Chapter 3, Problem 3.14P

(a)

Interpretation Introduction

Interpretation:

The mass of $0.68\text{ mol}$ of ${\text{KMnO}}_4$ is to be calculated.

Concept introduction:

The molar mass is defined as the mass of one mol of a substance in grams. The molar mass of a compound is calculated by adding the molar mass of each element multiplied by its number of atoms present in the chemical formula.

The mass of a compound in grams can be calculated from the given moles and the molar mass of that compound.

The expression to calculate the mass of a chemical substance in grams is as follows:

$\text{Mass }\left(\text{g}\right)=\left(\text{amount}\left(\text{mol}\right)\right)\left(\frac{\text{no}\text{. of grams}}{1\text{ mol}}\right)$

Answer to Problem 3.14P

The mass of $0.68\text{ mol}$ of ${\text{KMnO}}_4$ is $1.1\times {10}^2\text{ g}$.

Explanation of Solution

The expression to calculate the molar mass of ${\text{KMnO}}_4$ is as follows:

${\text{Molar mass of KMnO}}_4=\left(1\right)\left(\text{M of K}\right)+\left(1\right)\left(\text{M of Mn}\right)+\left(4\right)\left(\text{M of O}\right)$

Substitute $39.10\text{ g/mol}$ for $\text{M}$ of $\text{K}$, $54.94\text{ g/mol}$ for $\text{M}$ of $\text{Mn}$, and $16.00\text{ g/mol}$ for $\text{M of O}$ in the above equation as follows:

$\begin{array}{c}{\text{Molar mass of KMnO}}_4=\left(1\right)\left(39.10\text{ g/mol}\right)+\left(1\right)\left(54.94\text{ g/mol}\right)+\left(4\right)\left(16.00\text{ g/mol}\right)\\ =94.04\text{ g/mol}+\text{64 g/mol}\\ =\text{158}\text{.04 g/mol}\end{array}$

The molar mass of ${\text{KMnO}}_4$ is $\text{158}\text{.04 g/mol}$.

Calculate the mass of $0.68\text{ mol}$ of ${\text{KMnO}}_4$ as follows:

$\begin{array}{c}{\text{Mass ofKMnO}}_4\left(\text{g}\right)=\left(0.68{\text{ mol KMnO}}_4\right)\left(\frac{\text{158}{\text{.04 g KMnO}}_4}{1{\text{ mol KMnO}}_4}\right)\\ =1.1\times {10}^2{\text{ g KMnO}}_4\end{array}$

Conclusion

Therefore, the mass of $0.68\text{ mol}$ of ${\text{KMnO}}_4$ is $1.1\times {10}^2\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

The moles of oxygen atoms in $8.18\text{ g}$ of $\text{Ba}{\left({\text{NO}}_3\right)}_2$ is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in $12\text{ g}$ of $\text{C-12}$. This number is called Avogadro’s number. The value of Avogadro’s number is $\text{6}\text{.022}\times {\text{10}}^{23}\left(\text{entitites/mol}\right)$.

The amount of an element can be calculated from the given amount of compound, the moles of that element present in one mole of the compound and the Avogadro’ number. The expression to calculate the moles of atoms of an element in a compound is as follows:

$\text{Amount of element}\left(\text{mol}\right)=\left(\begin{array}{l}\text{amount of}\\ \text{compound}\\ \left(\text{mol}\right)\end{array}\right)\left(\frac{\begin{array}{l}\text{moles of element in}\\ \text{molecular formula}\end{array}}{1\text{ mol compound}}\right)$

Answer to Problem 3.14P

In $8.18\text{ g}$ of $\text{Ba}{\left({\text{NO}}_3\right)}_2$, $0.188\text{ mol O}$ atoms are present.

