Question

The empirical formula of the compound is $C{H_2}Cl$. If the molecular mass of the compound is $99g$, what is the molecular formula of the compound?

Answer 1

We have to know that the empirical formula of compound is the simple ratio of the amount of several atoms found. We have to know that the molecular formula is the multiple of empirical formula and we can calculate the molecular formula using the expression below,
$\text{Molecular Formula} = n \times \text{Empirical Formula}$

Complete answer:
Let us now see what is a molecular formula?
The molecular formula is the recipe obtained from particles and is illustrative of the total number of individual atoms present in a particle of a compound.
A molecular formula utilizes a subscript that reports the real number of each kind of atom in a particle of the compound.
Molecular formulas are related with gram sub-atomic masses that are basic entire number products of the comparing empirical formula mass.
We are provided with an empirical formula of the compound as $C{H_2}Cl$ and the molecular mass of the compound is $99g$. Let us now calculate the “n” factor.
We can calculate the “n” factor by dividing the given molecular mass to the empirical formula mass. Before obtaining the n-factor, let us calculate the empirical formula mass.
The atomic mass of carbon is $12g$.
The atomic mass of hydrogen is $1g$.
The atomic mass of chlorine is $35.5g$.
We have to the empirical formula mass (E.F.M) for the given empirical formula as,
$E.F.M = 12 + \left( {2 \times 1} \right) + 35.5 = 49.5g$
Empirical formula mass for the given empirical formula is $49.5g$.
We can calculate the n-factor as,
$\text{n - factor} = \dfrac{{\text{Empirical Formula Mass}}}{{\text{Molecular Mass}}}$
Now we can substitute the known values we get,
$\text{n - factor} = \dfrac{{99g}}{{49.5g}}$
On simplification we get,
$\text{n - factor} = 2$
Let us now multiply the empirical formula and “n” factor to get the molecular formula of the compound.
$\text{Molecular Formula} = n \times Empirical Formula$
Now we can substitute the known values we get,
$\text{Molecular Formula} = 2 \times C{H_2}Cl$
On simplification we get,
$\text{Molecular Formula} = {\left( {C{H_2}Cl} \right)_2}$
We have obtained the molecular formula of the compound as ${\left( {C{H_2}Cl} \right)_2}$.

Note:
We have to know the difference between empirical formula and molecular formula. In empirical formulas, the simplest whole number ratio of several atoms is found in a compound whereas in a molecular formula, the correct number of several kinds of atoms found in molecules of a compound are given.