Chapter F, Problem E.29E

(a)

Interpretation Introduction

Interpretation:

Number of moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present in $\text{8}\text{.61}\text{g}$ has to be calculated.

Answer to Problem E.29E

Number of moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is $0.0417\text{mol}$.

Explanation of Solution

The molar mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}\text{Molar}\text{mass}=1\times \text{M}(\text{Cu})+2\times M(Cl)+4\times \text{M}({\text{H}}_{\text{2}}\text{O})\\ =1(63.55\text{g}\cdot {\text{mol}}^{-1})+2(35.45\text{g}\cdot {\text{mol}}^{-1})+4(18.016\text{g}\cdot {\text{mol}}^{-1})\\ =63.55\text{g}\cdot {\text{mol}}^{-1}+70.90\text{g}\cdot {\text{mol}}^{-1}+72.064\text{g}\cdot {\text{mol}}^{-1}\\ =206.5\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is given as $\text{8}\text{.61}\text{g}$.  The number of moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O})}=\frac{8.61\text{g}}{206.5\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0417\text{mol}\end{array}$

Thus the moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present in $\text{8}\text{.61}\text{g}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is $0.0417\text{mol}$.

(b)

Interpretation Introduction

Interpretation:

Number of moles of ${\text{Cl}}^{-}$ present in $\text{8}\text{.61}\text{g}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ has to be calculated.

Answer to Problem E.29E

Number of moles of ${\text{Cl}}^{-}$ is $0.0834\text{mol}$.

Explanation of Solution

The molar mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}\text{Molar}\text{mass}=1\times \text{M}(\text{Cu})+2\times M(Cl)+4\times \text{M}({\text{H}}_{\text{2}}\text{O})\\ =1(63.55\text{g}\cdot {\text{mol}}^{-1})+2(35.45\text{g}\cdot {\text{mol}}^{-1})+4(18.016\text{g}\cdot {\text{mol}}^{-1})\\ =63.55\text{g}\cdot {\text{mol}}^{-1}+70.90\text{g}\cdot {\text{mol}}^{-1}+72.064\text{g}\cdot {\text{mol}}^{-1}\\ =206.5\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is given as $\text{8}\text{.61}\text{g}$.  The number of moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O})}=\frac{8.61\text{g}}{206.5\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0417\text{mol}\end{array}$

Thus the moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present in $\text{8}\text{.61}\text{g}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is $0.0417\text{mol}$.

One mole of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ contains two moles of ${\text{Cl}}^{-}$ ion.  Therefore, the amount of ${\text{Cl}}^{-}$ ion present in $\text{0}\text{.0417}\text{mol}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{Cl}}^{-})}=0.0417\text{mol}{\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}\times \frac{\text{2}\text{mol}{\text{Cl}}^{-}}{1\text{mol}{\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}}\\ =0.0834\text{mol}\end{array}$

Therefore, the moles of ${\text{Cl}}^{-}$ ions present is $0.0834\text{mol}$.

(c)

Interpretation Introduction

Interpretation:

Number of water molecules present in $\text{8}\text{.61}\text{g}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ has to be calculated.

Answer to Problem E.29E

Number of water molecules is $1.00\times {10}^{23}$.

Explanation of Solution

The molar mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}\text{Molar}\text{mass}=1\times \text{M}(\text{Cu})+2\times M(Cl)+4\times \text{M}({\text{H}}_{\text{2}}\text{O})\\ =1(63.55\text{g}\cdot {\text{mol}}^{-1})+2(35.45\text{g}\cdot {\text{mol}}^{-1})+4(18.016\text{g}\cdot {\text{mol}}^{-1})\\ =63.55\text{g}\cdot {\text{mol}}^{-1}+70.90\text{g}\cdot {\text{mol}}^{-1}+72.064\text{g}\cdot {\text{mol}}^{-1}\\ =206.5\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is given as $\text{8}\text{.61}\text{g}$.  The number of moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O})}=\frac{8.61\text{g}}{206.5\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0417\text{mol}\end{array}$

Thus the moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present in $\text{8}\text{.61}\text{g}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is $0.0417\text{mol}$.

