Determine the molecular formula of a substance in which the mass fraction of carbon is 39.97%, hydrogen 6.73%, oxygen 53.3%.

Determine the molecular formula of a substance in which the mass fraction of carbon is 39.97%, hydrogen 6.73%, oxygen 53.3%. The vapor density of this substance in terms of carbon dioxide is 4.091

Let the mass of the substance = 100 g
Then:
m (C) = 39.97 gr n (C) = 39.97 / 12 = 3.33 mol
m (H) = 6.73 gr n (H) = 6.73 = 6.73 mol
m (O) = 53.3 gr n (O) = 53.3 / 16 = 3.33 mol
Ratio: n (C): n (H): n (O) = 1: 2: 1
Primary formula CH2O
M (CH2O) = 12 + 2 + 16 = 30 g / mol
since CO2 vapor density = 4.091, then
x / 44 = 4.091 44g / mol-molar mass of CO2
Hence x = 180 g / mol is the paint mass of the desired substance
180/30 = 6
6 times you need to multiply all the coefficients in the primary formula
We get: C6H12O6 (glucose) – the desired formula