Chapter 3, Problem 3.19P

(a)

Interpretation Introduction

Interpretation:

The mass $\left(\text{g}\right)$ of $\text{8}\text{.42 mol}$ of $\text{chromium}\left(\text{III}\right)\text{sulfate decahydrate}$ 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.

Answer to Problem 3.19P

The mass $\left(\text{g}\right)$ of $\text{8}\text{.42 mol}$ $\text{chromium}\left(\text{III}\right)\text{sulfate decahydrate}$ is $4.82\times {10}^3\text{ g}$.

Explanation of Solution

$\text{Chromium}\left(\text{III}\right)\text{sulfate decahydrate}$ is a chemical compound with the chemical formula ${\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}$.

The formula to calculate the molar mass of ${\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}$ is,

$\text{Molar Mass}=\left[\begin{array}{l}\left(\text{number of Cr atoms}\right)\left(\text{atomic mass}\text{of Cr}\right)\\ +\left(\text{number of}\text{S atoms}\right)\left(\text{atomic}\text{mass}\text{of S}\right)\\ +\left(\text{number of}\text{H atoms}\right)\left(\text{atomic}\text{mass}\text{of}\text{H}\right)\\ +\left(\text{number of}\text{O atoms}\right)\left(\text{atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right]$ (1)

Substitute $2$ for number of $\text{Cr}$ atoms, $\text{52}\text{.00 g/mol}$ for the atomic mass of $\text{Cr}$, $3$ for number of atoms of $\text{S}$, $32.06\text{g/mol}$ for the atomic mass of $\text{S}$, $20$ for number of atoms of $\text{H}$, $1.008\text{g/mol}$ for the atomic mass of $\text{H}$, $22$ for number of atoms of $\text{O}$ and $16.00\text{g/mol}$ for the atomic mass of $\text{O}$ in equation (1).

$\begin{array}{c}{\text{Molar Mass of Cu}}_{\text{2}}{\text{CO}}_{\text{3}}=\left[\begin{array}{l}\left(2\right)\left(\text{52}\text{.00 g/mol}\right)+\left(3\right)\left(32.06\text{g/mol}\right)+\left(20\right)\left(1.008\text{g/mol}\right)\\ +\left(22\right)\left(16.00\text{g/mol}\right)\end{array}\right]\\ =104.00\text{ g/mol}+96.18\text{g/mol}+20.16\text{g/mol}+352.00\text{g/mol}\\ =572.34\text{g/mol}\end{array}$

The expression to calculate the mass $\left(\text{g}\right)$ of ${\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}$ is:

$\text{Mass (g)}=\left[\begin{array}{l}\left({\text{mol of Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}\right)\\ \left(\frac{\text{molecular mass}\left(\text{g}\right){\text{ Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O }}{{\text{1 mol Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}}\right)\end{array}\right]$ (2)

Substitute $\text{8}\text{.42 mol}$ for ${\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}$ and $572.34\text{g/mol}$ for the molecular mass of ${\text{Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}$ in equation (2).

$\begin{array}{c}{\text{Mass (g) of Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}=\left(\text{8}\text{.42 mol}\right)\left(\frac{\text{572}{\text{.34 g Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O }}{{\text{1 mol Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O}}\right)\\ =4819.103\text{ g}\\ =\text{4}\text{.82}\times {\text{10}}^{\text{3}}{\text{ g Cr}}_{\text{2}}{\left({\text{SO}}_{\text{4}}\right)}_{\text{3}}\cdot 10{\text{H}}_{\text{2}}\text{O }\end{array}$

Conclusion

The mass $\left(\text{g}\right)$ of $\text{8}\text{.42 mol}$$\text{chromium}\left(\text{III}\right)\text{sulfate decahydrate}$ is $4.82\times {10}^3\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

The mass $\left(\text{g}\right)$ of $\text{1}\text{.83}\times {\text{10}}^{\text{24}}\text{ molecules}$ of dichlorine heptoxide 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 a number of molecules are given.

