Chapter 3, Problem 3.17P

(a)

Interpretation Introduction

Interpretation:

Total number of ions in $38.1\text{ g}$ of $\text{Sr}\text{}{\text{F}}_2$ is to be calculated.

Concept introduction:

The molar mass of a substance is defined as the mass of one mole of a chemical entity. It is calculated as the summation of the product of the number of atoms of each element and the atomic mass of the respective element.

For example, the molar mass of $\text{NaCl}$ is calculated as follows.

$\text{Molar Mass of NaCl}=\left[\begin{array}{l}\left(\text{number of Na atoms}\right)\left(\text{atomic mass}\text{of Na}\right)\\ +\left(\text{number of Cl atoms}\right)\left(\text{atomic}\text{mass}\text{of Cl }\right)\end{array}\right]$ (1)

A mole is a unit of measurement that has the same number of particles as that in $12.000\text{ g}$ of $\text{carbon}-\text{12}$. One mole of a chemical substance (atoms/molecules/ions) has $6.022\times {10}^{23}$ number of entities.

Answer to Problem 3.17P

Total number of ions in $38.1\text{ g}$ of $\text{Sr}\text{}{\text{F}}_2$ is $5.48\times {10}^{23}\text{}\text{ ions}$.

Explanation of Solution

The formula to calculate the molar mass of $\text{Sr}\text{}{\text{F}}_2$. is,

$\text{Molar Mass of Sr}\text{}{\text{F}}_2=\left[\begin{array}{l}\left(\text{number of Sr}\text{}\text{ atoms}\right)\left(\text{atomic mass}\text{of Sr}\right)\\ +\left(\text{number of}\text{F atoms}\right)\left(\text{atomic}\text{mass}\text{of F }\right)\end{array}\right]$ (2)

Substitute $1$ for number of $\text{Sr}$ atoms and $87.62\text{ g/mol}$ for atomic mass of $\text{Sr}$. Similarly, substitute $2$ for number of $\text{F}$ atoms and $19.00\text{g/mol}$ for atomic mass of $\text{F}$ in equation (2).

$\begin{array}{c}\text{Molar Mass of Sr}\text{}{\text{F}}_2=\left[\left(1\right)\left(87.62\text{ g/mol}\right)+\left(2\right)\left(19.00\text{g/mol }\right)\right]\\ =87.62\text{ g/mol}+38.00\text{ g/mol}\\ =\text{125}\text{.62 g/mol}\end{array}$

The expression to calculate moles of $\text{Sr}\text{}{\text{F}}_2$ is:

$\text{Moles}\text{of}{\text{SrF}}_{\text{2}}=\left({\text{given mass of SrF}}_{\text{2}}\left(\text{g}\right)\right)\left(\frac{1{\text{ mol SrF}}_{\text{2}}}{\text{molecular mass}{\text{of SrF}}_{\text{2}}\left(\text{g}\right)}\right)$ (3)

Substitute $38.1\text{ g}$ for given mass of $\text{Sr}\text{}{\text{F}}_2$ and $\text{125}\text{.62 g/mol}$ for the molecular mass of $\text{Sr}\text{}{\text{F}}_2$ in equation (3).

$\begin{array}{c}\text{Moles of}\text{}\text{}{\text{SrF}}_{\text{2}}=\left(38.1\text{g}{\text{SrF}}_{\text{2}}\right)\left(\frac{{\text{1mol SrF}}_{\text{2}}}{125.62\text{g}{\text{SrF}}_{\text{2}}}\right)\\ =0.303296\text{}\text{}\text{mol}{\text{SrF}}_{\text{2}}\end{array}$

The expression to calculate moles of ions of $\text{Sr}\text{}{\text{F}}_2$ is:

$\text{Moles}\text{of}\text{ions}\text{of}{\text{SrF}}_{\text{2}}=\left({\text{mol of SrF}}_{\text{2}}\right)\left(\frac{{\text{mol of ions in SrF}}_{\text{2}}}{{\text{1 mol SrF}}_{\text{2}}}\right)$ (4)

Substitute $0.303296\text{}\text{}\text{mol}$ for $\text{Sr}\text{}{\text{F}}_2$ and $3$ for mol of ions in $\text{Sr}\text{}{\text{F}}_2$ in equation (4).

$\begin{array}{c}\text{Moles}\text{of}\text{ions}\text{of}{\text{SrF}}_{\text{2}}=\left(\text{0}{\text{.303296 mol SrF}}_{\text{2}}\right)\left(\frac{{\text{3 mol of ions in SrF}}_{\text{2}}}{{\text{1 mol SrF}}_{\text{2}}}\right)\\ =0.909888\text{ mol}\text{ions}\end{array}$

The expression to calculate number of ions of $\text{Sr}\text{}{\text{F}}_2$ is:

$\text{Number}\text{of}\text{ions}\text{of}{\text{SrF}}_{\text{2}}=\left({\text{mol ions of SrF}}_{\text{2}}\right)\left(\frac{6.022\times {10}^{23}\text{ ions}}{{\text{1 mol ions of SrF}}_{\text{2}}}\right)$ (5)

Substitute $0.909888\text{mol ions}$ for $\text{Sr}\text{}{\text{F}}_2$ in equation (5).

