Chapter 3, Problem 3.93P

(a)

Interpretation Introduction

Interpretation:Moles in $\text{0}\text{.588 g}$ of ${\text{NH}}_4\text{Br}$ is to be calculated.

Concept introduction: The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.93P

Moles in $\text{0}\text{.588 g}$ of ${\text{NH}}_4\text{Br}$ are $0.006\text{ mol}$ .

Explanation of Solution

The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

For ${\text{NH}}_4\text{Br}$

Given mass is $\text{0}\text{.588 g}$ .

Molar mass is $\text{97}\text{.94 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}\\ =\frac{\text{0}\text{.588 g}}{\text{97}\text{.94 g/mol}}\\ =0.006\text{ mol}\end{array}$

(b)

Interpretation Introduction

Interpretation:Number of potassium ions in $\text{88}{\text{.5 g KNO}}_3$ is to be calculated.

Concept introduction: The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.93P

Number of potassium ions in $\text{88}{\text{.5 g KNO}}_3$ is $5.27\times {10}^{23}$ .

Explanation of Solution

The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

For ${\text{KNO}}_3$

Given mass is $\text{88}\text{.5 g}$ .

Molar mass is $\text{101}\text{.1032 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}\\ =\frac{\text{88}\text{.5 g}}{\text{101}\text{.1032 g/mol}}\\ =0.8753\text{ mol}\end{array}$

The formula to convert moles to molecules is as follows:

  $\text{Number of ions}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)$

For ${\text{KNO}}_3$

Total moles is $0.8753\text{ mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of ions}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)\\ =\left(0.8753\text{ mol}\right)\left(6.022\times {10}^{23}\right)\\ =5.27\times {10}^{23}\end{array}$

(c)

Interpretation Introduction

Interpretation: Mass in $\text{5}\text{.85 mol}$ of ${\text{C}}_3{\text{H}}_8{\text{O}}_3$ is to be calculated.

Concept introduction: The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.93P

Mass in $\text{5}\text{.85 mol}$ of ${\text{C}}_3{\text{H}}_8{\text{O}}_3$ is $538.72\text{ g}$ .

Explanation of Solution

The formula to convert moles to mass in grams is as follows:

  ${\text{Mass of C}}_3{\text{H}}_8{\text{O}}_3=\left({\text{Moles of C}}_3{\text{H}}_8{\text{O}}_3\right)\left({\text{molar mass of C}}_3{\text{H}}_8{\text{O}}_3\right)$

Moles of ${\text{C}}_3{\text{H}}_8{\text{O}}_3$ is $\text{5}\text{.85 mol}$ .

Molar mass of ${\text{C}}_3{\text{H}}_8{\text{O}}_3$ is $\text{92}\text{.09 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{Mass of C}}_3{\text{H}}_8{\text{O}}_3=\left({\text{Moles of C}}_3{\text{H}}_8{\text{O}}_3\right)\left({\text{molar mass of C}}_3{\text{H}}_8{\text{O}}_3\right)\\ =\left(\text{5}\text{.85 mol}\right)\left(\text{92}\text{.09 g/mol}\right)\\ =538.72\text{ g}\end{array}$

(d)

Interpretation Introduction

Interpretation:Volume of $\text{2}\text{.85 mol}$ of ${\text{CH}}_3\text{Cl}$ is to be calculated.

Concept introduction: The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.93P

Volume of $\text{2}\text{.85 mol}$ of ${\text{CH}}_3\text{Cl}$ is $97.227\text{ mL}$ .

Explanation of Solution

The formula to convert moles to mass in grams is as follows:

  ${\text{Mass of CH}}_3\text{Cl}=\left({\text{Moles of CH}}_3\text{Cl}\right)\left({\text{molar mass of CH}}_3\text{Cl}\right)$

Moles of ${\text{CH}}_3\text{Cl}$ is $\text{2}\text{.85 mol}$ .

Molar mass of ${\text{CH}}_3\text{Cl}$ is $\text{119}\text{.37 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}{\text{Mass of CH}}_3\text{Cl}=\left({\text{Moles of CH}}_3\text{Cl}\right)\left({\text{molar mass of CH}}_3\text{Cl}\right)\\ =\left(\text{2}\text{.85 mol}\right)\left(\text{119}\text{.37 g/mol}\right)\\ =340.2045\text{ g}\end{array}$

The formula to calculate volume from density is as follows:

  $\text{Volume}=\frac{\text{Mass}}{\text{Density}}$

For ${\text{CH}}_3\text{Cl}$

Mass is $340.2045\text{ g}$ .