Explanation of Solution

The expression to calculate the molar mass of $\text{Ba}{\left({\text{NO}}_3\right)}_2$ is as follows:

$\text{Molar mass of Ba}{\left({\text{NO}}_3\right)}_2=\left(1\right)\left(\text{M of Ba}\right)+\left(2\right)\left(\text{M of N}\right)+\left(6\right)\left(\text{M of O}\right)$

Substitute $137.3\text{ g/mol}$ for $\text{M}$ of $\text{Ba}$, $14.01\text{ g/mol}$ for $\text{M}$ of $\text{N}$, and $16.00\text{ g/mol}$ for $\text{M of O}$ in the above equation as follows:

$\begin{array}{c}\text{Molar mass of Ba}{\left({\text{NO}}_3\right)}_2=\left(1\right)\left(137.3\text{ g/mol}\right)+\left(2\right)\left(14.01\text{ g/mol}\right)+\left(6\right)\left(16.00\text{ g/mol}\right)\\ =165.32\text{ g/mol}+9\text{6 g/mol}\\ =\text{261}\text{.3 g/mol}\end{array}$

The molar mass of $\text{Ba}{\left({\text{NO}}_3\right)}_2$ is $\text{261}\text{.3 g/mol}$.

Calculate the amount of $8.18\text{ g}$ of $\text{Ba}{\left({\text{NO}}_3\right)}_2$ as follows:

$\begin{array}{c}\text{Amount of Ba}{\left({\text{NO}}_3\right)}_2\left(\text{mol}\right)=\left(8.18\text{ g Ba}{\left({\text{NO}}_3\right)}_2\right)\left(\frac{1\text{ mol Ba}{\left({\text{NO}}_3\right)}_2}{\text{261}\text{.3 g Ba}{\left({\text{NO}}_3\right)}_2}\right)\\ =0.031305\text{ mol Ba}{\left({\text{NO}}_3\right)}_2\end{array}$

Therefore, $0.031305\text{ mol}$ of $\text{Ba}{\left({\text{NO}}_3\right)}_2$ are present in $8.18\text{ g}$ of $\text{Ba}{\left({\text{NO}}_3\right)}_2$.

One mole of $\text{Ba}{\left({\text{NO}}_3\right)}_2$ contains $\text{6 mol}$ of oxygen atoms. Calculate the moles of $\text{O}$ atoms in $0.031305\text{ mol}$ of $\text{Ba}{\left({\text{NO}}_3\right)}_2$ as follows:

$\begin{array}{c}\text{Moles of O atoms}=\left(0.031305\text{ mol Ba}{\left({\text{NO}}_3\right)}_2\right)\left(\frac{\text{6 mol O atoms}}{\text{1 mol Ba}{\left({\text{NO}}_3\right)}_2}\right)\\ =0.18783\text{ mol O atoms}\\ \approx 0.188\text{ mol O atoms}\end{array}$

Conclusion

In $8.18\text{ g}$ of $\text{Ba}{\left({\text{NO}}_3\right)}_2$, $0.188\text{ mol O}$ atoms are present.

(c)

Interpretation Introduction

Interpretation:

The number of oxygen atoms in $7.3\times {10}^{-3}\text{ g}$ of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

One mole is defined as the amount of substance that contains the same number of entities such as molecules, ions, atoms as the number of atoms in $12\text{ g}$ of $\text{C-12}$. This number is called Avogadro’s number. The value of Avogadro’s number is $\text{6}\text{.022}\times {\text{10}}^{23}\left(\text{entitites/mol}\right)$.

The number of an atom of an element can be calculated from the given amount of compound, the moles of that element present in one mole of the compound and the Avogadro’ number.

The expression to calculate the amount of a chemical substance in moles is as follows:

$\text{Amount}\left(\text{mol}\right)=\left(\text{Mass }\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{no}\text{. of grams}}\right)$

The expression to calculate the amount of atoms of an element in a compound is as follows:

$\text{Amount of element}\left(\text{mol}\right)=\left(\begin{array}{l}\text{amount of}\\ \text{compound}\\ \left(\text{mol}\right)\end{array}\right)\left(\frac{\begin{array}{l}\text{moles of element in}\\ \text{molecular formula}\end{array}}{1\text{ mol compound}}\right)$

The expression to calculate the number of atoms of an element is as follows:

$\text{Number of atoms}=\left(\begin{array}{l}\text{amount}\\ \left(\text{mol}\right)\end{array}\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}}{1\text{ mol}}\right)$

Answer to Problem 3.14P

The number of oxygen atoms in $7.3\times {10}^{-3}\text{ g}$ of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is $\text{1}\text{.5}\times {\text{10}}^{20}\text{ O atoms}$.

