Chapter 11, Problem 75E

Interpretation Introduction

(a)

Interpretation:

The percentage of water in ${\text{MnSO}}_{\text{4}}\cdot {\text{H}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

A compound in which water molecules are chemically attached to an atom in a compound is known as a hydrate. In inorganic hydrates, the water molecules can be removed by heating the compound. The percentage of water in a hydrate can be calculated by the formula given below.

$\text{Percentage of water}=\frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}$

Answer to Problem 75E

The percentage of water in ${\text{MnSO}}_{\text{4}}\cdot {\text{H}}_{\text{2}}\text{O}$ is $10.66\text{\%}$.

Explanation of Solution

The given hydrate is ${\text{MnSO}}_{\text{4}}\cdot {\text{H}}_{\text{2}}\text{O}$.

The molar mass of hydrogen is $\text{1}\text{.01}\text{g}/\text{mol}$.

The molar mass of oxygen is $\text{16}\text{g}/\text{mol}$.

The molar mass of sulphur is $\text{32}\text{g}/\text{mol}$.

The molar mass of manganese is $\text{54}\text{.93}\text{g}/\text{mol}$.

The molar mass of water is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{2}\left(\text{Mass}\text{of}\text{H}\right)+\text{Mass}\text{of}\text{O}$

Substitute the values of mass of hydrogen and mass of oxygen in the above expression.

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{2}\left(\text{Mass}\text{of}\text{H}\right)+\text{Mass}\text{of}\text{O}\\ & =& 2\left(\text{1}\text{.01}\text{g}/\text{mol}\right)+\text{16}\text{g}/\text{mol}\\ & =& 2.02\text{g}/\text{mol}+\text{16}\text{g}/\text{mol}\\ & =& 18.02\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of water is $18.02\text{g}/\text{mol}$.

The molar mass of ${\text{MnSO}}_{\text{4}}$ is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{Mass}\text{of}\text{Mn}+\text{Mass}\text{of}\text{S}+4\left(\text{Mass}\text{of}\text{O}\right)$

Substitute the values of masses of $\text{Mn}$, $\text{S}$and $\text{O}$ in the above expression.

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{Mass}\text{of}\text{Mn}+\text{Mass}\text{of}\text{S}+\text{4}\left(\text{Mass}\text{of}\text{O}\right)\\ & =& \text{54}\text{.9}\text{g}/\text{mol}+32\text{g}/\text{mol}+4\left(\text{16}\text{g}/\text{mol}\right)\\ & =& 150.9\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of ${\text{MnSO}}_{\text{4}}$ is $150.9\text{g}/\text{mol}$.

The percentage of water in a hydrate can be calculated by the formula given below.

$\text{Percentage of water}=\frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}$

Substitute the values of mass of water and mass of hydrate in the above expression.

$\begin{array}{rll}\text{Percentage of water}& =& \frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}\\ & =& \frac{18.02}{150.9\text{g}/\text{mol}+18.02\text{g}/\text{mol}}\times \text{100}\text{\%}\\ & =& 10.66\text{\%}\end{array}$

Therefore, the percentage of water in ${\text{MnSO}}_{\text{4}}{\text{.H}}_{\text{2}}\text{O}$ is $10.66\text{\%}$.

Conclusion

The percentage of water in ${\text{MnSO}}_{\text{4}}{\text{.H}}_{\text{2}}\text{O}$ is $10.66\text{\%}$.

Interpretation Introduction

(b)

Interpretation:

The percentage of water in $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_2\cdot 6{\text{H}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

A compound in which water molecules are chemically attached to an atom in a compound is known as a hydrate. In inorganic hydrates, the water molecules can be removed by heating the compound. The percentage of water in a hydrate can be calculated by the formula given below.

$\text{Percentage of water}=\frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}$

Answer to Problem 75E

The percentage of water in $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_2\cdot 6{\text{H}}_{\text{2}}\text{O}$ is $\text{33}\text{.81}\text{\%}$.

Explanation of Solution

The given hydrate is $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_2\cdot 6{\text{H}}_{\text{2}}\text{O}$.

The molar mass of hydrogen is $\text{1}\text{.01}\text{g}/\text{mol}$.

The molar mass of oxygen is $\text{16}\text{g}/\text{mol}$.

The molar mass of nitrogen is $\text{14}\text{g}/\text{mol}$.

The molar mass of strontium is $\text{87}\text{.62}\text{g}/\text{mol}$.

