Chapter 4, Problem 40QP

(a)

Interpretation Introduction

Interpretation:

The number of moles of the given substance is to be calculated.

Explanation of Solution

The given compound is ${\text{K}}_2{\text{SO}}_4$ . Number of moles is determined using the molar mass and mass of the substance given. Molar mass of ${\text{K}}_2{\text{SO}}_4$ is $174.26\text{ g/mol}$ . The number of moles is calculated as shown below.

$\begin{array}{c}\text{moles}=\frac{\text{mass}}{\text{molar mass}}\\ =72.2\cancel{{\text{g K}}_2{\text{SO}}_4}\times \frac{1{\text{ mol K}}_2{\text{SO}}_4}{174.26\cancel{{\text{g K}}_2{\text{SO}}_4}}\\ =0.414{\text{ mol K}}_2{\text{SO}}_4\end{array}$

Hence, the number of moles in ${\text{K}}_2{\text{SO}}_4$ is $0.414\text{ mol}$ .

(b)

Interpretation Introduction

Interpretation:

The number of moles of the given substance is to be calculated.

Explanation of Solution

The given compound is ${\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_4$ . Number of moles is determined using the molar mass and mass of the substance given. The molar mass of ${\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_4$ is $315.36\text{ g/mol}$ and the mass is given to be $1\text{ mg}\text{(}={10}^{-3}\text{ g)}$ . The number of moles is calculated as shown below.

$\begin{array}{c}\text{moles}=\frac{\text{mass}}{\text{molar mass}}\\ =160.0\cancel{{\text{mg C}}_{18}{\text{H}}_{21}{\text{NO}}_4}\times \frac{{10}^{-3}\cancel{{\text{g C}}_{18}{\text{H}}_{21}{\text{NO}}_4}}{1\cancel{{\text{mg C}}_{18}{\text{H}}_{21}{\text{NO}}_4}}\frac{1{\text{ mol C}}_{18}{\text{H}}_{21}{\text{NO}}_4}{315.36\cancel{{\text{g C}}_{18}{\text{H}}_{21}{\text{NO}}_4}}\\ =5.074\times {10}^{-4}{\text{ mol C}}_{18}{\text{H}}_{21}{\text{NO}}_4\end{array}$

Hence, the number of moles of ${\text{C}}_{18}{\text{H}}_{21}{\text{NO}}_4$ is $5.074\times {10}^{-4}\text{ mol}$ .

(c)

Interpretation Introduction

Interpretation:

The number of moles of the substance given is to be calculated.

Explanation of Solution

The given compound is ${\text{Fe}}_3{\text{O}}_4$ . The number of moles is determined using the molar mass and mass of the substance given. Molar mass of ${\text{Fe}}_3{\text{O}}_4$ is $\text{231}\text{.55 g/mol}$ and the mass is given to be $1\text{ kg}\text{(}={10}^3\text{ g)}$ . The number of moles is calculated as shown below.

$\begin{array}{c}\text{moles}=\frac{\text{mass}}{\text{molar mass}}\\ =2.82\cancel{{\text{kg Fe}}_3{\text{O}}_4}\times \frac{{10}^3\cancel{{\text{g Fe}}_3{\text{O}}_4}}{1\cancel{{\text{kg Fe}}_3{\text{O}}_4}}\times \frac{1{\text{ mol Fe}}_3{\text{O}}_4}{231.55\cancel{{\text{g Fe}}_3{\text{O}}_4}}\\ =12.2{\text{ mol Fe}}_3{\text{O}}_4\end{array}$

Hence, the number of moles of ${\text{Fe}}_3{\text{O}}_4$ is $12.2\text{ mol}$ .

(d)

Interpretation Introduction

Interpretation:

The number of moles of the substance given is to be calculated.

Explanation of Solution

The given compound is ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ . The number of moles is determined using the molar mass and mass of the substance given. Molar mass of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ is $\text{342}\text{.30 g/mol}$ and the mass given is to be $1\text{ μg}\text{(}={10}^{-6}\text{ g)}$ . The number of moles are calculated as shown below.

$\begin{array}{c}\text{moles}=\frac{\text{mass}}{\text{molar mass}}\\ =5.00\cancel{{\text{μg C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}\times \frac{{10}^{-6}\cancel{{\text{g C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}}{1\cancel{{\text{μg C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}}\times \frac{1{\text{ mol C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}{342.30\cancel{{\text{g C}}_{12}{\text{H}}_{22}{\text{O}}_{11}}}\\ =1.46\times {10}^{-8}{\text{ mol C}}_{12}{\text{H}}_{22}{\text{O}}_{11}\end{array}$

Hence, the number of moles of ${\text{C}}_{12}{\text{H}}_{22}{\text{O}}_{11}$ is $1.46\times {10}^{-8}\text{ mol}$ .