Question

The empirical formula of a monobasic acid is $C{H_2}O$. The vapour density of its ethyl ester is ${\text{44}}$. The molecular formula of the acid?
A) ${C_2}{H_4}{O_2}$
B) $C{H_2}{O_2}$
C) ${C_2}{H_2}{O_2}$
D) ${C_3}{H_6}{O_3}$

Answer 1

The empirical formula of a compound is simply the positive integer ratio of all the atoms present in that molecule. One can relate the value given of the vapour density of ester and the empirical formula to find out the molecular formula of the acid.

Complete step by step answer:

1. First of all let's understand the terms given in the question one by one. The vapour density of a molecule can be defined as the mass of a certain volume of a substance or molecule divided by the mass of the same volume of hydrogen.
2. Now the empirical formula of the desired acid compound is given which means we can calculate the empirical weight as follows,
   $C{H_2}O = 12 + 2 \times 1 + 16 = 30$
3. Now as we know the molecular mass of a compound is twice the value of the vapour density. Hence, the molecular mass of an unknown acid compound is as follows,
   Molecular mass $= 2 \times {\text{ Vapour density = }}2 \times 44 = 88$
   So, the molecular mass of the unknown acid is ${\text{88}}$.
4. Now that we know that the given compound in the question is of the ethyl ester which means we can calculate the mass of just the ethyl ester and relate it with the value of molecular mass we just calculated. The formula for ethyl ester is $COO({C_2}{H_5})$ and the molecular weight is as follows,
   $COO({C_2}{H_5}) = 12 + 16 + 16 + (2 \times 12 + 5 \times 1)$
   $COO({C_2}{H_5}) = 73$
   Now that the value of the mass of the ethyl ester is ${\text{73}}$ and the molecular mass of unknown acid is $88$ we can find out the alkane formula by calculating the difference between these two values as follows,
   $Alkane = 88 - 73 = 15$
   Hence, the value ${\text{15}}$ represents the mass of the alkane group.
5. Now that we need the mass of acid,
   ${\text{Mass of acid = Mass of Alkane + Mass of ester group(COOH)}}$
   ${\text{Mass of acid = 15 + 45}}$
   ${\text{Mass of acid = 60}}$
6. Now let's calculate the molecular formula by using the known values by the following formula,
   ${\text{Molecular formula of unknown acid = }}\dfrac{{{\text{Molecular weight}}}}{{{\text{Empirical weight}}}} \times {\text{Empirical formula}}$
   ${\text{Molecular formula of unknown acid = }}\dfrac{{60}}{{30}} \times C({H_2}O)$
   ${\text{Molecular formula of unknown acid = 2}} \times C({H_2}O)$
   ${\text{Molecular formula of unknown acid}} = {C_2}{H_4}{O_2}$
   Therefore, the molecular formula of the unknown acid is ${C_2}{H_4}{O_2}$ which shows option A as the correct choice.

Note:
The molecular formula which is of acid ${C_2}{H_4}{O_2}$ represents the formula having the acid functional group as in a simple way as $C{H_3} - COOH$ which is acetic acid. The molecular formula given can’t be converted into structures unless we know the functional group present in it which in this case is acid.