Question

An organic compound with vapor density$=94$ contains $\mathrm{C}=12.67\%,\mathrm{H}=2.13\%\mathrm{and}\mathrm{Br}=85.11\%$. Find the molecular formula.

$\left[\mathrm{Atomic}\mathrm{mass}:\mathrm{C}=12,\mathrm{H}=1,\mathrm{Br}=80\right]$

Answer 1

Step 1: Given information

• The vapor density of chlorine of compound$=94$
• The percentage composition of carbon $\left(\mathrm{C}\right)=12.67\%$ and its atomic weight is $12$.
• The percentage composition of hydrogen $\left(\mathrm{H}\right)=2.13\%$ and its atomic weight is $1$.
• The percentage composition of bromine $\left(\mathrm{Br}\right)=85.11\%$and its atomic weight is $80$.

Step 2: Formulas for the calculation of atomic ratio and simplest ratio

• The atomic ratio is calculated by dividing the percentage composition of each element by its atomic weight.

$\mathrm{Atomic}\mathrm{ratio}=\frac{\mathrm{Percentage}\mathrm{composition}\mathrm{of}\mathrm{an}\mathrm{element}}{\mathrm{Atomic}\mathrm{weight}\mathrm{of}\mathrm{an}\mathrm{element}}$

• The simplest ratio is calculated by dividing the atomic ratio of each element by the smallest atomic ratio.

$\mathrm{Simplest}\mathrm{ratio}=\frac{\mathrm{Atomic}\mathrm{ratio}\mathrm{of}\mathrm{an}\mathrm{element}}{\mathrm{Smallest}\mathrm{atomic}\mathrm{ratio}}$

Step 3: Determination of the empirical formula

Element	Percentage composition	Atomic weight	Atomic ratio	Simplest ratio
C	$12.67$	$12$	$\frac{12.67}{12}=1$	$\frac{1}{1}=1$
H	$2.13$	$1$	$\frac{2.13}{1}=2.13$	$\frac{2.13}{1}\simeq 2$
Br	$85.11$	$80$	$\frac{85.11}{80}=1$	$\frac{1}{1}=1$

• The simplest ratio of $\mathrm{C}:\mathrm{H}:\mathrm{Br}=1:2:1$, which is a whole number ratio.
• Therefore, the empirical formula of the compound is ${\mathrm{CH}}_2\mathrm{Br}$.

Step 4: Calculate relative molecular mass or molecular weight

• The calculation for molecular weight is as follows:

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{weight}& =& 2\times \mathrm{Vapor}\mathrm{density}\\ & =& 2\times 94\\ & =& 188\end{array}$

Step 5: Calculation of value of $\mathit{n}$

• The empirical formula weight is calculated as follows:

$\begin{array}{rcl}\mathrm{Empirical}\mathrm{formula}\mathrm{weight}& =& 12+1\times 2+80\\ & =& 94\end{array}$

• Determine the value of $n$ as follows:

$\begin{array}{rcl}n& =& \frac{\mathrm{Molecular}\mathrm{weight}}{\mathrm{Empirical}\mathrm{formula}\mathrm{weight}}\\ & =& \frac{188}{94}\\ & =& 2\end{array}$

Step 6: Determination of the molecular formula

• The molecular formula is determined as follows:

$\begin{array}{rcl}\mathrm{Molecular}\mathrm{formula}& =& {\left(\mathrm{Empirical}\mathrm{formula}\right)}_n\\ & =& {\left({\mathrm{CH}}_2\mathrm{Br}\right)}_2\\ & =& {\mathrm{C}}_2{\mathrm{H}}_4{\mathrm{Br}}_2\end{array}$

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