$\begin{array}{c}\text{Number}\text{of}\text{ions}\text{of}{\text{SrF}}_{\text{2}}=\left(\text{0}\text{.909888 mol ions}\right)\left(\frac{6.022\times {10}^{23}\text{ ions}}{{\text{1 mol ions of SrF}}_{\text{2}}}\right)\\ =5.47935\times {10}^{23}\text{ions}\\ \simeq 5.48\times {10}^{23}\text{}\text{ ions}\end{array}$

Conclusion

Total number of ions in $38.1\text{ g}$ of $\text{Sr}\text{}{\text{F}}_2$ is $5.48\times {10}^{23}\text{}\text{ ions}$.

(b)

Interpretation Introduction

Interpretation:

Mass (kg) of $\text{3}\text{.58 mol}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

The molar mass of a substance is defined as the mass of one mole of a chemical entity. It is calculated as the summation of the product of the number of atoms of each element and the atomic mass of the respective element.

For example, the molar mass of $\text{NaCl}$ is calculated as follows.

$\text{Molar Mass of NaCl}=\left[\begin{array}{l}\left(\text{number of Na atoms}\right)\left(\text{atomic mass}\text{of Na}\right)\\ +\left(\text{number of Cl atoms}\right)\left(\text{atomic}\text{mass}\text{of Cl }\right)\end{array}\right]$ (1)

The conversion factor for $\text{g}$ into $\text{kg}$ is given as.

$\text{Mass }\left(\text{kg}\right)=\left(\text{Mass}\left(\text{g}\right)\right)\left(\frac{\text{1 kg}}{{\text{10}}^{\text{3}}\text{g}}\right)$ (6)

Answer to Problem 3.17P

Massof $\text{3}\text{.58 mol}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is $0.610\text{ kg}$.

Explanation of Solution

The formula to calculate the molar mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is,

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

Substitute $1$ for number of $\text{Cu}$ atoms and $\text{63}\text{.55 g/mol}$ for atomic mass of $\text{Cu}$, $2$ for number of atoms of $\text{Cl}$ and $35.45\text{g/mol}$ for atomic mass of $\text{Cl}$, $\text{4}$ for number of atoms of $\text{H}$ and $1.008\text{g/mol}$ for atomic mass of $\text{H}$, $2$ for number of atoms of $\text{O}$ and $16.00\text{g/mol}$ for atomic mass of $\text{O}$ in equation(7).

$\begin{array}{c}{\text{Molar Mass of CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}=\left[\begin{array}{l}\left(1\right)\left(\text{63}\text{.55 g/mol}\right)+\left(2\right)\left(35.45\text{g/mol }\right)\\ +\left(4\right)\left(1.008\text{g/mol }\right)+\left(2\right)\left(16.00\text{g/mol }\right)\end{array}\right]\\ =63.55\text{ g/mol}+70.9\text{g/mol}+4.032\text{g/mol}+32.00\text{g/mol}\\ =170.48\text{g/mol}\end{array}$

The expression to calculate the mass $\left(\text{g}\right)$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is,

$\text{Mass (g)}=\left({\text{mol of CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}\right)\left(\frac{\text{molecular mass}\left(\text{g}\right){\text{ of CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O }}{{\text{1 mol CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}}\right)$ (8)

Substitute $\text{3}\text{.58 mol}$ for ${\text{CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}$ and $170.48\text{ g/mol}$ for molecular mass of ${\text{CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}$ in equation (8).

$\begin{array}{c}{\text{Mass (g) of CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}=\left(\text{3}\text{.58 mol}\right)\left(\frac{\text{170}{\text{.48 g CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O }}{{\text{1 mol CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}}\right)\\ =\left(610.32{\text{ g CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}\right)\left(\frac{\text{1 kg}}{{\text{10}}^{\text{3}}\text{g}}\right)\\ =0.61032\text{ kg}\\ =\text{0}{\text{.610 kg of CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}\end{array}$

Conclusion

Mass of $\text{3}\text{.58 mol}$ of ${\text{CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}$ is $0.610\text{ kg}$.

(c)

Interpretation Introduction

Interpretation:

Mass (mg) of $\text{2}\text{.88}\times {\text{10}}^{\text{22}}$ formula units of $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

The molar mass of a substance is defined as the mass of one mole of a chemical entity. It is calculated as the summation of the product of the number of atoms of each element and the atomic mass of the respective element.

