Chapter F, Problem E.24E

(a)

Interpretation Introduction

Interpretation:

The amount of ${\text{CN}}^{-}$ in moles has to be calculated that is present in $\text{4}\text{.00}\text{g}$ of $\text{NaCN}$.

Answer to Problem E.24E

Moles of ${\text{CN}}^{-}$ present in $\text{4}\text{.00}\text{g}$ of $\text{NaCN}$ is $0.082\text{mol}$.

Explanation of Solution

The mass of sodium cyanide is given as $\text{4}\text{.00}\text{g}$.  The molar mass of $\text{NaCN}$ is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=\text{M(Na)}+\text{M(C)}+\text{M(N)}\\ =22.99\text{g}\cdot {\text{mol}}^{-1}+12.01\text{g}\cdot {\text{mol}}^{-1}+14.00\text{g}\cdot {\text{mol}}^{-1}\\ =49\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for sodium cyanide, $49\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of sodium cyanide present in $\text{4}\text{.00}\text{g}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}\text{NaCN}}{\text{Molar}\text{mass}\text{of}\text{NaCN}}\\ =\frac{4.00\text{g}}{49\text{g}\cdot {\text{mol}}^{-1}}\\ =0.082\text{mol}\end{array}$

One mole of sodium cyanide contains one mole of ${\text{CN}}^{-}$ ions.  Therefore, the moles of ${\text{CN}}^{-}$ ions in $0.082\text{mol}$ of sodium cyanide is calculated as shown below;

    $\begin{array}{c}\text{Moles}\text{of}{\text{CN}}^{-}=\frac{0.082\text{mol}\text{NaCN×1}\text{mol}{\text{CN}}^{-}}{\text{1}\text{mol}\text{NaCN}}\\ =0.082\text{mol}\end{array}$

Thus the moles of ${\text{CN}}^{-}$ present in $\text{4}\text{.00}\text{g}$ of sodium cyanide is $0.082\text{mol}$.

(b)

Interpretation Introduction

Interpretation:

The amount of $\text{O}$ atoms in moles has to be calculated that is present in $\text{4}\text{.00}\times {\text{10}}^{\text{2}}\text{ng}$ of ${\text{H}}_{\text{2}}\text{O}$.

Answer to Problem E.24E

Moles of $\text{O}$ present in $\text{4}\text{.00}\times {\text{10}}^{\text{2}}\text{ng}$ of ${\text{H}}_{\text{2}}\text{O}$ is $0.222\times {10}^{-7}\text{mol}$.

Explanation of Solution

The mass of ${\text{H}}_{\text{2}}\text{O}$ is given as $\text{4}\text{.00}\times {\text{10}}^{\text{2}}\text{ng}$.  The molar mass of ${\text{H}}_{\text{2}}\text{O}$ is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=2\times \text{M(H)}+1\times \text{M(O)}\\ =2(1.008\text{g}\cdot {\text{mol}}^{-1})+1(16.00\text{g}\cdot {\text{mol}}^{-1})\\ =2.016\text{g}\cdot {\text{mol}}^{-1}+16.00\text{g}\cdot {\text{mol}}^{-1}\\ =18.016\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for ${\text{H}}_{\text{2}}\text{O}$, $18.016\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of ${\text{H}}_{\text{2}}\text{O}$ present in $\text{4}\text{.00}\times {\text{10}}^{\text{2}}\text{ng}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}{\text{H}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\text{H}}_{\text{2}}\text{O}}\\ =\frac{\text{4}\text{.00}\times {\text{10}}^{\text{-7}}\text{g}}{18.016\text{g}\cdot {\text{mol}}^{-1}}\\ =0.222\times {10}^{-7}\text{mol}\end{array}$

Thus the moles of ${\text{H}}_{\text{2}}\text{O}$ present in $\text{4}\text{.00}\times {\text{10}}^{\text{2}}\text{ng}$ of ${\text{H}}_{\text{2}}\text{O}$ is $0.222\times {10}^{-7}\text{mol}$.

One mole of ${\text{H}}_{\text{2}}\text{O}$ contains one mole of $\text{O}$ atom.  Therefore, the number of moles of $\text{O}$ present in $\text{4}\text{.00}\times {\text{10}}^{\text{2}}\text{ng}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}\text{O}=1\times 0.222\times {10}^{-7}\text{mol}\\ =0.222\times {10}^{-7}\text{mol}\end{array}$

Therefore, the number of moles of $\text{O}$ atoms present in $\text{4}\text{.00}\times {\text{10}}^{\text{2}}\text{ng}$ of ${\text{H}}_{\text{2}}\text{O}$ is $0.222\times {10}^{-7}\text{mol}$.

