Question

The empirical formula of a compound is $C{H_2}$. The mass of $1L$ this gas is exactly equal to that of $1L$ Nitrogen Gas under similar conditions. The molecular formula for the gas is:
A.${C_2}{H_4}$
B.${C_3}{H_6}$
C.${C_6}{H_{12}}$
D.${C_4}{H_12}$

Answer 1

By equating the mass of Nitrogen gas with the mass of the given gas, the molecular mass of the gas can be determined which can be further used to compare with the mass of molecules given in the options.

Complete step by step answer:
Let us first understand what Empirical Formula and Molecular formula means and how they can be interconverted.
Empirical formula is the simplest possible ratio between the atoms of the molecule. It can only determine the percentage composition of the atoms with respect to each other or the mass of that atom in the molecule. It does not represent the true molecule.

Molecular formula is the true representative of the number of atoms with respect to each other. It can be used to deduce structures also, and for simple hydrocarbon in follows the ${C_n}{H_{2n + 2}},{C_n}{H_{2n}},{C_n}{H_{2n - 2}}$ rule depending on the degree of unsaturation.

Empirical formula can be converted into molecular formula by multiplying the empirical formula with a factor of $n$. where$n$is an integer.
Since the empirical formula for all the given molecular formula will be the same $[C{H_2}]$ , we cannot use this property to determine the right molecular formula.

If two gases occupy the same volume then they have the same number of moles and hence they have the same mass. Since the gas with empirical formulae $C{H_2}$ has the same mass as nitrogen gas, we can say the mass of the unknown gas will be $28g$ since the mass of ${N_2}$ gas is $28g$ $[N = 14g]$.

Out of the above options, the mass of
${C_2}{H_4} = 28g$ and hence the molecular formula of the gas is ${C_2}{H_4}$.
Hence option A is correct.

Note:
When calculating the empirical formula from the molecular formula, find out the HCF of the number of individual atoms in the molecule. For example; ${C_2}{H_4}$ has $2$ Carbon atoms and $4$ Hydrogen atoms and hence the HCF of $2$and $4$ will be $2$. So by diving ${C_2}{H_4}$ by $2$, we get $C{H_2}$.