Chapter 7, Problem 32E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $\text{85}\text{.5}\text{g}$ of beryllium nitrate is to be calculated.

Concept introduction:

A mole is a basic unit that is used in the International system of units (SI). It is abbreviated as $\text{mol}$. It can be defined as $6.022\times {10}^{23}$ units of some quantity which can be atoms, molecules or ions. The quantity $6.022\times {10}^{23}$ is called the Avogadro’s number.

Answer to Problem 32E

The number of formula units in $\text{85}\text{.5}\text{g}$ of beryllium nitrate is $\text{3}{\text{.87×10}}^{\text{23}}\text{formula units}$.

Explanation of Solution

The molecular formula for beryllium nitrate is $\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2$.

The molar mass of nitrogen is $14.01\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of beryllium is $\text{9}\text{.012}\text{g}{\text{mol}}^{-1}$.

The molar mass of $\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& \text{9}\text{.012}+\left(2\times 14.01\right)+\left(6\times 16.00\right)\\ & =& 133.032\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of $\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $133.032\text{g}{\text{mol}}^{-1}$.

Thus one mole of $\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $\text{133}\text{.032}\text{g}$.

The number of moles in $\text{85}\text{.5}\text{g}$ of $\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2$ is given below.

$\begin{array}{rll}\text{Moles}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2& =& \frac{\text{1}\text{mole}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2}{\text{133}\text{.032}\text{g}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2}\times \text{weight}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2}{\text{133}\text{.032}\text{g}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2}\times \text{85}\text{.5}\text{g}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2\\ & =& 0.643\text{moles}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2\end{array}$

The number of formula units in $\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2$ is calculated as,

$\text{Formula units}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2=& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2\times \\ \text{moles}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2\end{array}}{\text{1}\text{mole}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2}\right)$

Substitute the number of moles of $\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2$ in above equation,

$\begin{array}{rll}\text{Formula units}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2\times \\ \text{0}\text{.643}\text{moles}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2\end{array}}{\text{1}\text{mole}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2}\right)\\ & =& 3.87\times {\text{10}}^{\text{23}}\text{formula units}\text{of}\text{Be}{\left({\text{NO}}_{\text{3}}\right)}_2\end{array}$

Conclusion

The number of formula units in $\text{85}\text{.5}\text{g}$ of beryllium nitrate is $\text{3}\text{.87}\times {\text{10}}^{\text{23}}\text{formula units}$$3.87{\text{×10}}^{\text{23}}$.

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $9.42\text{g}$ manganese is to be calculated.

Concept introduction:

A mole is a basic unit that is used in the International system of units (SI). It is abbreviated as $\text{mol}$ It can be defined as $6.022\times {10}^{23}$ units of some quantity which can be atoms, molecules or ions. The quantity $6.022\times {10}^{23}$ is called the Avogadro’s number.

Answer to Problem 32E

The number of formula units in $9.42\text{g}$ manganese is $\text{1}{\text{.03×10}}^{\text{23}}\text{formula units}$.

Explanation of Solution

The molar mass of manganese atom is $\text{54}\text{.94}\text{g}{\text{mol}}^{-1}$.

Thus, one mole of manganese is $\text{54}\text{.94}\text{g}$.

The number of moles in $9.42\text{g}$ of manganese is given below.

$\begin{array}{rll}\text{Moles}\text{of}\text{manganese}& =& \frac{\text{1}\text{mole}\text{of}\text{manganese}}{\text{54}\text{.94}\text{g}\text{of}\text{manganese}}\times \text{weight}\text{of}\text{manganese}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}\text{manganese}}{\text{54}\text{.94}\text{g}\text{of}\text{manganese}}\times \text{9}\text{.42}\text{g}\text{of}\text{manganese}\\ & =& 0.171\text{moles}\text{of}\text{manganese}\end{array}$

The number of formula units in manganese is calculated below.

$\text{Formula units}\text{of}\text{manganese}=& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}\text{manganese}\times \\ \text{moles}\text{of}\text{manganese}\end{array}}{\text{1}\text{mole}\text{of}\text{manganese}}\right)$

Substitute the number of moles of manganese in the above equation.

$\begin{array}{rll}\text{Formula units}\text{of}\text{manganese}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{formula units}\text{of}\text{manganese}\times \\ \text{0}\text{.171}\text{moles}\text{of}\text{manganese}\end{array}}{\text{1}\text{mole}\text{of}\text{manganese}}\right)\\ & =& 1.03\times {\text{10}}^{\text{23}}\text{formula units}\text{of}\text{manganese}\end{array}$

Conclusion

The number of formula units in $9.42\text{g}$ manganese is $\text{1}\text{.03}\times {\text{10}}^{\text{23}}\text{formula units}$.

Interpretation Introduction

(c)

Interpretation:

The number of molecules in $\text{0}\text{.0948}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ is to be calculated.

Concept introduction:

A mole is a basic unit that is used in the International system of units (SI). It is abbreviated as $\text{mol}$. It can be defined as $6.022\times {10}^{23}$ units of some quantity which can be atoms, molecules or ions. The quantity $6.022\times {10}^{23}$ is called the Avogadro’s number.

Answer to Problem 32E

The number of molecules in $\text{0}\text{.0948}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ is $9.49\times {\text{10}}^{\text{20}}\text{molecules}$.

Explanation of Solution

The molecular formula is ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$.

The molar mass of carbon is $12.01\text{g}{\text{mol}}^{-1}$.

The molar mass of oxygen is $16.00\text{g}{\text{mol}}^{-1}$.

The molar mass of hydrogen is $\text{1}\text{.008}\text{g}{\text{mol}}^{-1}$.

The molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ is calculated below.

$\begin{array}{rll}\text{Total}\text{molar}\text{mass}& =& 3\times \text{12}\text{.01}+\left(8\times 1.008\right)+16.00\\ & =& 60.094\text{g}{\text{mol}}^{-1}\end{array}$

Therefore, the molar mass of ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ is $60.094$.

Thus one mole of ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ is $60.0094$.

The number of moles in $\text{0}\text{.0948}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ is given below.

$\begin{array}{rll}\text{Moles}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}& =& \frac{\text{1}\text{mole}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}}{60.094\text{g}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}}\times \text{weight}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\text{in}\text{g}\\ & =& \frac{\text{1}\text{mole}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}{\text{OH}}_{\text{3}}}{60.094\text{g}\text{g}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}}\times \text{0}\text{.0948}\text{g}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\\ & =& 0.001577\text{moles}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\end{array}$

The number of molecules in ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ is calculated below.

$\text{Molecules}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}=& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\times \\ \text{moles}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\end{array}}{\text{1}\text{mole}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}}\right)$

Substitute the number of moles of ${\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ in above equation.

$\begin{array}{rll}\text{Molecules}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}& =& \left(\frac{\begin{array}{l}\text{6}{\text{.022×10}}^{\text{23}}\text{molecules}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\times \\ \text{0}\text{.001577}\text{moles}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\end{array}}{\text{1}\text{mole}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}}\right)\\ & =& 9.49\times {\text{10}}^{\text{20}}\text{molecules}\text{of}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}\end{array}$

Conclusion

The number of molecules in $\text{0}\text{.0948}\text{g}{\text{C}}_{\text{3}}{\text{H}}_{\text{7}}\text{OH}$ is $9.49\times {\text{10}}^{\text{20}}\text{molecules}$.