Chapter 10.4, Problem 66PP

Interpretation Introduction

Interpretation:

The empirical formula of morphine 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.

Answer to Problem 66PP

The empirical formula of morphine is ${\text{C}}_{17}{\text{H}}_{19}{\text{O}}_3\text{N}$.

Explanation of Solution

Given information:

The mass of C, H, O and N element is 17.90 g, 1.68 g, 4.225 g and 1.228 g respectively.

Calculation: The number of moles of each element can be calculated as follows:

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

The molar mass of C, H, O and N element is 12 g/mol, 1 g/mol, 16 g/mol and 14 g/mol respectively.

The number of moles will be:

$\begin{array}{l}{n}_C=\frac{17.90\text{ g}}{12.0\text{ g/mol}}=1.5\text{ mol}\\ n_H=\frac{\text{1}\text{.68 g}}{1\text{ g/mol}}=1.68\text{ mol}\\ n_O=\frac{4.225\text{ g}}{16\text{ g/mol}}=0.26\text{ mol}\\ n_N=\frac{1.228\text{ g}}{14\text{ g/mol}}=0.088\text{ mol}\end{array}$

The ratio of number of moles of each element will be:

$\begin{array}{l}C:H:O:N=1.5:1.68:0.26:0.088\\ =\frac{1.5}{0.088}:\frac{1.68}{0.088}:\frac{0.26}{0.088}:\frac{0.088}{0.088}\\ =17:19:3:1\end{array}$

Thus, the empirical formula is ${\text{C}}_{17}{\text{H}}_{19}{\text{O}}_3\text{N}$.

Conclusion

Therefore, the empirical formula of morphine is ${\text{C}}_{17}{\text{H}}_{19}{\text{O}}_3\text{N}$.