One mole of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ contains four moles of water molecule.  Therefore, the number of moles of water present in $\text{0}\text{.0417}\text{mol}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is calculated as follows;

    $\begin{array}{c}{n}_{({\text{H}}_{\text{2}}\text{O})}=0.0417\text{mol}{\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}\times \frac{4\text{mol}\text{of}\text{H2O}}{1\text{mol}\text{of}{\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}}\\ =0.1668\text{mol}\end{array}$

One mole of any substance contains $\text{6}{\text{.0221×10}}^{\text{23}}\text{molecules}$.  Therefore, the amount of water molecule present in $\text{0}\text{.1668}\text{mol}$ of ${\text{H}}_{\text{2}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{water}\text{molecule}=0.1668\text{mol}{\text{H}}_{\text{2}}\text{O}\times \frac{6.0221\times {10}^{23}\text{molecules}}{1\text{mol}{\text{H}}_{\text{2}}\text{O}}\\ =1.00\times {10}^{23}\text{molecules}\end{array}$

Therefore, the number of water molecules present is $1.00\times {10}^{23}$.

(d)

Interpretation Introduction

Interpretation:

Fraction of total mass of the sample that is due to oxygen has to be calculated.

Answer to Problem E.29E

Fraction of total mass that is due to oxygen is $0.310$.

Explanation of Solution

The molar mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is calculated as shown below;

    $\begin{array}{c}\text{Molar}\text{mass}=1\times \text{M}(\text{Cu})+2\times M(Cl)+4\times \text{M}({\text{H}}_{\text{2}}\text{O})\\ =1(63.55\text{g}\cdot {\text{mol}}^{-1})+2(35.45\text{g}\cdot {\text{mol}}^{-1})+4(18.016\text{g}\cdot {\text{mol}}^{-1})\\ =63.55\text{g}\cdot {\text{mol}}^{-1}+70.90\text{g}\cdot {\text{mol}}^{-1}+72.064\text{g}\cdot {\text{mol}}^{-1}\\ =206.5\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is given as $\text{8}\text{.61}\text{g}$.  The number of moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present is calculated as shown below;

    $\begin{array}{c}{n}_{({\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O})}=\frac{8.61\text{g}}{206.5\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0417\text{mol}\end{array}$

Thus the moles of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ present in $\text{8}\text{.61}\text{g}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is $0.0417\text{mol}$.

One mole of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ contains four moles of oxygen atom.  Therefore, the number of moles of oxygen atom present in $\text{0}\text{.0417}\text{mol}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is calculated as follows;

    $\begin{array}{c}{n}_{(\text{O})}=0.0417\text{mol}{\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}\times \frac{4\text{mol}\text{of}\text{O}}{1\text{mol}\text{of}{\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}}\\ =0.1668\text{mol}\end{array}$

The mass of oxygen atom in $\text{0}\text{.1668}\text{mol}$ is calculated as follows;

    $\begin{array}{c}\text{Mass}\text{of}\text{oxygen}=\text{Amount}\text{in}\text{moles×Molar}\text{mass}\\ =0.1668\text{mol}\times 16.00\text{g}\cdot {\text{mol}}^{-1}\\ =2.6688\text{g}\\ \text{=2}\text{.67}\text{g}\end{array}$

The mass of oxygen atoms in $\text{8}\text{.61}\text{g}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}$ is $\text{2}\text{.67}\text{g}$.  Therefore, the fraction of mass that is due to oxygen can be calculated as follows;

    $\begin{array}{c}\text{Fraction}=\frac{\text{Mass}\text{of}\text{O}}{\text{Mass}\text{of}{\text{CuCl}}_{\text{2}}\cdot {\text{4H}}_{\text{2}}\text{O}}\\ =\frac{2.67\text{g}}{8.16\text{g}}\\ =0.310\end{array}$

Therefore, the fraction of mass that is due to oxygen atom is $0.310$.