Chapter 11, Problem 77E

Interpretation Introduction

(a)

Interpretation:

The water of hydration for the compound ${\text{SrCl}}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is to be calculated which contains $18.5\%$ water. The chemical formula of the compound ${\text{SrCl}}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

The water of crystallization is the number of water molecules present in crystal compounds. The water of crystallization is also known as water of hydration. The water molecules present in the crystal helps the crystal to impart color. These water molecules can be easily removed from the crystal by heating them.

Answer to Problem 77E

The water of hydration for the compound ${\text{SrCl}}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is $2$. The chemical formula of compound is ${\text{SrCl}}_{\text{2}}\cdot 2{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

The compound ${\text{SrCl}}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ contains $18.5\%$ water. Therefore, $100\text{ g}$ of the compound will have $18.5\text{ g}$ of water in it.

The mass of ${\text{SrCl}}_{\text{2}}$ unit in $100\text{ g}$ of the compound is calculated as shown below.

$\begin{array}{rll}{m}_{{\text{SrCl}}_{\text{2}}}& =& 100\text{ g}-18.5\text{ g}\\ & =& 81.5\text{ g}\end{array}$

The molar mass of water is $18.02\text{g}/\text{mol}$.

The molar mass of ${\text{SrCl}}_{\text{2}}$ is $158.53 \text{g}/\text{mol}$.

The number of moles of a substance is given by the expression as shown below.

$n=\frac{m}{M}$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the values of mass and molar mass of water in the equation (1).

$\begin{array}{rll}n& =& \frac{18.5\text{ g}}{18.02\text{g}/\text{mol}}\\ & =& 1.027\text{ mol}\end{array}$

Substitute the values of mass and molar mass of ${\text{SrCl}}_{\text{2}}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{81.5\text{ g}}{158.53 \text{g}/\text{mol}}\\ & =& 0.514\text{ mol}\end{array}$

The molar ratio of ${\text{SrCl}}_{\text{2}}:{\text{H}}_2\text{O}$ can be obtained by dividing the number of moles with the lowest number of mole that is $0.514\text{ mol}$.

$\begin{array}{rll}{\text{SrCl}}_{\text{2}}:{\text{H}}_2\text{O}& =& \frac{0.514\text{ mol}}{0.514\text{ mol}}:\frac{1.027\text{ mol}}{0.514\text{ mol}}\\ {\text{SrCl}}_{\text{2}}:{\text{H}}_2\text{O}& =& 1:1.998\\ {\text{SrCl}}_{\text{2}}:{\text{H}}_2\text{O}\approx 1:2\end{array}$

Therefore, the water of hydration for the compound ${\text{SrCl}}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is $2$ and the chemical formula of the compound is ${\text{SrCl}}_{\text{2}}\cdot 2{\text{H}}_{\text{2}}\text{O}$.

Conclusion

The water of hydration for the compound ${\text{SrCl}}_{\text{2}}\cdot 2{\text{H}}_{\text{2}}\text{O}$ is calculated as $2$. The chemical formula of compound is ${\text{SrCl}}_{\text{2}}\cdot 2{\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

(b)

Interpretation:

The water of hydration for the compound $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is to be calculated which contains $33.8\%$ water. The chemical formula of the compound $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

The water of crystallization is the number of water molecules present in crystals compounds. The water of crystallization is also known as the water of hydration. The water molecules present in the crystal helps the crystal to impart color. These water molecules can be easily removed from the crystal by heating them.

Answer to Problem 77E

The water of hydration for the compound $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}$ is $6$ and the chemical formula of the compound is $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot 6{\text{H}}_2\text{O}$.

Explanation of Solution

The compound $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ contains $33.8\%$ water. Therefore, $100\text{ g}$ of the compound will have $33.8\text{ g}$ of water in it.

The mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ unit in $100\text{ g}$ of the compound is calculated as shown below.

$\begin{array}{rll}{m}_{\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}& =& 100\text{ g}-33.8\text{ g}\\ & =& 66.2\text{ g}\end{array}$

The molar mass of water is $18.02\text{g}/\text{mol}$.

The molar mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $211.6298 \text{g}/\text{mol}$.

The number of moles of a substance is given by the expression as shown below.

$n=\frac{m}{M}$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the values of mass and molar mass of water in the equation (1).

$\begin{array}{rll}n& =& \frac{33.8\text{ g}}{18.02\text{g}/\text{mol}}\\ & =& 1.876\text{ mol}\end{array}$

Substitute the values of mass and molar mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{66.2\text{ g}}{211.6298\text{g}/\text{mol}}\\ & =& 0.313\text{ mol}\end{array}$

The molar ratio of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}$ can be obtained by dividing the number of moles with the lowest number of mole that is $0.313\text{ mol}$.

$\begin{array}{rll}\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}& =& \frac{0.313\text{ mol}}{0.313\text{ mol}}:\frac{1.876\text{ mol}}{0.313\text{ mol}}\\ \text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}& =& 1:5.994\\ \text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}\approx 1:6\end{array}$

Therefore, the water of hydration for the compound $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}$ is $6$ and the chemcial formula of the compound is $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot 6{\text{H}}_2\text{O}$.

