Chapter F, Problem E.25E

(a)

Interpretation Introduction

Interpretation:

The number of formula units present in $\text{0}\text{.750}\text{mol}$ of ${\text{KNO}}_{\text{3}}$ has to be determined.

Answer to Problem E.25E

Number of formula units is $4.52\times {10}^{23}$.

Explanation of Solution

The amount of ${\text{KNO}}_{\text{3}}$ in moles is given as $\text{0}\text{.750}\text{mol}$.  One mole of a compound contains $\text{6}{\text{.0221×10}}^{\text{23}}\text{formula}\text{units}$.  Therefore, the number of formula units present in $\text{0}\text{.750}\text{mol}$ of ${\text{KNO}}_{\text{3}}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{formula}\text{units}=\text{0}\text{.750}\text{mol}\times \frac{\text{6}{\text{.0221×10}}^{\text{23}}\text{formula}\text{units}}{\text{1}\text{mol}}\\ =4.52\times {10}^{23}\text{formula}\text{units}\end{array}$

Therefore, the number of formula units present in $\text{0}\text{.750}\text{mol}$ of ${\text{KNO}}_{\text{3}}$ is $4.52\times {10}^{23}$.

(b)

Interpretation Introduction

Interpretation:

The mass of $\text{2}{\text{.39×10}}^{\text{20}}\text{formula}\text{units}$ of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ has to be determined.

Answer to Problem E.25E

Mass of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{124}\text{mg}$.

Explanation of Solution

Number of formula units of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is given as $2.39\times {10}^{20}$.  The molar mass of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated as shown below;

    $\begin{array}{c}\text{Molar}\text{mass}=2\times \text{M(Ag)}+1\times \text{M}(\text{S})+4\times \text{M(O)}\\ =2(107.87\text{g}\cdot {\text{mol}}^{-1})+1(32.06\text{g}\cdot {\text{mol}}^{-1})+4(16.00\text{g}\cdot {\text{mol}}^{-1})\\ =215.74\text{g}\cdot {\text{mol}}^{-1}+32.06\text{g}\cdot {\text{mol}}^{-1}+64.00\text{g}\cdot {\text{mol}}^{-1}\\ =311.8\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$, $311.8\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ present in $2.39\times {10}^{20}$ formula units can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{2}{\text{.39×10}}^{\text{20}}\text{formula}\text{units×1}\text{mol}}{\text{6}{\text{.0221×10}}^{\text{23}}\text{formula}\text{units}}\\ =0.3967\times {10}^{-3}\text{mol}\\ =3.967\times {10}^{-4}\text{mol}\end{array}$

Therefore, the mass of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is calculated as shown below;

    $\begin{array}{c}\text{Mass}\text{of}{\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}=\text{Amount}\text{in}\text{moles}\times \text{Molar}\text{mass}\\ =3.967\times {10}^{-4}\text{mol}\times \text{311}\text{.8}\text{g}\cdot {\text{mol}}^{-1}\\ =1236\times {10}^{-4}\text{g}\\ =124\text{mg}\end{array}$

Thus the mass of ${\text{Ag}}_{\text{2}}{\text{SO}}_{\text{4}}$ is $\text{124}\text{mg}$.

(c)

Interpretation Introduction

Interpretation:

Number of formula units of ${\text{NaHCO}}_{\text{2}}$ has to be estimated that is present in $\text{3}\text{.429}\text{g}$ of ${\text{NaHCO}}_{\text{2}}$.

Answer to Problem E.25E

Formula units of ${\text{NaHCO}}_{\text{2}}$ is $\text{3}{\text{.01×10}}^{\text{22}}$.

Explanation of Solution

The mass of ${\text{NaHCO}}_{\text{2}}$ is given as $\text{3}\text{.429}\text{g}$.  The molar mass of ${\text{NaHCO}}_{\text{2}}$ is calculated as follows;

    $\begin{array}{c}\text{Molar}\text{mass}=\text{M(Na)}+\text{M(H)}+\text{M(C)}+3\times \text{M(O)}\\ =22.99\text{g}\cdot {\text{mol}}^{-1}+1.008\text{g}\cdot {\text{mol}}^{-1}+12.01\text{g}\cdot {\text{mol}}^{-1}+2(16.00\text{g}\cdot {\text{mol}}^{-1})\\ =68.00\text{g}\cdot {\text{mol}}^{-1}\end{array}$

Therefore, for ${\text{NaHCO}}_{\text{2}}$, $68.00\text{g}\cdot {\text{mol}}^{-1}$ is the molar mass.  Number of moles of ${\text{NaHCO}}_{\text{2}}$ present in $\text{3}\text{.429}\text{g}$ can be calculated as follows;

    $\begin{array}{c}\text{Number}\text{of}\text{moles}=\frac{\text{Mass}\text{of}{\text{NaHCO}}_{\text{2}}}{\text{Molar}\text{mass}\text{of}{\text{NaHCO}}_{\text{2}}}\\ =\frac{3.429\text{g}}{68.00\text{g}\cdot {\text{mol}}^{-1}}\\ =0.0504\text{mol}\end{array}$

Thus the moles of ${\text{NaHCO}}_{\text{2}}$ present in $\text{3}\text{.429}\text{g}$ of ${\text{NaHCO}}_{\text{2}}$ is $0.0504\text{mol}$.

One mole of any compound contains $\text{6}{\text{.0221×10}}^{\text{23}}\text{formula}\text{units}$.  Therefore, the number of formula units in $\text{0}\text{.0504}\text{mol}$ of ${\text{NaHCO}}_{\text{2}}$ is calculated as shown below;

    $\begin{array}{c}\text{Number}\text{of}\text{formula}\text{units}=\text{0}\text{.0504}\text{mol}\times \frac{\text{6}{\text{.0221×10}}^{\text{23}}\text{formula}\text{units}}{\text{1}\text{mol}}\\ =\text{0}{\text{.3035×10}}^{\text{23}}\text{formula}\text{units}\\ =\text{3}{\text{.035×10}}^{\text{22}}\text{formula}\text{units}\end{array}$

Thus the total number of formula units in $\text{3}\text{.429}\text{g}$ of sodium formate is $\text{3}{\text{.035×10}}^{\text{22}}$.