The molar mass of water is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{2}\left(\text{Mass}\text{of}\text{H}\right)+\text{Mass}\text{of}\text{O}$

Substitute the values of mass of hydrogen and mass of oxygen in the above expression.

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{2}\left(\text{Mass}\text{of}\text{H}\right)+\text{Mass}\text{of}\text{O}\\ & =& 2\left(\text{1}\text{.01}\text{g}/\text{mol}\right)+\text{16}\text{g}/\text{mol}\\ & =& 2.02\text{g}/\text{mol}+\text{16}\text{g}/\text{mol}\\ & =& 18.02\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of water is $18.02\text{g}/\text{mol}$.

The molar mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_2$ is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{Mass}\text{of}\text{Sr}+2\left(\text{Mass}\text{of}\text{N}\right)+6\left(\text{Mass}\text{of}\text{O}\right)$

Substitute the values of masses of $\text{Sr}$, $\text{N}$, and $\text{O}$ in the above expression.

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{Mass}\text{of}\text{Sr}+2\left(\text{Mass}\text{of}\text{N}\right)+6\left(\text{Mass}\text{of}\text{O}\right)\\ & =& 87.62\text{g/mol}+2\left(14\text{g}/\text{mol}\right)+6\left(\text{16}\text{g}/\text{mol}\right)\\ & =& 211.62\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $273.62\text{g}/\text{mol}$.

The percentage of water in a hydrate can be calculated by the formula given below.

$\text{Percentage of water}=\frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}$

Substitute the values of mass of water and mass of hydrate in the above expression.

$\begin{array}{rll}\text{Percentage of water}& =& \frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}\\ & =& \frac{6\left(18.02\text{g}/\text{mol}\right)}{211.62\text{g}/\text{mol}+6\left(18.02\text{g}/\text{mol}\right)}\times \text{100}\text{\%}\\ & =& \text{33}\text{.81}\text{\%}\end{array}$

Therefore, the percentage of water in $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_2\cdot 6{\text{H}}_{\text{2}}\text{O}$ is $\text{33}\text{.81}\text{\%}$.

Conclusion

The percentage of water in $\text{Sr}{\left({\text{NO}}_{\text{3}}\right)}_2\cdot 6{\text{H}}_{\text{2}}\text{O}$ is $\text{33}\text{.81}\text{\%}$.

Interpretation Introduction

(c)

Interpretation:

The percentage of water in $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

A compound in which water molecules are chemically attached to an atom in a compound is known as a hydrate. In inorganic hydrates, the water molecules can be removed by heating the compound. The percentage of water in a hydrate can be calculated by the formula given below.

$\text{Percentage of water}=\frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}$

Answer to Problem 75E

The percentage of water in $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ is $\text{28}\text{.93}\text{\%}$.

Explanation of Solution

The given hydrate is $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$.

The molar mass of hydrogen is $\text{1}\text{.01}\text{g}/\text{mol}$.

The molar mass of oxygen is $\text{16}\text{g}/\text{mol}$.

The molar mass of cobalt is $\text{58}\text{.93}\text{g}/\text{mol}$.

The molar mass of carbon is $\text{12}\text{.01}\text{g}/\text{mol}$.

The molar mass of water is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{2}\left(\text{Mass}\text{of}\text{H}\right)+\text{Mass}\text{of}\text{O}$

Substitute the values of mass of hydrogen and mass of oxygen in the above expression.

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{2}\left(\text{Mass}\text{of}\text{H}\right)+\text{Mass}\text{of}\text{O}\\ & =& 2\left(\text{1}\text{.01}\text{g}/\text{mol}\right)+\text{16}\text{g}/\text{mol}\\ & =& 2.02\text{g}/\text{mol}+\text{16}\text{g}/\text{mol}\\ & =& 18.02\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of water is $18.02\text{g}/\text{mol}$.

The molar mass of $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{Mass}\text{of}\text{Co}+4\left(\text{Mass}\text{of}\text{C}\right)+6\left(\text{Mass}\text{of}\text{H}\right)+4\left(\text{Mass}\text{of}\text{O}\right)$

Substitute the values of masses of $\text{Co}$, $\text{C}$ and $\text{N}$ in the above expression.

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{Mass}\text{of}\text{Co}+4\left(\text{Mass}\text{of}\text{C}\right)+6\left(\text{Mass}\text{of}\text{H}\right)+4\left(\text{Mass}\text{of}\text{O}\right)\\ & =& \text{58}\text{.93}\text{g}/\text{mol}+4\left(\text{12}\text{.01}\text{g}/\text{mol}\right)+6\left(\text{1}\text{.01}\text{g}/\text{mol}\right)+4\left(16\text{g}/\text{mol}\right)\\ & =& 177\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}$ is $177\text{g}/\text{mol}$$136.99\text{g}/\text{mol}$.

