Chapter 11, Problem 76E

Interpretation Introduction

(a)

Interpretation:

The percentage of water in ${\text{SrCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$ 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 76E

The percentage of water in ${\text{SrCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$ is $\text{40}\text{.55}\text{\%}$.

Explanation of Solution

The given hydrate is ${\text{SrCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$.

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 chlorine is $\text{35}\text{.45}\text{g}/\text{mol}$.

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 ${\text{SrCl}}_{\text{2}}$ is calculated by the formula given below.

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

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

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{Mass}\text{of}\text{Sr}+\text{2}\left(\text{Mass}\text{of}\text{Cl}\right)\\ & =& \text{87}\text{.62}\text{g}/\text{mol}+2\left(\text{35}\text{.45}\text{g}/\text{mol}\right)\\ & =& 158.52\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of ${\text{SrCl}}_{\text{2}}$ is $158.52\text{g}/\text{mol}$.

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.

$\begin{array}{rll}\text{Percentage of water}& =& \frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}\\ & =& \frac{6\left(18.02\text{g}/\text{mol}\right)}{158.52\text{g}/\text{mol}+6\left(18.02\text{g}/\text{mol}\right)}\times \text{100}\text{\%}\\ & =& \text{40}\text{.55}\text{\%}\end{array}$

Therefore, the percentage of water in ${\text{SrCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$ is $\text{40}\text{.55}\text{\%}$.

Conclusion

The percentage of water in ${\text{SrCl}}_{\text{2}}\cdot {\text{6H}}_{\text{2}}\text{O}$ is $\text{40}\text{.55}\text{\%}$.

Interpretation Introduction

(b)

Interpretation:

The percentage of water in ${\text{K}}_2{\text{Cr}}_{\text{2}}{\text{O}}_7\cdot {\text{2H}}_{\text{2}}\text{O}$ 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 76E

The percentage of water in ${\text{K}}_2{\text{Cr}}_{\text{2}}{\text{O}}_7\cdot {\text{2H}}_{\text{2}}\text{O}$ is $\text{10}\text{.91}\text{\%}$.

Explanation of Solution

The given hydrate is ${\text{K}}_2{\text{Cr}}_{\text{2}}{\text{O}}_7\cdot {\text{2H}}_{\text{2}}\text{O}$.

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 potassium is $\text{39}\text{.098}\text{g}/\text{mol}$.

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 ${\text{K}}_2{\text{Cr}}_{\text{2}}{\text{O}}_7$ is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{2}\left(\text{Mass}\text{of}\text{K}\right)+\text{2}\left(\text{Mass}\text{of}\text{Cr}\right)+7\left(\text{Mass}\text{of}\text{O}\right)$

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

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{2}\left(\text{Mass}\text{of}\text{K}\right)+\text{2}\left(\text{Mass}\text{of}\text{Cr}\right)+7\left(\text{Mass}\text{of}\text{O}\right)\\ & =& 2\left(\text{39}\text{.098}\text{g}/\text{mol}\right)+2\left(\text{51}\text{.996}\text{g}/\text{mol}\right)+7\left(\text{16}\text{g}/\text{mol}\right)\\ & =& 294.188\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of ${\text{K}}_2{\text{Cr}}_{\text{2}}{\text{O}}_7$ is $294.188\text{g}/\text{mol}$.

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.

$\begin{array}{rll}\text{Percentage of water}& =& \frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}\\ & =& \frac{2\left(18.02\text{g}/\text{mol}\right)}{294.188\text{g}/\text{mol}+2\left(18.02\text{g}/\text{mol}\right)}\times \text{100}\text{\%}\\ & =& \text{10}\text{.91}\text{\%}\end{array}$

Therefore, the percentage of water in ${\text{K}}_2{\text{Cr}}_{\text{2}}{\text{O}}_7\cdot {\text{2H}}_{\text{2}}\text{O}$ is $\text{10}\text{.91}\text{\%}$.