Step 1: Determine the amount of substance in moles 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 moles with the molar mass of the chemical substance to obtain the mass of chemical substance in grams. The formula to calculate the mass of a substance in grams is as follows:

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

Answer to Problem 3.19P

The mass $\left(\text{g}\right)$ of $\text{1}\text{.83}\times {\text{10}}^{\text{24}}\text{ molecules}$ of dichlorine heptoxide is $5.56\times {10}^2\text{ g}$.

Explanation of Solution

Dichlorine heptoxide is an inorganic compound with the chemical formula ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$.

The formula to calculate the molar mass of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ is,

${\text{Molar Mass of Cl}}_{\text{2}}{\text{O}}_{\text{7}}=\left[\begin{array}{l}\left(\text{number of Cl atoms}\right)\left(\text{atomic mass}\text{of Cl}\right)\\ +\left(\text{number of}\text{O atoms}\right)\left(\text{atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right]$ (3)

Substitute $2$ for number of $\text{Cl}$ atoms, $\text{35}\text{.45 g/mol}$ for the atomic mass of $\text{Cl}$, $7$ for number of atoms of $\text{O}$ and $16.00\text{g/mol}$ for the atomic mass of $\text{O}$ in equation (3).

$\begin{array}{c}{\text{Molar Mass of Cl}}_{\text{2}}{\text{O}}_7=\left[\left(2\right)\left(\text{35}\text{.45 g/mol}\right)+\left(7\right)\left(16.00\text{g/mol }\right)\right]\\ =70.9\text{ g/mol}+112.00\text{g/mol}\\ =182.9\text{g/mol}\end{array}$

The expression to calculate moles of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ is:

$\text{Moles}\text{of}{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}=\left({\text{molecules of Cl}}_{\text{2}}{\text{O}}_{\text{7}}\right)\left(\frac{1{\text{ mol Cl}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}{\text{of Cl}}_{\text{2}}{\text{O}}_{\text{7}}}\right)$ (4)

Substitute $\text{1}\text{.83}\times {\text{10}}^{\text{24}}\text{ molecules}$ for ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ in equation (4).

$\begin{array}{c}\text{Moles}\text{of}{\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}=\left(1.83\times {\text{10}}^{\text{24}}{\text{molecules of Cl}}_{\text{2}}{\text{O}}_{\text{7}}\right)\left(\frac{1{\text{ mol Cl}}_{\text{2}}{\text{O}}_{\text{7}}}{\text{6}\text{.022}\times {\text{10}}^{\text{23}}{\text{of Cl}}_{\text{2}}{\text{O}}_{\text{7}}}\right)\\ =3.038858{\text{ mol Cl}}_{\text{2}}{\text{O}}_{\text{7}}\end{array}$

The expression to calculate the mass $\left(\text{g}\right)$ of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ is:

$\text{Mass (g)}=\left({\text{mol of Cl}}_{\text{2}}{\text{O}}_{\text{7}}\right)\left(\frac{\text{molecular mass}\left(\text{g}\right){\text{ of Cl}}_{\text{2}}{\text{O}}_{\text{7}}}{{\text{1 mol Cl}}_{\text{2}}{\text{O}}_{\text{7}}}\right)$ (5)

Substitute $3.038858\text{ mol}$ for mol of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ and $182.9\text{g/mol}$ for the molecular mass of ${\text{Cl}}_{\text{2}}{\text{O}}_{\text{7}}$ in equation (5).

$\begin{array}{c}{\text{Mass (g) of Cl}}_{\text{2}}{\text{O}}_{\text{7}}=\left(\text{3}\text{.038858 mol}\right)\left(\frac{\text{ 182}{\text{.9 g Cl}}_{\text{2}}{\text{O}}_{\text{7}}}{{\text{1 mol Cl}}_{\text{2}}{\text{O}}_{\text{7}}}\right)\\ =555.807\text{ g}\\ =\text{5}\text{.56}\times {\text{10}}^{\text{2}}{\text{ g Cl}}_{\text{2}}{\text{O}}_{\text{7}}\end{array}$

Conclusion

The mass $\left(\text{g}\right)$ of $\text{1}\text{.83}\times {\text{10}}^{\text{24}}\text{ molecules}$ of dichlorine heptoxide is $5.56\times {10}^2\text{ g}$.

