Chapter 8, Problem 51QAP

For each of the following samples of ionic substances, calculate the number of moles and mass of the positive ions present in each sample.

l type="a">

4.25 g of ammonium iodide, ${\text{NH}}_4\text{I}$

i>6.31 moles of ammonium sulfide, ${\left({\text{NH}}_4\right)}_2\text{S}$

i>9.71 g of barium phosphide, ${\text{Ba}}_3{\text{P}}_2$

i>7.63 moles of calcium phosphate, ${\text{Ca}}_3{\left({\text{PO}}_4\right)}_2$

Interpretation Introduction

(a)

Interpretation:

The number of moles and mass of positive ions present in the sample should be calculated.

Concept Introduction:

Number of moles is related to mass and molar mass as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Here, m is mass and M is molar mass.

According to Avogadro’s law, 1 mole of a substance contains$6.023\times {10}^{23}$ atoms. This is known as Avogadro’s number and denoted by symbol${\text{N}}_{\text{A}}$.

Thus, number of molecules can be calculated from number of moles using the following conversion factor:

$\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)$.

Answer to Problem 51QAP

Number of moles and mass of ammonium ion is 0.03 mol and 0.53 g respectively.

Explanation of Solution

The mass of ammonium iodide${\text{NH}}_{\text{4}}\text{I}$ is 4.25 g.

From mass, number of moles can be calculated as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Molar mass of${\text{NH}}_{\text{4}}\text{I}$ is 144.94 g/mol thus, number of moles will be:

$\begin{array}{c}\text{n}=\frac{4.25\text{ g}}{144.94\text{ g/mol}}\\ =0.03\text{ mol}\end{array}$

Thus, number of moles of${\text{NH}}_{\text{4}}\text{I}$ is 0.03 mol.

From the formula, 1 mol of${\text{NH}}_{\text{4}}\text{I}$ contains 1 mol of${\text{NH}}_4^{+}$ that is positive ion in${\text{NH}}_{\text{4}}\text{I}$. Thus, number of moles of${\text{NH}}_4^{+}$ in 0.03 mol of${\text{NH}}_{\text{4}}\text{I}$ is 0.03 mol.

From number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Molar mass of ammonium ion is 18.04 g/mol.

Putting the values,

$\begin{array}{c}\text{m}=0.03\text{ mol}\times 18.04\text{ g/mol}\\ =\text{0}\text{.53 g}\end{array}$

Therefore, number of moles and mass of ammonium ion is 0.03 mol and 0.53 g respectively.

Interpretation Introduction

(a)

Interpretation:

The number of moles and mass of positive ions present in the sample should be calculated.

Concept Introduction:

Number of moles is related to mass and molar mass as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Here, m is mass and M is molar mass.

According to Avogadro’s law, 1 mole of a substance contains$6.023\times {10}^{23}$ atoms. This is known as Avogadro’s number and denoted by symbol${\text{N}}_{\text{A}}$.

Thus, number of molecules can be calculated from number of moles using the following conversion factor:

$\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)$

Number of moles and mass of ammonium ion is 12.64 mol and 228 g respectively.

Number of moles of ammonium sulfide${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ is 6.31 mol

From the formula, 1 mol of${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ contains 2 mol of${\text{NH}}_4^{+}$ that is positive ion in${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$. Thus, number of moles of${\text{NH}}_4^{+}$ in 6.31 mol of${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ is$2\times 6.31\text{ mol}=12.62\text{ mol}$

From number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Molar mass of ammonium ion is 18.04 g/mol.

Putting the values,

$\begin{array}{c}\text{m}=12.64\text{ mol}\times 18.04\text{ g/mol}\\ =228\text{ g}\end{array}$

Therefore, number of moles and mass of ammonium ion is 12.64 mol and 228 g respectively.

(c)

Interpretation:

The number of moles and mass of positive ions present in the sample should be calculated.

Concept Introduction:

Number of moles is related to mass and molar mass as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Here, m is mass and M is molar mass.

According to Avogadro’s law, 1 mole of a substance contains$6.023\times {10}^{23}$ atoms. This is known as Avogadro’s number and denoted by symbol${\text{N}}_{\text{A}}$.

Thus, number of molecules can be calculated from number of moles using the following conversion factor:

$\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)$.

Number of moles and mass of barium ion is 0.0606 mol and 8.322 g respectively.

Mass of barium phosphide${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ is 9.71 g

From mass, number of moles can be calculated as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Molar mass of${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ is 479.977 g/mol thus, number of moles will be:

$\begin{array}{c}\text{n}=\frac{\text{9}\text{.71 g}}{\text{479}\text{.977 g/mol}}\\ =0.0202\text{ mol}\end{array}$

Thus, number of moles of${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ is 0.0202 mol.

From the formula, 1 mol of${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ contains 3 mol of${\text{Ba}}^{2+}$ that is positive ion in${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$. Thus, number of moles of${\text{Ba}}^{2+}$ in 0.0202 mol of${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ is$3\times 0.0202\text{ mol}=0.\text{0606 mol}$

From number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Molar mass of barium ion is 137.327 g/mol.

Putting the values,

$\begin{array}{c}\text{m}=0.0606\text{ mol}\times 137.327\text{ g/mol}\\ =8.322\text{ g}\end{array}$

Therefore, number of moles and mass of barium ion is 0.0606 mol and 8.322 g respectively.

(d)

Interpretation:

The number of moles and mass of positive ions present in the sample should be calculated.