The percentage of water in a hydrate can be calculated by the formula given below.

$\text{Percentage of water}=\frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}$

Substitute the values of mass of water and mass of hydrate in the above expression.

$\begin{array}{rll}\text{Percentage of water}& =& \frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}\\ & =& \frac{4\left(18.02\text{g}/\text{mol}\right)}{177\text{g}/\text{mol}+4\left(18.02\text{g}/\text{mol}\right)}\times \text{100}\text{\%}\\ & =& \text{28}\text{.93}\text{\%}\end{array}$

Therefore, the percentage of water in $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ is $\text{28}\text{.93}\text{\%}$.

Conclusion

The percentage of water in $\text{Co}{\left({\text{C}}_{\text{2}}{\text{H}}_{\text{3}}{\text{O}}_{\text{2}}\right)}_{\text{2}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ is $\text{28}\text{.93}\text{\%}$.

Interpretation Introduction

(d)

Interpretation:

The percentage of water in $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{.9H}}_{\text{2}}\text{O}$ is to be calculated.

Concept introduction:

A compound in which water molecules are chemically attached to an atom in a compound is known as a hydrate. In inorganic hydrates, the water molecules can be removed by heating the compound. The percentage of water in a hydrate can be calculated by the formula given below.

$\text{Percentage of water}=\frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}$

Answer to Problem 75E

The percentage of water in $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{.9H}}_{\text{2}}\text{O}$ is $\text{40}\text{.52}\text{\%}$.

Explanation of Solution

The given hydrate is $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{.9H}}_{\text{2}}\text{O}$.

The molar mass of hydrogen is $\text{1}\text{.01}\text{g}/\text{mol}$.

The molar mass of oxygen is $\text{16}\text{g}/\text{mol}$.

The molar mass of nitrogen is $\text{14}\text{g}/\text{mol}$.

The molar mass of chromium is $\text{51}\text{.996}\text{g}/\text{mol}$.

The molar mass of water is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{2}\left(\text{Mass}\text{of}\text{H}\right)+\text{Mass}\text{of}\text{O}$

Substitute the values of mass of hydrogen and mass of oxygen in the above expression.

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{2}\left(\text{Mass}\text{of}\text{H}\right)+\text{Mass}\text{of}\text{O}\\ & =& 2\left(\text{1}\text{.01}\text{g}/\text{mol}\right)+\text{16}\text{g}/\text{mol}\\ & =& 2.02\text{g}/\text{mol}+\text{16}\text{g}/\text{mol}\\ & =& 18.02\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of water is $18.02\text{g}/\text{mol}$.

The molar mass of $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{Mass}\text{of}\text{Cr}+3\left(\text{Mass}\text{of}\text{N}\right)+9\left(\text{Mass}\text{of}\text{O}\right)$

Substitute the values of masses of $\text{Na}$, $\text{Cr}$ and $\text{O}$ in the above expression.

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{Mass}\text{of}\text{Cr}+3\left(\text{Mass}\text{of}\text{N}\right)+9\left(\text{Mass}\text{of}\text{O}\right)\\ & =& \text{51}\text{.996}\text{g}/\text{mol}+3\left(14\text{g}/\text{mol}\right)+9\left(\text{16}\text{g}/\text{mol}\right)\\ & =& 237.996\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is $237.996\text{g}/\text{mol}$.

The percentage of water in a hydrate can be calculated by the formula given below.

$\text{Percentage of water}=\frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}$

Substitute the values of mass of water and mass of hydrate in the above expression.

$\begin{array}{rll}\text{Percentage of water}& =& \frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}\\ & =& \frac{9\left(18.02\text{g}/\text{mol}\right)}{237.996\text{g}/\text{mol}+9\left(18.02\text{g}/\text{mol}\right)}\times \text{100}\text{\%}\\ & =& \text{40}\text{.52}\text{\%}\end{array}$

Therefore, the percentage of water in $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{.9H}}_{\text{2}}\text{O}$ is $\text{40}\text{.52}\text{\%}$.

Conclusion

The percentage of water in $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}{\text{.9H}}_{\text{2}}\text{O}$ is $\text{40}\text{.52}\text{\%}$.