Conclusion

The water of hydration for the compound $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot 6{\text{H}}_2\text{O}$ is calculated as $6$. The chemcial formula of the compound is $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot 6{\text{H}}_2\text{O}$.

Interpretation Introduction

(c)

Interpretation:

The water of hydration for the compound $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is to be calculated which contains $30.5\%$ water. The chemical formula of the compound $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

The water of crystallization is the number of water molecules present in crystals compounds. The water of crystallization is also known as the water of hydration. The water molecules present in the crystal helps the crystal to impart color. These water molecules can be easily removed from the crystal by heating them.

Answer to Problem 77E

The water of hydration for the compound $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is $4$ and the chemical formula of the compound is $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

The compound $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ contains $30.5\%$ water. Therefore, $100\text{ g}$ of the compound will have $30.5\text{ g}$ of water in it.

The mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ unit in $100\text{ g}$ of the compound is calculated as shown below.

$\begin{array}{rll}{m}_{\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}& =& 100\text{ g}-30.5\text{ g}\\ & =& 69.5\text{ g}\end{array}$

The molar mass of water is $18.02\text{g}/\text{mol}$.

The molar mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $\text{164}\text{.0878 }\text{g}/\text{mol}$.

The number of moles of a substance is given by the expression as shown below.

$n=\frac{m}{M}$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the values of mass and molar mass of water in the equation (1).

$\begin{array}{rll}n& =& \frac{30.5\text{ g}}{18.02\text{g}/\text{mol}}\\ & =& 1.693\text{ mol}\end{array}$

Substitute the values of mass and molar mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ in the equation (1).

$\begin{array}{rll}n& =& \frac{69.5\text{ g}}{\text{164}\text{.0878 }\text{g}/\text{mol}}\\ & =& 0.4235\text{ mol}\end{array}$

The molar ratio of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}$ can be obtained by dividing the number of moles with the lowest number of mole that is $0.4235\text{ mol}$.

$\begin{array}{rll}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}& =& \frac{0.4235\text{ mol}}{0.4235\text{ mol}}:\frac{1.693\text{ mol}}{0.4235\text{ mol}}\\ \text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}& =& 1:3.9976\\ \text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}:{\text{H}}_2\text{O}\approx 1:4\end{array}$

Therefore, the water of hydration for the compound $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is $4$ and the chemical formula of the compound is $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$.

Conclusion

The water of hydration for the compound $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ is $4$. The chemical formula of the compound is $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$.

Interpretation Introduction

(d)

Interpretation:

The water of hydration for the compound ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is to be calculated which contains $30.9\%$ water. The chemical formula of the compound ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is to be stated.

Concept introduction:

The water of crystallization is the number of water molecules present in crystals compounds. The water of crystallization is also known as the water of hydration. The water molecules present in the crystal helps the crystal to impart color. These water molecules can be easily removed from the crystal by heating them.

Answer to Problem 77E

The water of hydration for the compound ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is $5$ and the chemical formula of the compound is ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot 5{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

The compound ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot X{\text{H}}_{\text{2}}\text{O}$ contains $30.9\%$ water. Therefore, $100\text{ g}$ of the compound will have $30.9\text{ g}$ of water in it.

The mass of ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}$ unit in $100\text{ g}$ of the compound is calculated as shown below.

$\begin{array}{rll}{m}_{{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}}& =& 100\text{ g}-30.9\text{ g}\\ & =& 69.1\text{ g}\end{array}$

The molar mass of water is $18.02\text{g}/\text{mol}$.

The molar mass of ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}$ is $201.22 \text{g}/\text{mol}$.

The number of moles of a substance is given by the expression as shown below.

$n=\frac{m}{M}$ …(1)

Where,

• $m$ is the mass of the substance.

• $M$ is the molar mass of the substance.

Substitute the values of mass and molar mass of water in the equation (1).

$\begin{array}{rll}n& =& \frac{30.9\text{ g}}{18.02\text{g}/\text{mol}}\\ & =& 1.715\text{ mol}\end{array}$

Substitute the values of mass and molar mass of ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}$ water in the equation (1).

$\begin{array}{rll}n& =& \frac{69.1\text{ g}}{201.22 \text{g}/\text{mol}}\\ & =& 0.343\text{ mol}\end{array}$

The molar ratio of ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}:{\text{H}}_2\text{O}$ can be obtained by dividing the number of moles with the lowest number of mole that is $0.343\text{ mol}$.

$\begin{array}{rll}{\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}:{\text{H}}_2\text{O}& =& \frac{0.343\text{ mol}}{0.343\text{ mol}}:\frac{1.715\text{ mol}}{0.343\text{ mol}}\\ {\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}:{\text{H}}_2\text{O}& =& 1:5\end{array}$

Therefore, the water of hydration for the compound ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot X{\text{H}}_{\text{2}}\text{O}$ is $5$ and the chemical formula of the compound is ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot 5{\text{H}}_{\text{2}}\text{O}$.

Conclusion

The water of hydration for the compound ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot 5{\text{H}}_{\text{2}}\text{O}$ is $5$. The chemical formula of the compound is ${\text{Na}}_{\text{2}}{\text{B}}_{\text{4}}{\text{O}}_{\text{7}}\cdot 5{\text{H}}_{\text{2}}\text{O}$.