(c)

Interpretation Introduction

Interpretation:

The number of moles and formula units in $6.2\text{ g}$ of lithium sulfate 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.

A formula unit is used for the ionic compound to represent their empirical formula. The steps to determine the formula unit of an ionic compound from the given mass are as follows:

Step 1: Divide the given mass of the ionic compound with the molar mass to calculate the moles. The expression to calculate the moles of an ionic compound when the mass is given is as follows:

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

Step 2: Multiply the calculated moles with the Avogadro’s number to determine the formula units of an ionic compound. The expression to determine the formula unit is as follows:

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

Answer to Problem 3.19P

The number of moles and formula unit in $\text{6}\text{.2 g}$ of lithium sulfate is $\text{0}\text{.056 mol}$ and $3.4\times {10}^{22}\text{ FU}$ respectively.

Explanation of Solution

Lithium sulfate is an ionic compound with the chemical formula ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$.

The formula to calculate the molar mass of ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ is,

${\text{Molar Mass of Li}}_{\text{2}}{\text{SO}}_{\text{4}}=\left[\begin{array}{l}\left(\text{number of Li atoms}\right)\left(\text{atomic mass}\text{of Li}\right)\\ +\left(\text{number of}\text{S atoms}\right)\left(\text{atomic}\text{mass}\text{of S}\right)\\ +\left(\text{number of}\text{O atoms}\right)\left(\text{atomic}\text{mass}\text{of}\text{O}\right)\end{array}\right]$ (6)

Substitute $2$ for number of $\text{Li}$ atoms, $\text{6}\text{.941 g/mol}$ for the atomic mass of $\text{Li}$, $1$ for number of atoms of $\text{S}$, $32.06\text{g/mol}$ for the atomic mass of $\text{S}$, $4$ for number of atoms of $\text{O}$ and $16.00\text{g/mol}$ for the atomic mass of $\text{O}$ in equation (2).

$\begin{array}{c}{\text{Molar Mass of Li}}_{\text{2}}{\text{SO}}_{\text{4}}=\left[\begin{array}{l}\left(2\right)\left(\text{6}\text{.941 g/mol}\right)+\left(1\right)\left(32.06\text{g/mol }\right)\\ +\left(4\right)\left(16.00\text{g/mol }\right)\end{array}\right]\\ =13.882\text{ g/mol}+32.06\text{g/mol}+64.00\text{g/mol}\\ =109.94\text{g/mol}\end{array}$

The expression to calculate moles of ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ is:

$\text{Moles}\text{of}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}=\left({\text{given mass of Li}}_{\text{2}}{\text{SO}}_{\text{4}}\left(\text{g}\right)\right)\left(\frac{1{\text{ mol Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{\text{molecular mass}\left(\text{g}\right)}\right)$ (7)

Substitute $\text{6}\text{.2 g}$ for given mass of ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $109.94\text{g/mol}$ for the molecular mass of ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ in equation (7).

$\begin{array}{c}\text{Moles of}\text{}\text{}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}=\left(\text{6}\text{.2 g}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\left(\frac{{\text{1mol Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{109.94{\text{ g Li}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\\ =0.056394\text{mol}\\ =0.056{\text{ mol Li}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

The expression to calculate formula units (FU) of ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ is:

${\text{FU of Li}}_{\text{2}}{\text{SO}}_{\text{4}}=\left({\text{mol of Li}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\left(\frac{6.022\times {10}^{23}{\text{ FU Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{{\text{1mol Li}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)$ (8)

Substitute $0.056394\text{mol}$ for ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ in equation (8).

$\begin{array}{c}{\text{FU of Li}}_{\text{2}}{\text{SO}}_{\text{4}}=\left(\text{0}{\text{.056394 mol Li}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\left(\frac{6.022\times {10}^{23}{\text{ FU Li}}_{\text{2}}{\text{SO}}_{\text{4}}}{{\text{1 mol Li}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\\ =3.3960\times {10}^{22}\text{ FU}\\ =\text{3}\text{.4}\times {10}^{22}{\text{ FU Li}}_{\text{2}}{\text{SO}}_{\text{4}}\end{array}$

Conclusion

The number of moles and formula unit in $\text{6}\text{.2 g}$ of lithium sulfate is $\text{0}\text{.056 mol}$ and $3.4\times {10}^{22}\text{ FU}$ respectively.

