Find the molecular formula of the limiting one-volume alcohol having a vapor density

Find the molecular formula of the limiting one-volume alcohol having a vapor density in air d = 1.5862, if the mass fraction of oxygen in it is 0.3478.

Let us determine the molar mass of alcohol, if its density by air is known:
M (ROH) = M (air) * D (air) = 29 * 1b5862 = 45.99 = 46 g / mol;
Let’s calculate the number of moles of oxygen by the formula:
Y (O) = m / M = 34.78 / 16 = 2.17 mol;
Hence, the mass of oxygen is equal to: m (O) = 2.17 * 16 = 34.72 g;
Thus, the mass of hydrogen and carbon is:
m (H) + m (C) = 46 – 34.72 = 11.2 g.
The sum of the masses of carbon, hydrogen and oxygen is: 34.12 + 11.2 = 45.9 g, which corresponds to the molar mass of alcohol.
Let’s make calculations based on the general formula of the limiting monohydric alcohol: СnH2n + 1OH;
Find the value: n
12n +2 ​​+1 + 16 + 1 = 46;
12n = 26;
n = 2;
Alcohol formula: C2H5OH, M (C2H5OH) = 12 * 2 +1 * 6 + 16 = 46 g / mol;
Answer: the molecular formula of alcohol: C2H5OH – ethanol.