Chapter 10, Problem 194A

Interpretation Introduction

Interpretation:

Percentage composition, empirical formula and molecular formula of the compound must be found for which there are 6 g C and 1 g H with molar mass of 42 g/mol.

Concept introduction:

Empirical formula of a molecule is the smallest unit which is repeated in the molecule.

Molecular formula ${\text{=(empirical formula)}}_{\text{n}}$

Where n is the number of times the empirical formula is present in the molecular formula, n can be 1, 2, 3 etc.

Answer to Problem 194A

There is 85.71 % C and 14.29 % H.

The empirical formula is CH₂ and molecular formula is C₃H₆.

Explanation of Solution

% of C and H are 85.71 and 14.29 respectively.

Considering 100 g of compound, mass of C and H is 85.71 g and 14.29 g respectively.

Also, molar mass of C and H is 12 g/mol and 1 g/mol respectively. Thus, number of moles of C and H will be:

Moles of C $\text{=}\frac{\text{85}\text{.71 g}}{\text{12 g/mol}}\text{=7}\text{.14 mol}$

Moles of H $=\frac{14.29\text{ g}}{1\text{ g/mol}}=14.29\text{ mol}$

Smallest mole ratio C: H

$\begin{array}{l}=\frac{7.14}{7.14}:\frac{14.29}{7.14}\\ =1:0:2.001\approx 1:2.\end{array}$

Thus empirical formula is CH₂.

Now,

$\begin{array}{l}{{\text{(CH}}_{\text{2}}}_{\text{)}}\text{=42}\\ \text{Or, (12+2)×n=42}\\ \text{Or, n=}\frac{\text{42}}{\text{14}}\text{=3}\text{.}\end{array}$

Conclusion

There is 85.71 % C and 14.29 % H. Empirical formula is CH₂ and molecular formula is C₃H₆.