Question

A gaseous compound of $$85.7\% $$ by mass carbon and $$14.3\% $$ by mass hydrogen. Its density is $$2.28{\text{ }}glitr{e^{ - 1}}$$ at $$300K$$ and $$1.0atm$$ pressure. Determine the molecular formula of the compound.
$(A){C_2}{H_2}$
$(B){C_2}{H_4}$
$(C){C_4}{H_8}$
$(D){C_4}{H_{10}}$

Answer 1

Hint: We know that a gaseous compound will follow the gas equation in the form of pressure and density. To solve this question, we will first find out the molecular mass of the given compound from the equation of relation between pressure and density.

Complete answer:
We know that the given compound is gas and hence, it will follow the gas equation in the form of pressure, density and molecular mass.
We will use the following formula:
$P \times M = d \times R \times T$
We can write this formula with respect to Molecular mass:
$M = \dfrac{{d \times R \times T}}{P}\_\_\_\_\_(1)$
Where d is the density of the compound, M is the molecular mass of the compound, T is temperature, P is the pressure and R is the gas constant.
$$d = 2.28{\text{ }}glitr{e^{ - 1}}$$
$$T = 300K$$
$$P = 1.0atm$$
$$R = 0.0821L(atm)mo{l^{ - 1}}{K^{ - 1}}$$
Now, put the given values in formula:
$M = \dfrac{{2.28 \times 0.0821 \times 300}}{{1.0}}$
$M = 56.15gmo{l^{ - 1}}$
Hence, the molecular mass is found to be $56.15gmo{l^{ - 1}}.$
Now, we will find out the empirical formula by the ratio method:
$\dfrac{{85.7}}{{12}}:\dfrac{{14.3}}{1}$
Hence, the simplest ratio is found to be $1:2$ .
The empirical formula will be: ${(C{H_2})_n}$
Where, n is given as:
$n = \dfrac{{56.15}}{{14}}$
$n = 4$
Hence, the molecular formula of given compound will be ${C_4}{H_8}.$
Therefore, the correct option is $(C){C_4}{H_8}$ .

Note:
In these types of questions, we should remember the concept of empirical formula and ratio. We should also know the gas equation in the form of pressure, temperature, molecular mass and density. Empirical formula is the simplest formula (with simplest ratios of the elements in the compound) of a compound.