(c)

Interpretation Introduction

Interpretation:

The amount of ${\text{CaSO}}_{\text{4}}$ molecules in moles has to be calculated that is present in $\text{4}\text{.00}\text{kg}$ of ${\text{CaSO}}_{\text{4}}$.

Answer to Problem E.24E

Moles of ${\text{CaSO}}_{\text{4}}$ present in $\text{4}\text{.00}\text{kg}$ of ${\text{CaSO}}_{\text{4}}$ is $29.4\text{mol}$.

Explanation of Solution

The mass of ${\text{CaSO}}_{\text{4}}$ is given as $\text{4}\text{.00}\text{kg}$.  The molar mass of ${\text{CaSO}}_{\text{4}}$ is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=\text{M(Ca)}+\text{M(S)}+4\times \text{M(O)}\\ =40.08\text{g}\cdot {\text{mol}}^{-1}+32.06\text{g}\cdot {\text{mol}}^{-1}+3(16.00\text{g}\cdot {\text{mol}}^{-1})\\ =136.14\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for ${\text{CaSO}}_{\text{4}}$, $136.14\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of ${\text{CaSO}}_{\text{4}}$ present in $\text{4}\text{.00}\text{kg}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}{\text{CaSO}}_{\text{4}}}{\text{Molar}\text{mass}\text{of}{\text{CaSO}}_{\text{4}}}\\ =\frac{\text{4}\text{.00}\times {\text{10}}^{\text{3}}\text{g}}{136.14\text{g}\cdot {\text{mol}}^{-1}}\\ =29.4\text{mol}\end{array}$

Thus the moles of ${\text{CaSO}}_{\text{4}}$ present in $\text{4}\text{.00}\text{kg}$ of ${\text{CaSO}}_{\text{4}}$ is $29.4\text{mol}$.

(d)

Interpretation Introduction

Interpretation:

The amount of ${\text{H}}_{\text{2}}\text{O}$ in moles has to be calculated that is present in $\text{4}\text{.00}\text{mg}$ of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$.

Answer to Problem E.24E

Moles of ${\text{H}}_{\text{2}}\text{O}$ present in $\text{4}\text{.00}\text{mg}$ of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$ is $\text{0}\text{.0656}\times {10}^{-3}\text{mol}$.

Explanation of Solution

The mass of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$ is given as $\text{4}\text{.00}\text{mg}$.  The molar mass of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$ is calculated as follows;

$\begin{array}{c}\text{Molar}\text{mass}=2\times \text{M(Al)}+3\times \text{M(S)}+12\times \text{M(O)}+8\times {\text{M(H}}_{\text{2}}\text{O)}\\ =2\times 26.98\text{g}\cdot {\text{mol}}^{-1}+3(32.06\text{g}\cdot {\text{mol}}^{-1})+12(16.00\text{g}\cdot {\text{mol}}^{-1})+8(18.00\text{g}\cdot {\text{mol}}^{-1})\\ =(53.96+96.18+192.00+144.00)\text{g}\cdot {\text{mol}}^{-1}\\ =486.14\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$, $486.14\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$ present in $\text{4}\text{.00}\text{mg}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}}{\text{Molar}\text{mass}\text{of}{\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}}\\ =\frac{4.00\times {10}^{-3}\text{g}}{486.14\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0082\times {10}^{-3}\text{mol}\end{array}$

Thus the moles of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$ present in $\text{4}\text{.00}\text{mg}$ of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$ is $0.0082\times {10}^{-3}\text{mol}$.

One mole of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$ contains eight moles of ${\text{H}}_{\text{2}}\text{O}$ molecules.  Therefore, the number of moles of ${\text{H}}_{\text{2}}\text{O}$ molecule present in $\text{4}\text{.00}\text{mg}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{H}}_{\text{2}}\text{O}=8\times 0.0082\times {10}^{-3}\text{mol}\\ \text{=0}\text{.0656}\times {10}^{-3}\text{mol}\end{array}$

Therefore, the number of moles of ${\text{H}}_{\text{2}}\text{O}$ molecules present in $\text{4}\text{.00}\text{mg}$ of ${\text{Al}}_{\text{2}}{{\text{(SO}}_{\text{4}}\text{)}}_{\text{3}}\cdot {\text{8H}}_{\text{2}}\text{O}$ is $\text{0}\text{.0656}\times {10}^{-3}\text{mol}$.