Chapter 3, Problem 3.15P

(a)

Interpretation Introduction

Interpretation:

The mass of $4.6\times {10}^{21}\text{ molecules}$ of ${\text{NO}}_2$ in $\text{kg}$ 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.

Following are the steps to calculate the mass of a chemical substance when number of molecules is given.

Step 1: Determine the amount of substance by using Avogadro’s number. The expression to calculate the moles of a chemical substance is as follows:

$\text{Amount }\left(\text{mol }\right)=\left(\text{Given molecules}\right)\left(\frac{1\text{ mol}}{6.022\times {10}^{23}\text{ molecules}}\right)$

Step 2: Multiply the moleswith the molar mass of the chemical substance to obtain the mass of chemical substance in grams. The mass can be converted to kilogram by using the relation between gram and kilogram. The formula to calculate the mass of a substance in kilogram is as follows:

$\text{Mass }\left(\text{kg }\right)=\left(\text{Amount }\left(\text{mol}\right)\right)\left(\frac{\text{No}\text{. of grams}}{\text{1 mol}}\right)\left(\frac{{10}^{-3}\text{ kg}}{1\text{ g}}\right)$

Answer to Problem 3.15P

The mass of $4.6\times {10}^{21}\text{ molecules}$ of ${\text{NO}}_2$ in $\text{kg}$ is $3.5\times {10}^{-4}{\text{ g NO}}_2$.

Explanation of Solution

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

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

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

$\begin{array}{c}{\text{Molar mass of NO}}_2=\left(1\right)\left(14.01\text{ g/mol}\right)+\left(2\right)\left(16.00\text{ g/mol}\right)\\ =\text{14}\text{.01 g/mol}+\text{32}\text{.00 g/mol}\\ =\text{46}\text{.01 g/mol}\end{array}$

Therefore, the molar mass of ${\text{NO}}_2$ is $\text{46}\text{.01 g/mol}$.

One mole of ${\text{NO}}_2$ contains $6.022\times {10}^{23}{\text{ NO}}_2\text{ molecules}$. To calculate the moles of ${\text{NO}}_2$, divide $4.6\times {10}^{21}\text{ molecules}$ of ${\text{NO}}_2$ by the Avogadro’s number as follows:

$\begin{array}{c}{\text{Amount of NO}}_2\left(\text{mol}\right)=\left(4.6\times {10}^{21}{\text{ molecules NO}}_2\right)\left(\frac{1{\text{ mol NO}}_2}{6.022\times {10}^{23}{\text{ NO}}_2\text{ molecules}}\right)\\ =7.63866\times {10}^{-3}{\text{ mol NO}}_2\end{array}$

Multiply $7.63866\times {10}^{-3}{\text{ mol NO}}_2$ with the molar mass to calculate the mass of ${\text{NO}}_2$ in gram as follows:

$\begin{array}{c}{\text{Mass of NO}}_2\left(\text{g}\right)=\left(7.63866\times {10}^{-3}{\text{ mol NO}}_2\right)\left(\frac{\text{46}{\text{.01 g NO}}_2}{{\text{1 mol NO}}_2}\right)\\ =3.514547\times {10}^{-1}{\text{ g NO}}_2\end{array}$

The relation between gram and kilogram is as follows:

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

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

Multiply $3.514547\times {10}^{-1}{\text{ g NO}}_2$ with the conversion factor to calculate the mass of ${\text{NO}}_2$ in kilogram as follows:

$\begin{array}{c}{\text{Mass of NO}}_2\left(\text{kg}\right)=\left(3.514547\times {10}^{-1}{\text{ g NO}}_2\right)\left(\frac{{\text{10}}^{-3}\text{ kg}}{\text{1 g}}\right)\\ =3.514547\times {10}^{-4}{\text{ g NO}}_2\\ \approx 3.5\times {10}^{-4}{\text{ g NO}}_2\end{array}$

Conclusion

The mass of $4.6\times {10}^{21}\text{ molecules}$ of ${\text{NO}}_2$ in $\text{kg}$ is $3.5\times {10}^{-4}{\text{ g NO}}_2$.

(b)

Interpretation Introduction

Interpretation:

The amount of chlorine atoms in $0.0615\text{ g}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ in $\text{mol}$ 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.

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

Following are the steps to calculate the amount of an element in a compound.

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

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

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 an element 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 3.15P

The amount of chlorine atoms in $0.0615\text{ g}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ in $\text{mol}$ is $1.2431\times {10}^{-3}\text{ mol}$.

Explanation of Solution

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

${\text{Molar mass of C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}=\left(2\right)\left(\text{M of C}\right)+\left(4\right)\left(\text{M of H}\right)+\left(2\right)\left(\text{M of Cl}\right)$

Substitute $12.01\text{ g/mol}$ for $\text{M}$ of $\text{C}$, $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ and $35.45\text{ g/mol}$ for $\text{M}$ of $\text{Cl}$ in the above equation as follows:

$\begin{array}{c}{\text{Molar mass of C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}=\left(2\right)\left(12.01\text{ g/mol}\right)+\left(4\right)\left(1.008\text{ g/mol}\right)+\left(2\right)\left(35.45\text{ g/mol}\right)\\ =\text{24}\text{.02 g/mol}+\text{4}\text{.032 g/mol}+\text{70}\text{.9 g/mol}\\ =\text{98}\text{.95 g/mol}\end{array}$