Explanation of Solution

The expression to calculate the molar mass of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is as follows:

$\begin{array}{c}{\text{Molar mass of CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}=\left(1\right)\left(\text{M of Ca}\right)+\left(1\right)\left(\text{M of S}\right)+\left(4\right)\left(\text{M of H}\right)\\ +\left(6\right)\left(\text{M of O}\right)\end{array}$

Substitute $40.08\text{ g/mol}$ for $\text{M of Mg}$, $1.008\text{ g/mol}$ for $\text{M of H}$, $16.00\text{ g/mol}$ for $\text{M of O}$, and $32.06\text{ g/mol}$ for $\text{M of S}$ in the above equation as follows:

$\begin{array}{c}{\text{Molar mass of CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}=\left[\begin{array}{c}\left(1\right)\left(40.08\text{ g/mol}\right)+\left(1\right)\left(32.06\text{ g/mol}\right)+\left(4\right)\left(1.008\text{ g/mol}\right)\\ +\left(6\right)\left(16.00\text{ g/mol}\right)\end{array}\right]\\ =40.08\text{ g/mol}+3\text{2}\text{.06 g/mol}+4.032\text{ g/mol}+96.00\text{ g/mol}\\ =172.17\text{ g/mol}\end{array}$

Therefore, the molar mass of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is $172.17\text{ g/mol}$.

Calculate the amount of $7.3\times {10}^{-3}\text{ g}$ of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ as follows:

$\begin{array}{c}{\text{Amount of CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}\left(\text{mol}\right)=\left(7.3\times {10}^{-3}\text{ g}\right)\left(\frac{1{\text{ mol CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}}{172.17{\text{ g CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}}\right)\\ =4.239995\times {10}^{-5}{\text{ mol CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}\end{array}$

Therefore, $4.239995\times {10}^{-5}\text{ mol}$ of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ are present in $7.3\times {10}^{-3}\text{ g}$ of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$.

One mole of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ contains $\text{6 mol}$ of oxygen atoms. Calculate the moles of $\text{O}$ atoms in $4.239995\times {10}^{-5}\text{ mol}$ of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ as follows:

$\begin{array}{c}\text{Moles of O atoms}=\left(4.239995\times {10}^{-5}{\text{ mol CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}\right)\left(\frac{\text{6 mol O atoms}}{{\text{1 mol CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}}\right)\\ =2.543997\times {10}^{-5}\text{ mol O atoms}\end{array}$

Therefore, $2.543997\times {10}^{-5}\text{ mol}$ of oxygen atoms are present in $4.239995\times {10}^{-3}\text{ mol}$ of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$.

Calculate the number of oxygen atoms in $2.543997\times {10}^{-5}\text{ mol}$ of oxygen atoms as follows:

$\begin{array}{c}\text{Number of O atoms}=\left(2.543997\times {10}^{-5}\text{ mol O atoms}\right)\left(\frac{\text{6}\text{.022}\times {\text{10}}^{23}\text{ O atoms}}{1\text{ mol O atom}}\right)\\ =\text{1}\text{.5320}\times {\text{10}}^{20}\text{ O atoms}\\ \approx \text{1}\text{.5}\times {\text{10}}^{20}\text{ O atoms}\end{array}$

Conclusion

In $7.3\times {10}^{-3}\text{ g}$ of ${\text{CaSO}}_{\text{4}}\cdot {\text{2H}}_{\text{2}}\text{O}$ , $\text{1}\text{.5}\times {\text{10}}^{20}\text{ O atoms}$ are present.