Question

A carbon compound contains $12.8\% $ of carbon; $21\% $ of hydrogen and $85.1\% $ of bromine. The molecular mass of the compound is $187.9$ . Calculate the molecular formula of the compound (${\text{Atomic mass H = 1}}{\text{.008,C = 12,Br = 79}}{\text{.9}}$ ).
A. ${\text{C}}{{\text{H}}_3}{\text{Br}}$
B. ${\text{C}}{{\text{H}}_2}{\text{B}}{{\text{r}}_2}$
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

Hint: From the given percentage of elements in the carbon compound we will calculate the number of moles and then empirical formula. Afterwards we will divide the molar mass of the compound with the molar mass of the empirical formula and we will get the factor to multiply with the empirical formula.

Complete step by step answer: In the problem it is provided that a carbon compound contains $12.8\% $ of carbon; $2.1\% $ of hydrogen and $85.1\% $ of bromine. It indicates that the compound has chemical formula ${{\text{(}}{{\text{C}}_x}{{\text{H}}_y}{\text{B}}{{\text{r}}_z})_n}$ where ${\text{x,y,z and n }}$ are integers and ${{\text{C}}_x}{{\text{H}}_y}{\text{B}}{{\text{r}}_z}$is the empirical formula of the compound.
So to calculate the values of ${\text{x,y and z}}$ we will use the percentage of elements in the compound.
We will find out the empirical formula using the percentage of elements.
The percentage of carbon in the compound is $12.8\% $.
$ \Rightarrow $ Moles of carbon= $\dfrac{{12.8}}{{12}} = 1.067 \approx 1$
$ \Rightarrow x = 1$
The percentage of hydrogen in the compound is $2.1\% $.
$ \Rightarrow $ Moles of hydrogen= $\dfrac{{2.1}}{{1.008}} = 2.08 \approx 2$
$ \Rightarrow y = 2$
The percentage of bromine in the compound is $85.1\% $.
$ \Rightarrow $ Moles of bromine=$\dfrac{{85.1}}{{79.9}} = 1.065 \approx 1$
$ \Rightarrow z = 1$
From these moles we can write down our empirical formula i.e. ${{\text{C}}_1}{{\text{H}}_2}{\text{B}}{{\text{r}}_1}$.
And the molecular formula is ${{\text{(}}{{\text{C}}_x}{{\text{H}}_y}{\text{B}}{{\text{r}}_z})_n}$ so we have to calculate the value of $n$ now.
We will divide the molar mass of the compound with the molar mass of the empirical formula of compound to get the value of $n$.
Empirical mass=$93.9$ and Molecular mass=$187.9$
$ \Rightarrow n = \dfrac{{187.9}}{{93.9}} = 2$
So the value of $n$ is 2.
Hence our molecular formula is ${{\text{(}}{{\text{C}}_1}{{\text{H}}_2}{\text{B}}{{\text{r}}_1})_2}$ which can also be represented as ${{\text{C}}_2}{{\text{H}}_4}{\text{B}}{{\text{r}}_2}$.

So, the correct answer is “Option C”.

Note: To solve such types of questions empirical formula plays an important role. To calculate empirical formula we will convert mass of each element to moles and then divide it by the smallest mole number among all of the elements and round off the number and then represent it as ${{\text{C}}_x}{{\text{H}}_y}{\text{B}}{{\text{r}}_z}$ where ${\text{x,y and z}}$are the calculated moles of elements.