Therefore, the molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ is $\text{98}\text{.95 g/mol}$. Hence, the mass of one mole of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ is $\text{98}\text{.95 g}$. Multiply $0.0615\text{ g}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ with the molar mass of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ to calculate the moles as follows:

$\begin{array}{c}{\text{Amount of C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}\left(\text{mol}\right)=\left(0.0615{\text{ g C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}\right)\left(\frac{1{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}{\text{98}{\text{.95 g C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}\right)\\ =6.21526\times {10}^{-4}{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}\end{array}$

From the molecular formula of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$, it is concluded that $2\text{ mol}$ of chlorine atoms are present in $1\text{ mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$. Calculate the moles of chlorine atoms in $6.21526\times {10}^{-4}{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ as follows:

$\begin{array}{c}\text{Amount of Cl atoms}\left(\text{mol}\right)=\left(6.21526\times {10}^{-4}{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}\right)\left(\frac{2\text{ mol Cl atoms}}{1{\text{ mol C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}}\right)\\ =1.2431\times {10}^{-3}\text{ mol Cl atoms}\end{array}$

Conclusion

The amount of chlorine atoms in $0.0615\text{ g}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{4}}{\text{Cl}}_{\text{2}}$ in $\text{mol}$ is $1.2431\times {10}^{-3}\text{ mol}$.

(c)

Interpretation Introduction

Interpretation:

The number of ${\text{H}}^{-}$ ions in $5.82\text{ g}$ of ${\text{SrH}}_2$ 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 ions 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 ions of an element from the molecular formula of the compound. The formula to calculate the moles of ions is as follows:

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

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

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

Answer to Problem 3.15P

In $5.82\text{ g}$ of ${\text{SrH}}_2$, $7.82\times {10}^{22}{\text{ H}}^{-}\text{ ions}$ are present.

Explanation of Solution

The expression to calculate the molar mass of ${\text{SrH}}_2$ is as follows:

${\text{Molar mass of SrH}}_2=\left(1\right)\left(\text{M of Sr}\right)+\left(2\right)\left(\text{M of H}\right)$

Substitute $87.62\text{ g/mol}$ for $\text{M}$ of $\text{Sr}$ and $1.008\text{ g/mol}$ for $\text{M}$ of $\text{H}$ in the above equation as follows:

$\begin{array}{c}{\text{Molar mass of SrH}}_2=\left(1\right)\left(87.62\text{ g/mol}\right)+\left(2\right)\left(1.008\text{ g/mol}\right)\\ =87.62\text{ g/mol}+\text{2}\text{.016 g/mol}+\text{70}\text{.9 g/mol}\\ =\text{89}\text{.636 g/mol}\\ \approx \text{89}\text{.64 g/mol}\end{array}$

Therefore, the molar mass of ${\text{SrH}}_2$ is $\text{89}\text{.64 g/mol}$. Hence, the mass of one mole of ${\text{SrH}}_2$ is $\text{89}\text{.64 g}$. Divide $5.82\text{ g}$ of ${\text{SrH}}_2$ with the molar mass of ${\text{SrH}}_2$ to calculate the moles of ${\text{SrH}}_2$ as follows:

$\begin{array}{c}{\text{Amount of SrH}}_2\left(\text{mol}\right)=\left(5.82{\text{ g SrH}}_2\right)\left(\frac{1{\text{ mol SrH}}_2}{\text{89}{\text{.64 g SrH}}_2}\right)\\ =0.064926372{\text{ mol SrH}}_2\\ \approx 0.0649264{\text{ mol SrH}}_2\end{array}$

From the molecular formula of ${\text{SrH}}_2$, it is concluded that $2\text{ mol}$ of hydride $\left({\text{H}}^{-}\right)$ ions are present in $1\text{ mol}$ of ${\text{SrH}}_2$. Calculate the moles of hydride ion in $0.0649264{\text{ mol SrH}}_2$ as follows:

$\begin{array}{c}{\text{Amount of H}}^{-}\text{ ions}\left(\text{mol}\right)=\left(0.0649264{\text{ mol SrH}}_2\right)\left(\frac{2{\text{ mol H}}^{-}\text{ ions}}{1{\text{ mol SrH}}_2}\right)\\ =0.1298528{\text{ mol H}}^{-}\text{ ions}\end{array}$

One mole of hydride ion has $6.022\times {10}^{23}{\text{ H}}^{-}\text{ ions}$. Multiply $0.1298528{\text{ mol H}}^{-}\text{ ions}$ with the Avogadro’s number to obtain the number of ${\text{H}}^{-}\text{ ions}$. Calculate the number of ${\text{H}}^{-}\text{ ions}$ as follows:

$\begin{array}{c}{\text{Number of H}}^{-}\text{ ions}=\left(0.1298528{\text{ mol H}}^{-}\text{ ions}\right)\left(\frac{6.022\times {10}^{23}{\text{ H}}^{-}\text{ ions}}{1{\text{ mol H}}^{-}\text{ ions}}\right)\\ =7.8187\times {10}^{22}{\text{ H}}^{-}\text{ ions}\\ \approx 7.82\times {10}^{22}{\text{ H}}^{-}\text{ ions}\end{array}$

Conclusion

In $5.82\text{ g}$ of ${\text{SrH}}_2$, $7.82\times {10}^{22}{\text{ H}}^{-}\text{ ions}$ are present.