2 = 0.01 mol
Divide everything by the least multiple
We get 0.01: 0.005 = 2
Simplest formula: CH2
M (CH2) = 14 grams / mol
But we know that the hydrocarbon has M = 42 grams / mol
42/14 = 3
True Formula: C3H6