Question

The empirical formula of a compound is ${\mathrm{CH}}_4\mathrm{N}$. The empirical formula mass of the compound is equal to its vapour density. Which among the following compounds possesses the same number of carbon atoms as present in the above compound?

• A. ${\mathrm{Na}}_2{\mathrm{CO}}_3$
• B. $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$
• C. ${\mathrm{Al}}_2{\left({\mathrm{CO}}_3\right)}_3$
• D. $\mathrm{Fe}{\left({\mathrm{HCO}}_3\right)}_3$

Answer 1

The correct option is B

$\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$

Explanation for the correct option:

The empirical formula of the compound is given as ${\mathrm{CH}}_4\mathrm{N}$.

The vapor density is same as the empirical formula mass, so the mass of ${\mathrm{CH}}_4\mathrm{N}$ is:

${\mathrm{CH}}_4\mathrm{N}=12+4+14=30$

The vapor density = $30$

As the Molecular formula mass = $\mathrm{Vapor}\mathrm{density}\times 2$

Therefore, the Molecular formula mass $=30\times 2=60$.

By dividing the Molecular formula mass by empirical formula mass, we can get the value of $\mathrm{n}$.

$=\frac{\mathrm{Molecular}\mathrm{formula}\mathrm{mass}}{\mathrm{Empirical}\mathrm{formula}\mathrm{mass}}=\mathrm{n}$

$=\frac{60}{30}=2$

Now, the Molecular formula will be $=\mathrm{n}\times {\mathrm{CH}}_4\mathrm{N}=2\times {\mathrm{CH}}_4\mathrm{N}={\mathrm{C}}_2{\mathrm{H}}_8{\mathrm{N}}_2$

So, the Molecular formula have two Carbon ($\mathrm{C}$) atoms. Hence, from the given options, $\mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2$ also has two Carbon atoms.

Therefore, the correct option is (B).