Question

An organic compound contains 49.3% carbon 6.84% hydrogen and its vapour density is 73. Molecular formula of the compound is:
A) ${C_3}{H_5}{O_2}$
B) ${C_6}{H_{10}}{O_4}$
C) ${C_3}{H_{10}}{O_2}$
D) ${C_4}{H_{10}}{O_2}$

Answer 1

Hint: We are given the percentage composition of two elements Carbon and Hydrogen. We can find the percentage composition by subtracting the whole my 100. On finding the no. of moles of each present, we can find the empirical formula and from that the molecular formula can be determined

Complete answer:
We are given an Organic compound, mostly the organic compounds consist of Carbon, Hydrogen and Oxygen as their constituent elements. We are given the percentage composition of Carbon and Hydrogen as 49.3% and 6.84% respectively.
Assume the total of the masses of the organic compound gives a total of 100. Considering the mass of Oxygen as ‘x’, we can determine it as:
$49.3 + 6.84 + x = 100$
$x = 43.86$
Therefore, out of the 100g of composition, we have 43.86 g as Oxygen, 4.93g as carbon and 6.84g as Hydrogen. Finding the no. of moles of each element present in the compound.
$Moles = \dfrac{{Mass(g)}}{{Molar{\text{ }}Mass(g/mol)}}$
[The molar masses of carbon, hydrogen and Oxygen are 12, 1 and 16g/mol respectively]
${n_{carbon}} = \dfrac{{49.3}}{{12}} = 4.10$mol
${n_{hydrogen}} = \dfrac{{6.84}}{1} = 6.84mol$
${n_{oxygen}} = \dfrac{{43.86}}{{16}} = 2.74$mol
Dividing throughout by common divisor to get a minimum ratio of Carbon:Hydrogen:Oxygen
${n_C}:{n_H}:{n_O} = 1.5:2.5:1 \to 3:5:2$
The empirical formula of the compound thus can be given as: ${C_3}{H_5}{O_2}$
The empirical formula gives us the minimum whole no. ratio of the elements present in the compound. The empirical formula mass of the compound will be $ = 3 \times 12 + 5 \times 1 + 2 \times 16 = 73g/mol$
We are given that the vapour density of the compound is 73. The vapour density (VD) can be given as:
$V.D = \dfrac{{Molecular{\text{ }}Mass}}{2}$
Substituting the values to get the Molecular Mass $ = 73 \times 2 = 146$
The relationship between Empirical Formula Mass and Molecular Formula mass is:
$Molecular{\text{ }}Mass = n \times Empirical{\text{ }}Formula{\text{ }}Mass$
$n = \dfrac{{146}}{{73}} = 2$
Therefore, the Molecular Formula $ = {({C_3}{H_5}{O_2})_n} = {({C_3}{H_5}{O_2})_2} = {C_6}{H_{10}}{O_4}$

The correct option is B .

Note:
The vapour density is defined as the ratio of the molecular mass of n molecules of the compound to the molecular mass of n molecules of Hydrogen atom. It can be given as: $Molecular{\text{ }}Mass = V.D \times Molar{\text{ }}Mas{s_{Hydrogen}}$