Chapter 4, Problem 73QP

What are the empirical formulas of the compounds with the following compositions?

$\begin{array}{l}\left(\text{a}\right)72.36\%\text{ Fe, }27.64\%\text{ O }\\ \left(\text{b}\right)58.53\%\text{ C, }4.09\%\text{ H, }11.38\%\text{ N, }25.99\%\text{ O}\end{array}$

Interpretation Introduction

Interpretation:

The empirical formula is to be determined.

Explanation of Solution

The simplest positive integer ratio of the number of constituent atoms of a compound is known as its empirical formula. In order to determine the empirical formula, first find the number of moles of an element present in the compound followed by the mole ratio. Let the total mass of the compound substance be $100\text{ g}$ .

Here, the percent composition of sample is given to be $72.36\text{ \% Fe}$ and $27.64\text{ \% O}$ . The total mass of the substance is $100\text{ g}$ . Therefore, it consists of $72.36\text{ g Fe}$ and $27.64\text{ g O}$ . The number of moles of each element is calculated as shown below.

$n_{\text{Fe}}=\frac{m}{\text{MM}}$ …… (1)

Here, $n_{\text{Fe}}$ is the number of moles of Fe, $m$ is the given mass and $\text{MM}$ is the molar mass.

$\begin{array}{c}{n}_{\text{Fe}}=72.36\cancel{\text{g Fe}}\times \frac{1\text{ mol Fe}}{55.85\cancel{\text{g Fe}}}\\ =1.296\text{ mol Fe}\end{array}$

The number of moles of oxygen is calculated as shown below.:

$\begin{array}{c}{n}_{\text{O}}=27.64\cancel{\text{g O}}\times \frac{1\text{ mol O}}{16.00\cancel{\text{g O}}}\\ =1.728\text{ mol O}\end{array}$

To determine the mole ratio, divide the number of moles of each element with the smallest amount in moles, which is $1.296\text{ mol Fe}$ .

$\begin{array}{c}\frac{n_{\text{Fe}}}{n_{\text{Fe}}}=\frac{1.296\text{ moles of Fe}}{\text{1}\text{.296 moles of Fe}}\\ =\frac{1\text{ mole of Fe}}{1\text{ mole of Fe}}\end{array}$

$\begin{array}{c}\frac{n_{\text{O}}}{n_{\text{Fe}}}=\frac{1.728\text{ mol of O}}{\text{1}\text{.296 mol of Fe}}\\ =\frac{1.333\text{ mol of O}}{1\text{ mol of Fe}}\end{array}$

The obtained ratio is not a whole number. In order to convert the ratio into a whole number each ratio is multiplied by $3$ . By multiplying $1.333\text{ mol of O}$ and $1.296\text{ mol of Fe}$ with $3$ , $4$ moles of O and $3\text{ mol of Fe}$ are obtained respectively.

Therefore, the empirical formula becomes ${\text{Fe}}_3{\text{O}}_4$ .

Interpretation Introduction

Interpretation:

The empirical formula is to analyzed.

Explanation of Solution

In order to determine the empirical formula, first calculate the number of moles of an element present in a compound and then find its mole ratio. Let the total amount of compound substance be $100\text{ g}$ .

The percent composition of the sample is given to be $58.53\text{ \% C, 4}\text{.09 \% H, 11}\text{.38 \% N}$ and $25.99\text{ \% O}$ . The total mass of the compound substance is $100\text{ g}$ . It consists $58.53\text{ g C, 4}\text{.09 g H, 11}\text{.38 g N}$ and $25.99\text{ g O}$ . The number of moles is to be calculated using equation $1$ as shown below.

$\begin{array}{c}{n}_{\text{C}}=58.53\cancel{\text{g C}}\times \frac{1\text{ mol C}}{12.01\cancel{\text{g C}}}\\ =4.873\text{ mol C}\end{array}$

The number of moles of hydrogen is calculated as follows:

$\begin{array}{c}{n}_{\text{H}}=4.09\cancel{\text{g H}}\times \frac{1\text{ mol H}}{1.008\cancel{\text{g H}}}\\ =4.06\text{ mol H}\end{array}$

The number of moles of nitrogen is calculated as follows:

$\begin{array}{c}{n}_{\text{N}}=11.38\cancel{\text{g N}}\times \frac{1\text{ mol N}}{14.01\cancel{\text{g N}}}\\ =0.8123\text{ mol N}\end{array}$

The number of moles of oxygen is calculated as follows:

$\begin{array}{c}{n}_{\text{O}}=25.99\cancel{\text{g O}}\times \frac{1\text{ mol O}}{16.00\cancel{\text{g O}}}\\ =1.624\text{ mol O}\end{array}$

To determine the mole ratio, divide the number of moles of each element with the smallest amount in moles which is $0.8123\text{ mol N}$ .

$\begin{array}{c}\frac{n_{\text{C}}}{n_{\text{N}}}=\frac{4.873\text{ moles of C}}{\text{0}\text{.8123 moles of N}}\\ =\frac{6.000\text{ mole of C}}{1\text{ mole of N}}\end{array}$

$\begin{array}{c}\frac{n_{\text{H}}}{n_{\text{N}}}=\frac{4.06\text{ moles of H}}{\text{0}\text{.8123 moles of N}}\\ =\frac{5.00\text{ mole of H}}{1\text{ mole of N}}\end{array}$

$\begin{array}{c}\frac{n_{\text{N}}}{n_{\text{N}}}=\frac{\text{0}\text{.8123 moles of N}}{\text{0}\text{.8123 moles of N}}\\ =\frac{1\text{ mole of N}}{1\text{ mole of N}}\end{array}$

$\begin{array}{c}\frac{n_{\text{O}}}{n_{\text{N}}}=\frac{\text{1}\text{.624 moles of O}}{\text{0}\text{.8123 moles of N}}\\ =\frac{2.00\text{ mole of O}}{1\text{ mole of N}}\end{array}$

From the mole ratios obtained , it is clear that for every $1$ mole of nitrogen- 6 moles of carbon, 5 moles of hydrogen and $2$ moles of oxygen are present.

Therefore, the empirical formula becomes ${\text{C}}_6{\text{H}}_5{\text{NO}}_2$ .