Chapter 10, Problem 6STP

Interpretation Introduction

Interpretation:

The empirical formula for the given compound should be predicted.

Concept introduction:

The representation of atoms of a compound in simple whole number ratio is known as empirical formula.

The rules for determining empirical formula is given as follows:

• Determine the mass of elements in given compound.
• Calculate number of moles using molar mass of each compound or element.
• Divide each number of moles by smallest number of mole value calculated in second step.
• Round the value calculated in step 3 to nearest whole number.

Answer to Problem 6STP

The empirical formula for given compound is ${\text{CH}}_{\text{3}}{\text{NO}}_2$.

Explanation of Solution

Assuming the percentage of the elements is given mass by mass which means specific amount of the element is present in 100 g of the solution.

1. Mass of carbon = 19.68 g

Mass of hydrogen = 4.96 g

Mass of oxygen = 52.42 g

Mass of nitrogen = 22.95 g

2. Now, calculating the moles of carbon, hydrogen, nitrogen and oxygen using their molar mass as follows:

$\begin{array}{c}\text{Moles}\text{of}\text{carbon}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}\\ =\frac{\text{19}\text{.68}\text{g}}{\text{12}\text{.0107}\text{g/mol}}\\ {\text{n}}_{\text{C}}=1.64\text{mol}\\ \text{Moles}\text{of}\text{hydrogen}=\frac{\text{4}\text{.96}\text{g}}{\text{1}\text{.008}\text{g/mol}}\\ {\text{n}}_{\text{H}}=4.92\text{mol}\\ \text{Moles}\text{of}\text{oxygen}=\frac{\text{52}\text{.42}\text{g}}{\text{15}\text{.999}\text{g/mol}}\\ {\text{n}}_{\text{O}}=3.28\text{mol}\\ \text{Moles}\text{of}\text{nitrogen}=\frac{\text{22}\text{.95}\text{g}}{14.0067\text{g/mol}}\\ {\text{n}}_{\text{O}}=1.64\text{mol}\end{array}$

3. Now, dividing each mole value by smallest mole value as follows:

$\begin{array}{c}{\text{n}}_{\text{c}}=\frac{1.64}{1.64}\\ =1\\ {\text{n}}_{\text{H}}=\frac{4.92}{1.64}\\ =3\\ {\text{n}}_{\text{O}}=\frac{3.28}{1.64}\\ =2\\ {\text{n}}_{\text{N}}=\frac{1.64}{1.64}\\ =1\end{array}$

Thus, the empirical formula of given compound is ${\text{CH}}_{\text{3}}{\text{NO}}_2$ .

Hence, the correct option is (C) that is ${\text{CH}}_{\text{3}}{\text{NO}}_2$.