Conclusion

The percentage of water in ${\text{K}}_2{\text{Cr}}_{\text{2}}{\text{O}}_7\cdot {\text{2H}}_{\text{2}}\text{O}$ is $\text{10}\text{.91}\text{\%}$.

Interpretation Introduction

(c)

Interpretation:

The percentage of water in $\text{Co}{\left(\text{CN}\right)}_{\text{3}}\cdot 3{\text{H}}_{\text{2}}\text{O}$ 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 76E

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

Explanation of Solution

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

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 nitrogen is $\text{14}\text{.01}\text{g}/\text{mol}$.

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 $\text{Co}{\left(\text{CN}\right)}_{\text{3}}$ is calculated by the formula given below.

$\text{Molar}\text{mass}=\text{Mass}\text{of}\text{Co}+\text{3}\left(\text{Mass}\text{of}\text{C}\right)+3\left(\text{Mass}\text{of}\text{N}\right)$

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

$\begin{array}{rll}\text{Molar}\text{mass}& =& \text{Mass}\text{of}\text{Co}+\text{3}\left(\text{Mass}\text{of}\text{C}\right)+3\left(\text{Mass}\text{of}\text{N}\right)\\ & =& \text{58}\text{.93}\text{g}/\text{mol}+3\left(\text{12}\text{.01}\text{g}/\text{mol}\right)+3\left(\text{14}\text{.01}\text{g}/\text{mol}\right)\\ & =& 136.99\text{g}/\text{mol}\end{array}$

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

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.

$\begin{array}{rll}\text{Percentage of water}& =& \frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}\\ & =& \frac{3\left(18.02\text{g}/\text{mol}\right)}{136.99\text{g}/\text{mol}+3\left(18.02\text{g}/\text{mol}\right)}\times \text{100}\text{\%}\\ & =& \text{28}\text{.3}\text{\%}\end{array}$

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

Conclusion

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

Interpretation Introduction

(d)

Interpretation:

The percentage of water in ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ 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 76E

The percentage of water in ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ is $\text{30}\text{.80}\text{\%}$.

Explanation of Solution

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

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 sodium is $\text{22}\text{.99}\text{g}/\text{mol}$.

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 ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is calculated by the formula given below.

$\text{Molar}\text{mass}=2\left(\text{Mass}\text{of}\text{Na}\right)+\text{Mass}\text{of}\text{Cr}+4\left(\text{Mass}\text{of}\text{O}\right)$

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

$\begin{array}{rll}\text{Molar}\text{mass}& =& 2\left(\text{Mass}\text{of}\text{Na}\right)+\text{Mass}\text{of}\text{Cr}+4\left(\text{Mass}\text{of}\text{O}\right)\\ & =& 2\left(\text{22}\text{.99}\text{g}/\text{mol}\right)+\text{51}\text{.996}\text{g}/\text{mol}+4\left(\text{16}\text{g}/\text{mol}\right)\\ & =& 161.976\text{g}/\text{mol}\end{array}$

Therefore, the molar mass of ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $161.976\text{g}/\text{mol}$.

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.

$\begin{array}{rll}\text{Percentage of water}& =& \frac{\text{Mass}\text{of}\text{water}}{\text{Mass}\text{of}\text{hydrate}}\times \text{100}\text{\%}\\ & =& \frac{4\left(18.02\text{g}/\text{mol}\right)}{161.976\text{g}/\text{mol}+4\left(18.02\text{g}/\text{mol}\right)}\times \text{100}\text{\%}\\ & =& \text{30}\text{.80}\text{\%}\end{array}$

Therefore, the percentage of water in ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ is $\text{30}\text{.80}\text{\%}$.

Conclusion

The percentage of water in ${\text{Na}}_{\text{2}}{\text{CrO}}_{\text{4}}\cdot 4{\text{H}}_{\text{2}}\text{O}$ is $\text{30}\text{.80}\text{\%}$.