Chapter 3, Problem 3.16P

(a)

Interpretation Introduction

Interpretation:

The mass of $6.44\times {10}^{-2}\text{ mol}$ of ${\text{MnSO}}_4$ in grams is to be calculated.

Concept introduction:

Molar mass is defined as the mass of $1\text{ mol}$ of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

Molar mass is used to convert the mass of a substance to moles. 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 3.16P

The mass of $6.44\times {10}^{-2}\text{ mol}$ of ${\text{MnSO}}_4$ in grams is $9.72\text{ g}$.

Explanation of Solution

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

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

Substitute $54.94\text{ g/mol}$ for $\text{M}$ of $\text{Mn}$, $32.06\text{ g/mol}$ for $\text{M}$ of $\text{S}$, and $16.00\text{ g/mol}$ for $\text{M of O}$ in the above equation as follows:

$\begin{array}{c}{\text{Molar mass of MnSO}}_4=\left(1\right)\left(54.94\text{ g/mol}\right)+\left(1\right)\left(32.06\text{ g/mol}\right)+\left(4\right)\left(16.00\text{ g/mol}\right)\\ =54.94\text{ g/mol}+32.06\text{ g/mol}+64.00\text{ g/mol}\\ =\text{151}\text{.00 g/mol}\end{array}$

Therefore, the molar mass of ${\text{MnSO}}_4$ is $\text{151}\text{.00 g/mol}$.

Calculate the mass of $6.44\times {10}^{-2}\text{ mol}$ of ${\text{MnSO}}_4$ as follows:

$\begin{array}{c}{\text{Mass of MnSO}}_4\left(\text{g}\right)=\left(6.44\times {10}^{-2}{\text{ mol MnSO}}_4\right)\left(\frac{\text{ 151}{\text{.00 g MnSO}}_4}{1{\text{ mol MnSO}}_4}\right)\\ =9.7244{\text{ g MnSO}}_4\\ \approx 9.72{\text{ g MnSO}}_4\end{array}$

Conclusion

The mass of $6.44\times {10}^{-2}\text{ mol}$ of ${\text{MnSO}}_4$ in grams is $9.72\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

The amount of $15.8\text{ kg}$ of $\text{Fe}{\left({\text{ClO}}_4\right)}_3$ in moles is to be calculated.

Concept introduction:

Molar mass is defined as the mass of $1\text{ mol}$ of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

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

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

Answer to Problem 3.16P

The amount of $15.8\text{ kg}$ of $\text{Fe}{\left({\text{ClO}}_4\right)}_3$ is $4.6\text{ mol}$.

Explanation of Solution

The expression to calculate the molar mass of $\text{Fe}{\left({\text{ClO}}_4\right)}_3$ is as follows:

$\text{Molar mass of Fe}{\left({\text{ClO}}_4\right)}_3=\left(1\right)\left(\text{M of Fe}\right)+\left(3\right)\left(\text{M of Cl}\right)+\left(12\right)\left(\text{M of O}\right)$

Substitute $55.85\text{ g/mol}$ for $\text{M}$ of $\text{Fe}$, $35.45\text{ g/mol}$ for $\text{M}$ of $\text{Cl}$, and $16.00\text{ g/mol}$ for $\text{M of O}$ in the above equation as follows:

$\begin{array}{c}\text{Molar mass of Fe}{\left({\text{ClO}}_4\right)}_3=\left(1\right)\left(55.85\text{ g/mol}\right)+\left(3\right)\left(35.45\text{ g/mol}\right)+\left(12\right)\left(16.00\text{ g/mol}\right)\\ =55.85\text{ g/mol}+106.35\text{ g/mol}+192.00\text{ g/mol}\\ =\text{354}\text{.20 g/mol}\end{array}$

Therefore, the molar mass of $\text{Fe}{\left({\text{ClO}}_4\right)}_3$ is $\text{354}\text{.20 g/mol}$.

The relation between $\text{kg}$ and $\text{g}$ is as follows:

$1\text{ kg}={\text{10}}^3\text{ g}$

Therefore, the conversion factor is $\left(\frac{{\text{10}}^3\text{ g}}{1\text{ kg}}\right)$.

Calculate the amount of $15.8\text{ kg}$ of $\text{Fe}{\left({\text{ClO}}_4\right)}_3$ as follows:

$\begin{array}{c}\text{Amount of Fe}{\left({\text{ClO}}_4\right)}_3\left(\text{mol}\right)=\left(15.8\text{ kg Fe}{\left({\text{ClO}}_4\right)}_3\right)\left(\frac{{\text{10}}^3\text{ g}}{1\text{ kg}}\right)\left(\frac{1\text{ mol Fe}{\left({\text{ClO}}_4\right)}_3}{\text{354}\text{.20 g Fe}{\left({\text{ClO}}_4\right)}_3}\right)\\ =44.60756635\text{ mol Fe}{\left({\text{ClO}}_4\right)}_3\\ \approx 4.6\text{ mol Fe}{\left({\text{ClO}}_4\right)}_3\end{array}$

Conclusion

The amount of $15.8\text{ kg}$ of $\text{Fe}{\left({\text{ClO}}_4\right)}_3$ is $4.6\text{ mol}$.

(c)

Interpretation Introduction

Interpretation:

The number of $\text{N}$ atoms in $92.6\text{ mg}$ of ${\text{NH}}_4{\text{NO}}_2$ 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 $\text{6}\text{.022}\times {\text{10}}^{23}\left(\text{entitites/mol}\right)$.

