Chapter 6, Problem 1ASA

A student is given a sample of a pink manganese $\left(\text{II}\right)$ chloride hydrate. She weighs the sample in a dry, covered crucible and obtains a mass of $26.742\text{ g}$ for the crucible, cover, and sample. Earlier she had found that the crucible and cover weighed $23.599\text{ g}$. She then heats the crucible to drive off the water of hydration, keeping the crucible at red heat for about $10$ minutes with the cover slightly ajar. She then lets the crucible cool and finds it has a lower mass; the crucible, cover and contents then weigh $25.598\text{ g}$. In the process the sample was converted to off-white anhydrous ${\text{MnCl}}_{\text{2}}$.

a. What was the mass of the hydrate sample?

$\underset{\_}{}\text{g}\text{hydrate}$

b. What is the mass of the anhydrous ${\text{MnCl}}_{\text{2}}$?

$\underset{\_}{}\text{g}{\text{MnCl}}_2$

c. How much water was driven off?

$\underset{\_}{}\text{g}{\text{H}}_{\text{2}}\text{O}$

d. What is the percent by mass of water in the hydrate?

$\%\text{water}=\frac{\text{mass}\text{of}\text{water}\text{in}\text{sample}}{\text{mass}\text{of}\text{hydrate}\text{in}\text{sample}}\times 100\%$

$\underset{\_}{}{\%}_{\text{mass}}{\text{H}}_2\text{O}$

e. How many grams of water would there be in $100.0\text{ g}$ hydrate? How many moles?

$\underset{\_}{}\text{g}{\text{H}}_{\text{2}}\text{O}$; $\underset{\_}{}\text{moles}{\text{H}}_{\text{2}}\text{O}$

f. How many grams of ${\text{MnCl}}_2$ are there in $100.0\text{ g}$ hydrate? How many moles? (What percentage of the hydrate is ${\text{MnCl}}_2$? Convert the mass of ${\text{MnCl}}_2$ to moles. The molar mass of ${\text{MnCl}}_2$ is $125.85\text{g}/\text{mol}$.)

$\underset{\_}{}\text{g}{\text{MnCl}}_{\text{2}}$; $\underset{\_}{}\text{moles}{\text{MnCl}}_2$

g. How many moles of water are present per mole ${\text{MnCl}}_2$?

$\underset{\_}{}$

h. What is the formula of the hydrate?

$\underset{\_}{}$

Interpretation Introduction

(a)

Interpretation:

The mass of the hydrated sample of pink manganese $\left(\text{II}\right)$ chloride hydrate is to be calculated.

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 1ASA

The mass of the hydrated sample of pink manganese $\left(\text{II}\right)$ chloride hydrate is $3.143\text{g}$.

Explanation of Solution

The sum of the masses of crucible, cover, and sample of a pink manganese $\left(\text{II}\right)$ chloride hydrate is $\left(m_1\right)$ $26.742\text{ g}$.

The sum of the masses of crucible and cover $\left(m_2\right)$ is $23.599\text{ g}$.

The mass of hydrate can be calculated by the difference between these two mass.

The mass of hydrate is given by the formula as shown below.

$\text{Mass of hydrate}=m_1-m_2$

Substitute the value of $m_1$ and $m_2$ in the above equation.

$\begin{array}{rll}\text{Mass of hydrate}& =& 26.742\text{ g}-23.599\text{ g}\\ & =& 3.143\text{g}\end{array}$

Therefore, the mass of the hydrate is $3.143\text{g}$.

Conclusion

The mass of the hydrated sample of pink manganese $\left(\text{II}\right)$ chloride hydrate is calculated as $3.143\text{g}$.

Interpretation Introduction

(b)

Interpretation:

The mass of the anhydrous ${\text{MnCl}}_{\text{2}}$ is to be calculated.

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 1ASA

The mass of the anhydrous ${\text{MnCl}}_{\text{2}}$ is $1.999\text{g}$.

Explanation of Solution

The sum of the masses of crucible, cover, and sample of an anhydrous manganese $\left(\text{II}\right)$ chloride is $\left(m_3\right)$ $25.598\text{ g}$.

The sum of the masses of crucible and cover $\left(m_2\right)$ is $23.599\text{ g}$.

The anhydrous ${\text{MnCl}}_{\text{2}}$ can be calculated by the difference between these two mass.

The mass of the anhydrous ${\text{MnCl}}_{\text{2}}$ is given by the formula as shown below.

$\text{Mass of}\text{anhydrous}=m_3-m_2$

Substitute the value of $m_3$ and $m_2$ in the above equation.

$\begin{array}{rll}\text{Mass of}\text{anhydrous}& =& 25.598\text{ g}-23.599\text{ g}\\ & =& 1.999\text{g}\end{array}$

Therefore, the mass of the anhydrous ${\text{MnCl}}_{\text{2}}$ is $1.999\text{g}$.

Conclusion

The mass of the anhydrous ${\text{MnCl}}_{\text{2}}$ is calculated as $1.999\text{g}$.

Interpretation Introduction

(c)

Interpretation:

The mass of water released from the hydrated sample of ${\text{MnCl}}_{\text{2}}$ is to be calculated.

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 1ASA

The mass of water released from the hydrated sample of ${\text{MnCl}}_{\text{2}}$ is $1.144\text{g}$.

Explanation of Solution

The mass of the anhydrous ${\text{MnCl}}_{\text{2}}$ is calculated as $1.999\text{g}$.

