Chapter 10.4, Problem 60PP

Interpretation Introduction

Interpretation:

The empirical formula of propane needs to be determined if it contains $81.82\%$ carbon and $18.18\%$ hydrogen.

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.

Answer to Problem 60PP

The empirical formula is ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$.

Explanation of Solution

Given Information:

The compound has $81.82\%$ C atom and $18.18\%$ H atom.

Calculation:

Assume the total mass of compound is 100 g thus, mass of C and H atom will be 81.82 g and 18.18 g respectively.

The molar mass of C and H is 12 g/mol and 1 g/mol respectively. Thus, moles of C and H can be calculated as follows:

$n_C=\frac{81.82\text{ g}}{12\text{ g/mol}}=6.81\text{ mol}$

And,

$n_H=\frac{18.18\text{ g}}{1\text{ g/mol}}=18.18\text{ mol}$

To determine the simplest ratio, divide number of moles of C and H with smaller number of moles that is number of moles of C.

Thus,

$\begin{array}{l}C:H=\frac{6.81}{6.81}:\frac{18.18}{6.81}=1:2.67\\ =1:\frac{8}{3}\\ =3:8\end{array}$

Thus, empirical formula of compound is ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$.

Conclusion

The empirical formula of propane is ${\text{C}}_{\text{3}}{\text{H}}_{\text{8}}$.