Question

The empirical formula of a compound is $C{{H}_{2}}$. The mass of 1 litre of this gas is exactly equal to that of 1 litre of nitrogen gas under similar conditions. The molecular formula of the gas is:
(A) ${{C}_{2}}{{H}_{4}}$
(B) ${{C}_{3}}{{H}_{6}}$
(C) ${{C}_{6}}{{H}_{12}}$
(D) ${{C}_{4}}{{H}_{8}}$

Answer 1

Hint: The empirical formula of a compound can be calculated with the molecular formula by converting it into the simplest whole-number ratio. The molecular formula of the compound can be calculated by equating the mass of nitrogen gas to the mass of the compound.

Complete step by step solution:
The empirical formula is the chemical formula that tells the simplest whole-number ratio of the atoms of all elements present in one molecule.
It is deduced by- (a) percentage composition of elements (b)- atomic masses.
The molecular formula of a compound is the chemical formula that tells the true formula of the compound or molecule. It expresses the actual number of atoms of various elements present in a molecule.
The molecular formula is calculated by multiplying n with the empirical formula, where n is an integer.
Here the empirical formula is given $C{{H}_{2}}$and the volume of this gas is exactly equal to the volume of nitrogen gas. So under similar conditions, if the volumes of the 2 gas are the same, then the number of moles of the gases will be equal. This means that the mass of both gases is the same. So, the molecular mass of the nitrogen gas will be equal to the molecular mass of the compound. We know that the molecular mass of nitrogen gas (${{N}_{2}}$) is $28\text{ g/mol}$. So, let us see the options:
(A) ${{C}_{2}}{{H}_{4}}$- Its molecular mass is 28 g/mol.
(B) ${{C}_{3}}{{H}_{6}}$- Its molecular mass is 42 g/mol.
(C) ${{C}_{6}}{{H}_{12}}$- Its molecular mass is 84 g/mol.
(D) ${{C}_{4}}{{H}_{8}}$- Its molecular mass is 56 g/mol.
So, the molecular formula of the compound is ${{C}_{2}}{{H}_{4}}$.

Hence, the correct answer is an option (a)- ${{C}_{2}}{{H}_{4}}$.

Note: The atomic mass of carbon is taken 12 g/mol and the atomic mass of hydrogen is taken 1 g/mol. ${{C}_{2}}{{H}_{4}}=2(12)+4(1)=28g/mol$, ${{C}_{3}}{{H}_{6}}=3(12)+6(1)=42g/mol$ , ${{C}_{6}}{{H}_{12}}=6(12)+12(1)=84g/mol$ , and ${{C}_{4}}{{H}_{8}}=4(12)+8(1)=56g/mol$.