Question

A carbon compound contains 12.8% of carbon, 2.1% of hydrogen and 85.1% of bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula of the compound. (Atomic wts: H = 1.008, C = 12.0, Br = 79.9)

• A. $\text{C}{\text{H}}_3\text{B}\text{r}$
• B. $\text{C}{\text{H}}_2\text{B}{\text{r}}_2\text{I}$
• C. ${\text{C}}_2{\text{H}}_4\text{B}{\text{r}}_2$
• D. ${\text{C}}_2{\text{H}}_3\text{B}{\text{r}}_3$

Answer 1

Answer: (C)

$\text{E}\text{l}\text{e}\text{m}\text{e}\text{n}\text{t}$	$\text{\%}$	$\text{A}\text{t}.\text{w}\text{t}$	$\text{R}\text{e}\text{l}\text{a}\text{t}\text{i}\text{v}\text{e}\text{n}\text{u}\text{m}\text{b}\text{e}\text{r}$	$\text{S}\text{i}\text{m}\text{p}\text{l}\text{e}\text{r}\text{a}\text{t}\text{i}\text{o}$
$\text{C}$	$12.8$	$12$	$\frac{12.8}{12}=1.06$	$\frac{106}{1.06}=1$
$\text{H}$	$2.1$	$1$	$\frac{2.1}{1}=2.1$	$\frac{2.1}{1.06}=2$
$\text{B}\text{r}$	$85.1$	$80$	$\frac{85.1}{80}=1.06$	$\frac{1.06}{1.06}=1$

Therefore, the Empirical Formula is $C{H}_{2}Br$
Weight of Empirical Formula $=12+2+80=94$
$\text{n}=\frac{\text{M}.\text{W}}{\text{w}\text{t}\text{o}\text{f}\text{E}\text{F}}=\frac{187.9}{94}=2$
$MF=(EF{)}_n={\left(C{H}_{2}Br\right)}_2$
Therefore, Molecular Formula $={\text{C}}_2{\text{H}}_4\text{B}{\text{r}}_2$