Vapor density is the density of vapor in relation to hydrogen gas, and can be expressed in the equation:
vapor density = molar mass of gas/molar mass of H2.
Using this, we can find the vapor density of pure oxygen to be 16, and the molar mass of compound B to be 32 g/mol.
Because Compound B is an alcohol, it must be the only alcohol with molar mass 32, which is methanol (CH3OH).
Compound D has a composition 60% C, 13.4% H, and 26.6 % O. Assuming 100g and dividing each of these by their molar masses, the formula appears to be C5H13.3O1.66. Simplifying this, the formula becomes C3H8O. Compound D is therefore n-propanol, because Compound C is a carboxylic acid.
This means that Compound C is propanoic acid.
With propanoic acid and methanol, the ester is methyl propanoate.
My final answers are:
Compound A: methyl propanoate, (CH3CH2COOCH3)
Compound B: methanol, (CH3OH)
Compound C: propanoic acid, (CH3CH2COOH)
Compound D: propanol, (CH3CH2COOH)