Molar mass is defined as the mass of $1\text{ mol}$ of a chemical substance in grams. The molar mass of a compound is calculated by the addition of the molar mass of each element multiplied by its number of atoms present in the chemical formula.

The molecular formula of a compound tells the number of atoms/ions of each element present in the compound.

Following are the steps to calculate the number of atoms in a compound.

Step 1: Convert the mass of the compound to moles by using the molar mass of that compound as follows:

$\text{Amount}\left(\text{mol}\right)=\left(\text{Given mass}\left(\text{g}\right)\right)\left(\frac{1\text{ mol}}{\text{No}\text{. of grams}}\right)$

Step 2: Determine the number of moles of atoms of an element from the molecular formula of the compound. The formula to calculate the moles of atoms is as follows:

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

Step 3: Determine the number of atoms by multiplying the moles of atoms with the Avogadro’s number. The expression to calculate the number of atoms of an element is as follows:

$\text{Number of atoms}=\left(\text{Amount }\left(\text{mol}\right)\right)\left(\frac{6.022\times {10}^{23}\text{ atoms}}{1\text{ mol}}\right)$

Answer to Problem 3.16P

In $92.6\text{ mg}$ of ${\text{NH}}_4{\text{NO}}_2$, $1.74\times {10}^{21}\text{ N atoms}$ are present.

Explanation of Solution

The expression to calculate the molar mass of ${\text{NH}}_4{\text{NO}}_2$ is as follows:

${\text{Molar mass of NH}}_4{\text{NO}}_2=\left(2\right)\left(\text{M of N}\right)+\left(4\right)\left(\text{M of H}\right)+\left(2\right)\left(\text{M of O}\right)$

Substitute $14.01\text{ g/mol}$ for $\text{M}$ of $\text{N}$ , $16.00\text{ g/mol}$ for $\text{M}$ of $\text{O}$ , and $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$  in the above equation as follows:

$\begin{array}{c}{\text{Molar mass of NH}}_4{\text{NO}}_2=\left(2\right)\left(14.01\text{ g/mol}\right)+\left(4\right)\left(1.008\text{ g/mol}\right)+\left(2\right)\left(16.00\text{ g/mol}\right)\\ =28.02\text{ g/mol}+\text{4}\text{.032 g/mol}+\text{32}\text{.00 g/mol}\\ =\text{64}\text{.052 g/mol}\\ \approx \text{64}\text{.05 g/mol}\end{array}$

Therefore, the molar mass of ${\text{NH}}_4{\text{NO}}_2$ is $\text{64}\text{.05 g/mol}$. Hence, the mass of one mole of ${\text{NH}}_4{\text{NO}}_2$ is $\text{64}\text{.05 g}$.

The relation between $\text{mg}$ and $\text{g}$ is as follows:

$1\text{ mg}={\text{10}}^{-3}\text{ g}$

Therefore, the conversion factor is $\left(\frac{{\text{10}}^{-3}\text{ g}}{1\text{ mg}}\right)$.

Multiply $92.6\text{ mg}$ with the conversion factor and then divide with the molar mass of ${\text{NH}}_4{\text{NO}}_2$ to calculate the moles of ${\text{NH}}_4{\text{NO}}_2$ as follows:

$\begin{array}{c}{\text{Amount of NH}}_4{\text{NO}}_2\left(\text{mol}\right)=\left(92.6{\text{ mg NH}}_4{\text{NO}}_2\right)\left(\frac{{\text{10}}^{-3}\text{ g}}{1\text{ mg}}\right)\left(\frac{1{\text{ mol NH}}_4{\text{NO}}_2}{\text{64}{\text{.05 g NH}}_4{\text{NO}}_2}\right)\\ =1.44575\times {10}^{-3}{\text{ mol NH}}_4{\text{NO}}_2\end{array}$

From the molecular formula of ${\text{NH}}_4{\text{NO}}_2$, it is concluded that $2\text{ mol}$ of nitrogen $\left(\text{N}\right)$ atoms are present in $1\text{ mol}$ of ${\text{NH}}_4{\text{NO}}_2$. Calculate the moles of nitrogen atom in $1.44575\times {10}^{-3}{\text{ mol NH}}_4{\text{NO}}_2$ as follows:

$\begin{array}{c}\text{Amount of N atoms }\left(\text{mol}\right)=\left(1.44575\times {10}^{-3}{\text{ mol NH}}_4{\text{NO}}_2\right)\left(\frac{2\text{ mol N atoms}}{1{\text{ mol NH}}_4{\text{NO}}_2}\right)\\ =2.8915\times {10}^{-3}\text{ mol N atoms}\end{array}$

One mole of nitrogen atom contains $6.022\times {10}^{23}\text{ N atoms}$. Multiply $2.8915\times {10}^{-3}\text{ mol N atoms}$ with the Avogadro’s number to obtain the number of $\text{N}$ atoms. Calculate the number of $\text{N}$ atoms as follows:

$\begin{array}{c}\text{Number of N atoms}=\left(2.8915\times {10}^{-3}\text{ mol N atoms}\right)\left(\frac{6.022\times {10}^{23}\text{ N atoms}}{1\text{ mol N atoms}}\right)\\ =1.74\times {10}^{21}\text{ N atoms}\end{array}$

Conclusion

In $92.6\text{ mg}$ of ${\text{NH}}_4{\text{NO}}_2$, $1.74\times {10}^{21}\text{ N atoms}$ are present.