During the combustion of 7.4 organic matter, 8.96 liters of carbon dioxide and 0.5 mol of water were released.

During the combustion of 7.4 organic matter, 8.96 liters of carbon dioxide and 0.5 mol of water were released. The relative density for hydrogen is 37. Determine the formula of the substance.

N (CO2 = 8.96 / 22.4 = 0.4 mol
nCO2 = n (C) = 0.4 mol
n (H2O) = 0.5 mol
n (H) = 0.5 * 2 = 1 mol
m (CO2 = 0.4 * 12 = 4.8 g
m (H) 1 * 1 = 1 gr
7.4-4.8-1 = 1.6 g (O)
n (O) = 1.6 / 16 = 0.1 mol
Now we determine M (org. Ve-va) 37 * 2 = 74 g / mol
C = 0.4 mol
H = 1 mol
O = 0.1 mol
Least multiple of 0.1
We get the resulting formula C4H10O
M (C4H10O) = 74 g / mol
CH3-CH2-CH2-CH2-OH