Chapter 10.2, Problem 19SP

Interpretation Introduction

Interpretation: The number of moles for $75.0\text{ g}$ of dinitrogen trioxide should be calculated.

Concept Introduction: The mass in grams of one mole of the compound is said to be its molar mass.

Answer to Problem 19SP

The number of moles for $75.0\text{ g}$ of dinitrogen trioxide is $\text{0}\text{.99 mol}$ .

Explanation of Solution

To determine number of moles for $75.0\text{ g}$ of dinitrogen trioxide, the molar mass of dinitrogen trioxide is calculated first. The molecular formula of dinitrogen trioxide is ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$ . The molar mass is calculated by adding up the sum of the masses of N and O elements.

The moles of N and O are converted to grams of N and O as:

  $\begin{array}{l}\text{2 mol N }\times \frac{\text{14}\text{.0 g N}}{1\text{ mol N}}\text{ = 28}\text{.0 g N}\\ \text{3 mol O }\times \frac{\text{16}\text{.0 g O}}{1\text{ mol O}}\text{ = 48}\text{.0 g O}\end{array}$

The mass of 1 mol of ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$ is:

  $\begin{array}{l}{\text{mass of 1 mol of N}}_{\text{2}}{\text{O}}_{\text{3}}\text{ = 28}\text{.0 g N + 48}\text{.0 g O}=\text{ 76}\text{.0 g}\\ {\text{molar mass of N}}_{\text{2}}{\text{O}}_{\text{3}}=76.\text{0 g/mol}\end{array}$

Thus, the molar mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{3}}$ is $\text{76}\text{.0 g/mol}$ .

Now, number of moles for $75.0\text{ g}$ of dinitrogen trioxide is:

  $\text{75}\text{.0 g × }\frac{\text{1 mol}}{\text{76}\text{.0 g}}\text{ = 0}\text{.99 mol}$

Hence, the number of moles for $75.0\text{ g}$ of dinitrogen trioxide is $\text{0}\text{.99 mol}$ .