Concept Introduction:

Number of moles is related to mass and molar mass as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Here, m is mass and M is molar mass.

According to Avogadro’s law, 1 mole of a substance contains$6.023\times {10}^{23}$ atoms. This is known as Avogadro’s number and denoted by symbol${\text{N}}_{\text{A}}$.

Thus, number of molecules can be calculated from number of moles using the following conversion factor:

$\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)$.

Number of moles and mass of calcium ion is 22.89 mol and$917.385\text{ g}$ respectively.

Number of moles of calcium phosphate${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is 7.63 mol

From the formula, 1 mol of${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ contains 3 mol of${\text{Ca}}^{2+}$ that is positive ion in${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$. Thus, number of moles of${\text{Ca}}^{2+}$ in 7.63 mol of${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is$3\times 7.63\text{ mol}=22.89\text{ mol}$

From number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Molar mass of calcium ion is 40.078 g/mol.

Putting the values,

$\begin{array}{c}\text{m}=22.89\text{ mol}\times 40.078\text{ g/mol}\\ =917.385\text{ g}\end{array}$

Therefore, number of moles and mass of calcium ion is 22.89 mol and$917.385\text{ g}$ respectively.

Answer to Problem 51QAP

Number of moles and mass of ammonium ion is 12.64 mol and 228 g respectively.

Explanation of Solution

Number of moles of ammonium sulfide${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ is 6.31 mol

From the formula, 1 mol of${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ contains 2 mol of${\text{NH}}_4^{+}$ that is positive ion in${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$. Thus, number of moles of${\text{NH}}_4^{+}$ in 6.31 mol of${\left({\text{NH}}_{\text{4}}\right)}_{\text{2}}\text{S}$ is$2\times 6.31\text{ mol}=12.62\text{ mol}$

From number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Molar mass of ammonium ion is 18.04 g/mol.

Putting the values,

$\begin{array}{c}\text{m}=12.64\text{ mol}\times 18.04\text{ g/mol}\\ =228\text{ g}\end{array}$

Therefore, number of moles and mass of ammonium ion is 12.64 mol and 228 g respectively.

Interpretation Introduction

(c)

Interpretation:

The number of moles and mass of positive ions present in the sample should be calculated.

Concept Introduction:

Number of moles is related to mass and molar mass as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Here, m is mass and M is molar mass.

According to Avogadro’s law, 1 mole of a substance contains$6.023\times {10}^{23}$ atoms. This is known as Avogadro’s number and denoted by symbol${\text{N}}_{\text{A}}$.

Thus, number of molecules can be calculated from number of moles using the following conversion factor:

$\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)$.

Answer to Problem 51QAP

Number of moles and mass of barium ion is 0.0606 mol and 8.322 g respectively.

Explanation of Solution

Mass of barium phosphide${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ is 9.71 g

From mass, number of moles can be calculated as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Molar mass of${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ is 479.977 g/mol thus, number of moles will be:

$\begin{array}{c}\text{n}=\frac{\text{9}\text{.71 g}}{\text{479}\text{.977 g/mol}}\\ =0.0202\text{ mol}\end{array}$

Thus, number of moles of${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ is 0.0202 mol.

From the formula, 1 mol of${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ contains 3 mol of${\text{Ba}}^{2+}$ that is positive ion in${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$. Thus, number of moles of${\text{Ba}}^{2+}$ in 0.0202 mol of${\text{Ba}}_{\text{3}}{\text{P}}_{\text{2}}$ is$3\times 0.0202\text{ mol}=0.\text{0606 mol}$

From number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Molar mass of barium ion is 137.327 g/mol.

Putting the values,

$\begin{array}{c}\text{m}=0.0606\text{ mol}\times 137.327\text{ g/mol}\\ =8.322\text{ g}\end{array}$

Therefore, number of moles and mass of barium ion is 0.0606 mol and 8.322 g respectively.

Interpretation Introduction

(d)

Interpretation:

The number of moles and mass of positive ions present in the sample should be calculated.

Concept Introduction:

Number of moles is related to mass and molar mass as follows:

$\text{n}=\frac{\text{m}}{\text{M}}$

Here, m is mass and M is molar mass.

According to Avogadro’s law, 1 mole of a substance contains$6.023\times {10}^{23}$ atoms. This is known as Avogadro’s number and denoted by symbol${\text{N}}_{\text{A}}$.

Thus, number of molecules can be calculated from number of moles using the following conversion factor:

$\left(\frac{6.023\times {10}^{23}\text{ molecules}}{1\text{ mol}}\right)$.

Answer to Problem 51QAP

Number of moles and mass of calcium ion is 22.89 mol and$917.385\text{ g}$ respectively.

Explanation of Solution

Number of moles of calcium phosphate${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is 7.63 mol

From the formula, 1 mol of${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ contains 3 mol of${\text{Ca}}^{2+}$ that is positive ion in${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$. Thus, number of moles of${\text{Ca}}^{2+}$ in 7.63 mol of${\text{Ca}}_{\text{3}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{2}}$ is$3\times 7.63\text{ mol}=22.89\text{ mol}$

From number of moles, mass can be calculated as follows:

$\text{m}=\text{n}\times \text{M}$

Molar mass of calcium ion is 40.078 g/mol.

Putting the values,

$\begin{array}{c}\text{m}=22.89\text{ mol}\times 40.078\text{ g/mol}\\ =917.385\text{ g}\end{array}$

Therefore, number of moles and mass of calcium ion is 22.89 mol and$917.385\text{ g}$ respectively.