Question

(ii) A compound has the following percentage composition by mass:

Carbon $14.4\%$, hydrogen $1.2\%$, and chlorine $84.5\%$

The relative molecular mass of this compound is $168$. What is its molecular formula?

Answer 1

The empirical formula of the compound calculated in part (i) is ${\mathrm{HCCl}}_2$.

Given Information

• The molecular weight of this compound is $168$.

Calculation of empirical formula weight

• The empirical formula weight is calculated as follows:

$$\begin{gathered} \mathrm{Empirical}\mathrm{formula}\mathrm{weight}=1+\left(1\times 12\right)+\left(35.5\times 2\right) \\ =84 \end{gathered}$$

Calculation of value of $\mathit{n}$

• Determine the value of $n$ as follows:

$$\begin{gathered} n=\frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}} \\ \mathit{}\mathit{}\mathit{}\mathrm{=}\frac{168}{84} \\ =2 \end{gathered}$$

Determination of the molecular formula

• The molecular formula is determined as follows:

$$\begin{gathered} \text{Molecular formula}={\left(\mathrm{Empirical}\mathrm{formula}\right)}_n \\ ={\left({\mathrm{HCCl}}_2\right)}_2 \\ ={\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{Cl}}_4 \end{gathered}$$

Therefore, the molecular formula of the compound is ${\mathrm{H}}_2{\mathrm{C}}_2{\mathrm{Cl}}_4$.