During the combustion of 7.5 g of oxygen-containing organic matter, 4.5 g of water vapor and 11 g of carbon monoxide

During the combustion of 7.5 g of oxygen-containing organic matter, 4.5 g of water vapor and 11 g of carbon monoxide are formed (4). Find the molecular formula of a substance if you know that the density of its vapors for hydrogen is 15.

Given.
m (o.v) = 7.5g
m (H2O) = 4.5g
m (Co2) = 11g
Find the molecular formula of a substance.
n (CO2) = m (CO2) \ M (CO2) = 11 \ 44 = 0.25 mol
m (C) = n (C) ⋅M (C) = 0.25⋅12 = 3g.
n (H2O) = m (H2O) \ M (H2O) = 4.5 \ 18 = 0.25 mol
n (H) = 2⋅n (H2O) = 2⋅0.25 = 0.5 mol
m (H) = n (H) ⋅M (H) = 0.5⋅1 = 0.5g
The mass of an organic compound is greater than the sum of the masses of carbon and hydrogen, which means that the compound must contain oxygen. Find the mass of oxygen:
m (O) = m (in-islands) −m (C) −m (H) = 7.5−3−0.5 = 4гm (O) = m (in-islands) -m (C) -m (H) = 7.5-3-0.5 = 4g
n (O) = m (O) \ M (O) = 4 \ 16 = 0.25 mol.
We find the ratio of carbon atoms, hydrogen, oxygen:
x: y: z = n (C): n (H): n (O) x: y: z = n (C): n (H): n (O)
x: y: z = 0.25: 0.5: 0.25
x: y: z = 1: 2: 1
Molar mass of a substance:
Dair. (In-va) = M (in-va) / M (air)
M (in-islands) = DH2 (in-islands) ⋅M (H2) = 15⋅2 = 30 g / mol
Substance formula: CH2OCH2O.
Answer: CH2OCH2O