Chapter 10.4, Problem 62PP

Interpretation Introduction

Interpretation:

The molecular formula of the compound needs to be determined.

Concept introduction:

The empirical formula tells about the simplest ratio of atoms of present in the molecule. It does not give the actual number of atoms present in the molecule. The molecular formula shows the actual number of atoms present in the molecule.

Answer to Problem 62PP

The molecular formula of the compound is ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$.

Explanation of Solution

Given information:

The mass of C and H in compound is $49.98$ g and $10.47$ g respectively. The molar mass of the compound is $58.12$ g/mol.

Calculation :

From given mass of C and H, the number of moles of C and H can be calculated as follows:

$n=\frac{m}{M}$

Molar mass of C and H is 12 g/mol and 1 g/mol respectively. Thus,

$n_C=\frac{49.98\text{ g}}{12\text{ g/mol}}=4.165\text{ mol}$

And,

$n_H=\frac{10.47\text{ g}}{1\text{ g/mol}}=10.4\text{7 mol}$

The ratio of number of C and H atom will be:

$\begin{array}{l}C:H=4.165:10.47\\ =1:2.5\\ =1:\frac{5}{2}\\ =2:5\end{array}$

Thus, the empirical formula of compound is ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}$.

The empirical mass can be calculated as follows:

$\text{E}\text{. M}=2\times 12+5\times 1=29\text{ g/mol}$

The molecular mass is given 58.12 g/mol thus, ratio of molecular mass to empirical mass is:

$x=\frac{58.12}{29}\sim 2$

Thus, molecular formula will be:

$\text{M}\text{. F}=\text{2}\times \text{E}\text{. F}={\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$

Therefore, molecular formula of compound is ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$.

Conclusion

The molecular formula of the compound is ${\text{C}}_{\text{4}}{\text{H}}_{\text{10}}$.