(d)

Interpretation Introduction

Interpretation:

The number of lithium ions, sulfate ions, sulfur atoms and oxygen atoms in the mass of the compound ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ is to be calculated.

Concept introduction:

A formula unit is used for the ionic compound to represent their empirical formula. The molecular formula of a compound tells the number of atoms/ions of each element present in the compound.

A number of ions in a chemical compound is directly linked to the formula unit of the compound.

Answer to Problem 3.19P

The number of lithium ions is $6.8\times {10}^{22}$, sulfate ions are $3.4\times {10}^{22}$, sulfur atoms are $3.4\times {10}^{22}$ and oxygen atoms are $1.4\times {10}^{23}$ in the mass of the compound ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$.

Explanation of Solution

The expression to calculate the number of ions/atoms in ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ is:

$\text{Number of ions}=\left({\text{FU of Li}}_{\text{2}}{\text{SO}}_{\text{4}}\right)\left(\frac{\text{Number of ions}/\text{atoms}}{\text{1 FU}\text{of}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)$ (9)

Substitute $3.3960\times {10}^{22}\text{ FU}$ for ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $2$ for ${\text{Li}}^{\text{+}}$ ion in equation (9).

$\begin{array}{c}{\text{Number of Li}}^{\text{+}}\text{ ions}=\left(\text{3}\text{.3960}\times {\text{10}}^{\text{22}}\text{ FU}\right)\left(\frac{1{\text{ Li}}^{\text{+}}\text{ ion}}{\text{1 FU}\text{of}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\\ =6.7920\times {10}^{22}\\ =\text{6}\text{.8}\times {\text{10}}^{\text{22}}{\text{ Li}}^{\text{+}}\text{ ions}\end{array}$

Substitute $3.3960\times {10}^{22}\text{ FU}$ for ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $1$ for ${\text{SO}}_{\text{4}}{}^{2-}$ ion in equation (9).

$\begin{array}{c}{\text{Number of SO}}_{\text{4}}{}^{2-}\text{ ions}=\left(\text{3}\text{.3960}\times {\text{10}}^{\text{22}}\text{ FU}\right)\left(\frac{1{\text{ SO}}_{\text{4}}{}^{2-}\text{ ion}}{\text{1 FU}\text{of}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\\ =3.3960\times {10}^{22}\\ =3.4\times {10}^{22}{\text{ SO}}_{\text{4}}{}^{2-}\text{ ions}\end{array}$

Substitute $3.3960\times {10}^{22}\text{ FU}$ for ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $1$ for $\text{S}$ atom in equation (9).

$\begin{array}{c}\text{Number of S atoms}=\left(\text{3}\text{.3960}\times {\text{10}}^{\text{22}}\text{ FU}\right)\left(\frac{1\text{ S atom}}{\text{1 FU}\text{of}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\\ =3.3960\times {10}^{22}\\ \text{=3}\text{.4}\times {\text{10}}^{\text{22}}\text{ S atoms}\end{array}$

Substitute $3.3960\times {10}^{22}\text{ FU}$ for ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$ and $4$ for $\text{O}$ atom in equation (9).

$\begin{array}{c}\text{Number of O atoms}=\left(\text{3}\text{.3960}\times {\text{10}}^{\text{22}}\text{ FU}\right)\left(\frac{\text{4 O atom}}{\text{1 FU}\text{of}{\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}}\right)\\ =1.3584\times {10}^{23}\\ \text{=1}\text{.4}\times {\text{10}}^{\text{23}}\text{ O atoms}\end{array}$

Conclusion

The number of lithium ions is $6.8\times {10}^{22}$, sulfate ions are $3.4\times {10}^{22}$, sulfur atoms are $3.4\times {10}^{22}$ and oxygen atoms are $1.4\times {10}^{23}$ in the mass of the compound ${\text{Li}}_{\text{2}}{\text{SO}}_{\text{4}}$.