Q. 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)
NTA AbhyasNTA Abhyas 2022
Solution:
$\text{E}\text{l}\text{e}\text{m}\text{e}\text{n}\text{t}$ $\text{\%}$ $\text{A}\text{t}.\text{w}\text{t}$ $\text{R}\text{e}\text{l}\text{a}\text{t}\text{i}\text{v}\text{e} \, \text{n}\text{u}\text{m}\text{b}\text{e}\text{r}$ $\text{S}\text{i}\text{m}\text{p}\text{l}\text{e} \, \text{r}\text{a}\text{t}\text{i}\text{o}$ $\text{C}$ $12.8$ $12$ $\frac{12.8}{12}=1.06$ $\frac{106}{1.06}=1$ $\text{H}$ $2.1$ $1$ $\frac{2.1}{1}=2.1$ $\frac{2.1}{1.06}=2$ $\text{B}\text{r}$ $85.1$ $80$ $\frac{85.1}{80}=1.06$ $\frac{1.06}{1.06}=1$
Therefore, the Empirical Formula is $CH_{2}Br$
Weight of Empirical Formula $=12+2+80=94$
$\text{n}=\frac{\text{M} . \text{W}}{\text{w} \text{t} \, \text{o} \text{f} \, \text{E} \text{F}}=\frac{187.9}{94}=2$
$MF =( EF )_{ n }=\left( CH _{2} Br \right)_{2}$
Therefore, Molecular Formula $=\text{C}_{2}\text{H}_{4}\text{B}\text{r}_{2}$
| $\text{E}\text{l}\text{e}\text{m}\text{e}\text{n}\text{t}$ | $\text{\%}$ | $\text{A}\text{t}.\text{w}\text{t}$ | $\text{R}\text{e}\text{l}\text{a}\text{t}\text{i}\text{v}\text{e} \, \text{n}\text{u}\text{m}\text{b}\text{e}\text{r}$ | $\text{S}\text{i}\text{m}\text{p}\text{l}\text{e} \, \text{r}\text{a}\text{t}\text{i}\text{o}$ |
| $\text{C}$ | $12.8$ | $12$ | $\frac{12.8}{12}=1.06$ | $\frac{106}{1.06}=1$ |
| $\text{H}$ | $2.1$ | $1$ | $\frac{2.1}{1}=2.1$ | $\frac{2.1}{1.06}=2$ |
| $\text{B}\text{r}$ | $85.1$ | $80$ | $\frac{85.1}{80}=1.06$ | $\frac{1.06}{1.06}=1$ |