Density is $1.48\text{ g/mL}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Volume}=\frac{\text{Mass}}{\text{Density}}\\ =\frac{340.2045\text{ g}}{1.48\text{ g/mL}}\\ =230\text{ mL}\end{array}$

(e)

Interpretation Introduction

Interpretation:Number of sodium ions in $\text{2}{\text{.11 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is to be calculated.

Concept introduction: The formula to convert moles to mass in grams is as follows:

  $\text{Mass}=\left(\text{Moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.93P

Number of sodium ions in $\text{2}{\text{.11 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $2.54\times {10}^{24}$ .

Explanation of Solution

Since ${\text{1 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ furnishes ${\text{2 mol Na}}^{+}$ ions, therefore, moles of ${\text{Na}}^{+}$ ions produced from $\text{2}{\text{.11 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}$ are calculated as follows:

  $\begin{array}{c}{\text{Moles of Na}}^{+}=\left(\text{2}{\text{.11 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}\right)\left(\frac{{\text{2 mol Na}}^{+}}{{\text{1 mol Na}}_{\text{2}}{\text{CO}}_{\text{3}}}\right)\\ =4.22{\text{ mol Na}}^{+}\end{array}$

The formula to convert moles to molecules is as follows:

  $\text{Number of ions}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)$

For ${\text{Na}}^{+}$

Total moles is $4.22\text{ mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of ions}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)\\ =\left(4.22\text{ mol}\right)\left(6.022\times {10}^{23}\right)\\ =2.54\times {10}^{24}\end{array}$

(f)

Interpretation Introduction

Interpretation:Number of atoms in $\text{25}\text{.0 μg Cd}$ is to be calculated.

Concept introduction:The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

The formula to convert moles to atoms is as follows:

  $\text{Number of atoms}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)$

Answer to Problem 3.93P

Number of atoms in $\text{25}\text{.0 μg Cd}$ is $1.336\times {10}^{17}$ .

Explanation of Solution

The conversion factor to convert $\text{kg}$ is $\text{g}$ as follows:

  $1\text{ kg}=1000\text{ g}$

Hence convert $\text{25}\text{.0 μg}$ to $\text{g}$ as follows:

  $\begin{array}{c}\text{Mass}=\left(\text{25}\text{.0 μg}\right)\left(\frac{{10}^{-6}\text{ g}}{1\text{ μg}}\right)\\ =25.0\times {10}^{-6}\text{ g}\end{array}$

The formula to convert mass in grams to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}$

For $\text{Cd}$

Given mass is $25.0\times {10}^{-6}\text{ g}$ .

Molar mass is $\text{112}\text{.41 g/mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}\\ =\frac{25.0\times {10}^{-6}\text{ g}}{\text{112}\text{.41 g/mol}}\\ =2.22\times {10}^{-7}\text{ mol}\end{array}$

The formula to convert moles to atoms is as follows:

  $\text{Number of atoms}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)$

For $\text{Cd}$

Total moles is $2.22\times {10}^{-7}\text{ mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of atoms}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)\\ =\left(2.22\times {10}^{-7}\text{ mol}\right)\left(6.022\times {10}^{23}\right)\\ =1.336\times {10}^{17}\end{array}$

(g)

Interpretation Introduction

Interpretation:Number of atoms in $\text{0}{\text{.0015 mol F}}_2$ is to be calculated.

Concept introduction: The formula to convert moles to atoms is as follows:

  $\text{Number of atoms}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\right)$

Answer to Problem 3.93P

Number of atoms in $\text{0}{\text{.0015 mol F}}_2$ is $1.8\times {10}^{21}\text{ atoms}$ .

Explanation of Solution

Chemical formula of fluorine gas is ${\text{F}}_{\text{2}}$ . $\text{1 mol}$ fluorine gas has $\text{2 mol}$

  $\text{F}$ atoms.Therefore, the number of fluorine atoms in $\text{0}\text{.0015 mol}$ of fluorine gas is calculated as,

  $\begin{array}{c}\text{Moles of F atoms}=\left(\text{0}\text{.0015 mol}\right)\left(\frac{2\text{ mol}}{1\text{ mol}}\right)\\ =3\times {10}^{-3}\text{ mol}\end{array}$

The formula to convert moles to atoms is as follows:

  $\text{Number of atoms}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\text{ atoms/mol}\right)$

Total moles is $3\times {10}^{-3}\text{ mol}$ .

Substitute the value in above formula.

  $\begin{array}{c}\text{Number of atoms}=\left(\text{Total moles}\right)\left(6.022\times {10}^{23}\text{ atoms/mol}\right)\\ =\left(3\times {10}^{-3}\text{ mol}\right)\left(6.022\times {10}^{23}\text{ atoms/mol}\right)\\ =1.8066\times {10}^{21}\text{ atoms}\\ \approx 1.8\times {10}^{21}\text{ atoms}\end{array}$