For example, the molar mass of $\text{NaCl}$ is calculated as follows.

$\text{Molar Mass of NaCl}=\left[\begin{array}{l}\left(\text{number of Na atoms}\right)\left(\text{atomic mass}\text{of Na}\right)\\ +\left(\text{number of Cl atoms}\right)\left(\text{atomic}\text{mass}\text{of Cl }\right)\end{array}\right]$ (1)

The conversion factor for $\text{g}$ into $\text{mg}$ is given as.

$\text{Mass }\left(\text{mg}\right)=\left(\text{Mass}\left(\text{g}\right)\right)\left(\frac{\text{1 mg}}{{\text{10}}^{-3}\text{g}}\right)$ (9)

Answer to Problem 3.17P

Mass of $\text{2}\text{.88}\times {\text{10}}^{\text{22}}$ formula units of $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is $\text{2}\text{.32}\times {\text{10}}^{\text{4}}\text{ mg}$.

Explanation of Solution

The formula to calculate the molar mass of $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$. is,

$\text{Molar Mass}=\left[\begin{array}{l}\left(\text{number of Bi atoms}\right)\left(\text{atomic mass}\text{of Bi}\right)\\ +\left(\text{number of N atoms}\right)\left(\text{atomic}\text{mass}\text{of N }\right)\\ +\left(\text{number of H atoms}\right)\left(\text{atomic}\text{mass}\text{of H }\right)\\ +\left(\text{number of}\text{O atoms}\right)\left(\text{atomic}\text{mass}\text{of}\text{O }\right)\end{array}\right]$ (10)

Substitute $1$ for number of $\text{Bi}$ atoms, $\text{209}\text{.0 g/mol}$ for atomic mass of $\text{Bi}$, $3$ for number of atoms of $\text{N}$, $14.01\text{g/mol}$ for atomic mass of $\text{N}$, $\text{10}$ for number of atoms of $\text{H}$, $1.008\text{g/mol}$ for atomic mass of $\text{H}$, $14$ for number of atoms of $\text{O}$ and $16.00\text{g/mol}$ for atomic mass of $\text{O}$ in equation (10).

$\begin{array}{c}\text{Molar Mass of Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}=\left[\begin{array}{l}\left(1\right)\left(\text{209}\text{.0 g/mol}\right)+\left(3\right)\left(14.01\text{g/mol }\right)\\ +\left(10\right)\left(1.008\text{g/mol }\right)+\left(14\right)\left(16.00\text{g/mol }\right)\end{array}\right]\\ =209.0\text{ g/mol}+42.03\text{g/mol}+10.08\text{g/mol}+224.00\text{g/mol}\\ =485.1\text{g/mol}\end{array}$

The expression to calculate moles of $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is.

$\text{Moles}\text{of Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}=\left(\text{number of FU}\right)\left(\frac{1\text{ mol}}{6.022\times {10}^{23}\text{ FU}}\right)$ (11)

Substitute $\text{2}\text{.88}\times {\text{10}}^{\text{22}}$ for formula units of $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$ in equation (11).

$\begin{array}{c}\text{Moles}\text{of Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}=\left(\text{2}\text{.88}\times {\text{10}}^{\text{22}}\text{ FU}\right)\left(\frac{1\text{ mol}}{6.022\times {10}^{23}\text{ FU}}\right)\\ =0.047825\text{ mol Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}\end{array}$

The expression to calculate the mass $\left(\text{g}\right)$ of $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is:

$\text{Mass (g)}=\left(\text{mol Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}\right)\left(\frac{\text{molecular mass}\left(\text{g}\right)\text{ Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O }}{{\text{1 mol CuCl}}_{\text{2}}\cdot {\text{2H}}_{\text{2}}\text{O}}\right)$ (12)

Substitute $0.047825\text{ mol}$ for $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$ and $485.1\text{ g/mol}$ for molecular mass of $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$ in equation (12).

$\begin{array}{c}\text{Mass (g)}=\left(\text{0}\text{.047825 mol Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}\right)\left(\frac{\text{485}\text{.1 g Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O }}{\text{1 mol Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}}\right)\\ =\left(23.1999\text{ g Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}\right)\left(\frac{\text{1 mg}}{{\text{10}}^{-3}\text{g}}\right)\\ =23199.9\text{ mg of Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}\\ =\text{2}\text{.32}\times {\text{10}}^{\text{4}}\text{ mg of Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}\end{array}$

Conclusion

Mass of $\text{2}\text{.88}\times {\text{10}}^{\text{22}}$ formula units of $\text{Bi}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}\cdot {\text{5H}}_{\text{2}}\text{O}$ is $\text{2}\text{.32}\times {\text{10}}^{\text{4}}\text{ mg}$.