Chapter 3, Problem 3.15P

(a)

Interpretation Introduction

Interpretation: The mass in grams in $\text{8}{\text{.42 mol Cu}}_2{\text{CO}}_3$ is to be calculated.

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

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)$

Answer to Problem 3.15P

Mass in grams in $\text{8}{\text{.42 mol Cu}}_2{\text{CO}}_3$ is $1562.4\text{ g}$ .

Explanation of Solution

Given Information:

Number of moles of ${\text{Cu}}_2{\text{CO}}_3$ is $\text{8}\text{.35 mol}$ .

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

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)$

Given number of moles of ${\text{Cu}}_2{\text{CO}}_3$ is $\text{8}\text{.35 mol}$ .

Molar mass of ${\text{Cu}}_2{\text{CO}}_3$ is $187.1\text{ g/mol}$ .

Substitute the values in the above formula.

  $\begin{array}{c}\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)\\ =\left(\text{8}\text{.35 mol}\right)\left(187.1\text{ g/mol}\right)\\ =1562.4\text{ g}\end{array}$

(b)

Interpretation Introduction

Interpretation: The mass in grams present in $4.04\times {\text{10}}^{20}$ molecules of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ is to be calculated.

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

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)$

The formula to convert number of atoms to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given number of molecules}}{6.022\times {10}^{23}}$

Answer to Problem 3.15P

Mass in $4.04\times {\text{10}}^{20}$ molecules of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ is $\text{0}\text{.0725 g}$ .

Explanation of Solution

The formula to convert number of atoms to moles is as follows:

  $\text{Number of moles}=\frac{\text{Given number of molecules}}{6.022\times {10}^{23}}$

Number molecules of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ is $4.04\times {\text{10}}^{20}$ .

Substitute the value in the above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{Given number of molecules}}{6.022\times {10}^{23}}\\ =\frac{4.04\times {\text{10}}^{20}}{6.022\times {10}^{23}}\\ =6.708\times {10}^{-4}\text{ mol}\end{array}$

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

  $\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)$

Molar mass of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ is $108.01\text{ g/mol}$ .

Number of moles of ${\text{N}}_{\text{2}}{\text{O}}_{\text{5}}$ is $6.708\times {10}^{-4}\text{ mol}$ .

Substitute the value in above equation.

  $\begin{array}{c}\text{Mass}=\left(\text{Number of moles}\right)\left(\text{molar mass}\right)\\ =\left(6.708\times {10}^{-4}\text{ mol}\right)\left(108.01\text{ g/mol}\right)\\ =0.07246\text{ g}\\ \approx \text{0}\text{.0725 g}\end{array}$

(c)

Interpretation Introduction

Interpretation: The number of moles and formula units in $78.9{\text{ g NaClO}}_4$ is to be calculated.

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

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

Answer to Problem 3.15P

The moles and formula units present in $78.9{\text{ g NaClO}}_4$ are $0.64439\text{ mol}$ and $3.8805\times {10}^{23}$ respectively.

Explanation of Solution

Given Information:

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

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

Given mass of ${\text{NaClO}}_4$ is $78.9\text{ g}$ .

Molar mass of ${\text{NaClO}}_4$ s $\text{122}\text{.44 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}\\ =\frac{78.9\text{ g}}{\text{122}\text{.44 g/mol}}\\ =0.64439\text{ mol}\end{array}$

Since $6.022\times {10}^{23}$ formula units are found in ${\text{1 mol NaClO}}_4$ so number of formula units present in $0.64439{\text{ mol NaClO}}_4$ is calculated as follows:

  $\begin{array}{c}\text{Formula units}=\left(\frac{6.022\times {10}^{23}\text{ units}}{1\text{ mol}}\right)\left(0.64439{\text{ mol NaClO}}_4\right)\\ =3.8805\times {10}^{23}{\text{ formula units NaClO}}_4\end{array}$

(d)

Interpretation Introduction

Interpretation: The number of ${\text{Na}}^{+}$ ions, perchlorate ions, $\text{Cl}$ atoms, and $\text{O}$ atoms present in $78.9{\text{ g NaClO}}_4$ is to be calculated.

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

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

Answer to Problem 3.15P

Number of sodium ions, perchlorate ions, chlorine atoms are $3.88\times {10}^{23}$ and oxygen atoms are $1.55\times {10}^{24}$ in the mass of compound ${\text{NaClO}}_{\text{4}}$ .

Explanation of Solution

Since one mole ${\text{NaClO}}_4$ has one ${\text{Na}}^{+}$ so molar mass of ${\text{Na}}^{+}$ present is $22.9898\text{ g/mol}$ .

Similarly molar mass of ${\text{ClO}}_4^{-}$ , $\text{Cl}$ and $\text{O}$ are $\text{99}\text{.45 g/mol}$ , $\text{35}\text{.4527 g/mol}$ and $\text{15}\text{.9994 g/mol}$ respectively.

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

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

Given mass of ${\text{NaClO}}_4$ is $78.9\text{ g}$ .

Molar mass of ${\text{NaClO}}_4$ s $\text{122}\text{.44 g/mol}$ .

Substitute the values in above formula.

  $\begin{array}{c}\text{Number of moles}=\frac{\text{Given mass}}{\text{molar mass}}\\ =\frac{78.9\text{ g}}{\text{122}\text{.44 g/mol}}\\ =0.64439\text{ mol}\end{array}$

The number of ions/atoms of ${\text{Na}}^{+}$ in ${\text{NaClO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}{\text{Number of Na}}^{\text{+}}\text{ ions}=\left(\text{3}\text{.88056}\times {\text{10}}^{\text{23}}\text{ FU}\right)\left(\frac{1{\text{ Na}}^{\text{+}}\text{ ion}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)\\ =3.88\times {10}^{23}{\text{ Na}}^{+}\text{ ions}\end{array}$

The number of ions/atoms of ${\text{Na}}^{+}$ in ${\text{NaClO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}{\text{Number of ClO}}_{\text{4}}{}^{-}\text{ ions}=\left(\text{3}\text{.88056}\times {\text{10}}^{\text{23}}\text{ FU}\right)\left(\frac{1{\text{ ClO}}_{\text{4}}{}^{-}\text{ ion}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)\\ =3.88\times {10}^{23}{\text{ ClO}}_{\text{4}}{}^{-}\text{ ions}\end{array}$

The number of ions/atoms of $\text{Cl}$ in ${\text{NaClO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}\text{Number of Cl atoms}=\left(\text{3}\text{.88056}\times {\text{10}}^{\text{23}}\text{ FU}\right)\left(\frac{1\text{ Cl atom}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)\\ =3.88\times {10}^{23}\text{ Cl atoms}\end{array}$

The number of ions/atoms of $\text{O}$ in ${\text{NaClO}}_{\text{4}}$ is calculated as follows:

  $\begin{array}{c}\text{Number of O atoms}=\left(\text{3}\text{.88056}\times {\text{10}}^{\text{23}}\text{ FU}\right)\left(\frac{\text{ 4 O atom}}{\text{1FU}\text{of}{\text{NaClO}}_{\text{4}}}\right)\\ =1.55\times {10}^{24}\text{ O atoms}\end{array}$

Number of sodium ions, perchlorate ions, chlorine atoms are $3.88\times {10}^{23}$ and oxygen atoms are $1.55\times {10}^{24}$ in the mass of compound ${\text{NaClO}}_{\text{4}}$ .