The mass of the hydrated sample of pink manganese $\left(\text{II}\right)$ chloride hydrate is calculated as $3.143\text{g}$.

The mass of water released on heating can be given by the formula as shown below.

$\text{Mass}\text{of}\text{water}=\text{Mass of hydrate}-\text{Mass of}\text{anhydrous}$

Substitute the mass of hydrate and mass of anhydrate in the above equation.

$\begin{array}{rll}\text{Mass}\text{of}\text{water}& =& 3.143\text{g}-1.999\text{g}\\ & =& 1.144\text{g}\end{array}$

Therefore, the mass of water released from the hydrated sample is $1.144\text{g}$.

Conclusion

The mass of water released from the hydrated sample is calculated as $1.144\text{g}$.

Interpretation Introduction

(d)

Interpretation:

The percent by mass of water in the hydrate of ${\text{MnCl}}_{\text{2}}$ is to be calculated.

Concept introduction:

The water of crystallization is the number of water molecules present in crystals 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 1ASA

The percent by mass of water in the hydrate of ${\text{MnCl}}_{\text{2}}$ is $36.3983\%$.

Explanation of Solution

The mass of water released from the hydrated sample is $1.144\text{g}$.

The mass of the hydrated sample of pink manganese $\left(\text{II}\right)$ chloride hydrate is calculated as $3.143\text{g}$.

The percent by mass of water in the hydrate is given by the expression as shown below.

$\%\text{water}=\frac{\text{mass}\text{of}\text{water}\text{in}\text{sample}}{\text{mass}\text{of}\text{hydrate}\text{in}\text{sample}}\times 100\%$

Substitute the values of the mass of water and mass of hydrate in the sample in the above equation.

$\begin{array}{rll}\%\text{water}& =& \frac{1.144\text{g}}{3.143\text{g}}\times 100\%\\ & =& 36.3983\%\end{array}$

Therefore, the percent by mass of water in the hydrate is $36.3983\%$.

Conclusion

The percent by mass of water in the hydrate is calculated as $36.3983\%$.

Interpretation Introduction

(e)

Interpretation:

The mass of water and the number of moles of water in $100.0\text{ g}$ hydrate are to be calculated.

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 1ASA

The mass of water and the number of moles of water in $100.0\text{ g}$ hydrate are $36.3983\text{g}$ and $2.02033\text{mol}$ respectively.

Explanation of Solution

The percent by mass of water in the hydrate is $36.3983\%$. Therefore, $100.0\text{ g}$ hydrate will have $36.3983\text{g}$ of water.

The molar mass of ${\text{H}}_2\text{O}$ is $18.016\text{g}/\text{mol}$.

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

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

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 above equation.

$\begin{array}{rll}n& =& \frac{36.3983\text{g}}{18.016\text{g}/\text{mol}}\\ & =& 2.02033\text{mol}\end{array}$

Therefore, the number of moles of water in $100.0\text{ g}$ hydrate is $2.02033\text{mol}$.

Conclusion

The mass of water and the number of moles of water in $100.0\text{ g}$ hydrate are calculated as $36.3983\text{g}$ and $2.02033\text{mol}$ respectively.

Interpretation Introduction

(f)

Interpretation:

The mass of water and the number of moles of ${\text{MnCl}}_2$ in $100.0\text{ g}$ hydrate are to be calculated. The percentage of ${\text{MnCl}}_{\text{2}}$ in the sample is to be calculated.

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 1ASA

The percent by mass of water in the ${\text{MnCl}}_{\text{2}}$ is $63.6017\%$. The mass of ${\text{MnCl}}_{\text{2}}$ in $100.0\text{ g}$ hydrate is $63.6017\text{g}$. The number of moles of ${\text{MnCl}}_2$ in $100.0\text{ g}$ hydrate is $0.50537\text{mol}$.

Explanation of Solution

The mass of the anhydrous ${\text{MnCl}}_{\text{2}}$ is calculated as $1.999\text{g}$.

The mass of the hydrated sample of pink manganese $\left(\text{II}\right)$ chloride hydrate is calculated as $3.143\text{g}$.

The percent by mass of ${\text{MnCl}}_{\text{2}}$ in the hydrate is given by the expression as shown below.

$\%{\text{MnCl}}_{\text{2}}=\frac{\text{mass}\text{of}{\text{MnCl}}_{\text{2}}\text{in}\text{sample}}{\text{mass}\text{of}\text{hydrate}\text{in}\text{sample}}\times 100\%$

Substitute the values of the mass of ${\text{MnCl}}_{\text{2}}$ and mass of hydrate in the sample in the above equation.

$\begin{array}{rll}\%\text{water}& =& \frac{1.999\text{g}}{3.143\text{g}}\times 100\%\\ & =& 63.6017\%\end{array}$

The percent by mass of water in the ${\text{MnCl}}_{\text{2}}$ is $63.6017\%$. Therefore, $100.0\text{ g}$ hydrate will have $63.6017\text{g}$ of ${\text{MnCl}}_{\text{2}}$.

The molar mass of ${\text{MnCl}}_2$ is $125.85\text{g}/\text{mol}$.

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

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

Where,

• $m$ is the mass of the substance.

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

Substitute the values of the mass and molar mass of ${\text{MnCl}}_2$ in the above equation.

$\begin{array}{rll}n& =& \frac{63.6017\text{g}}{125.85\text{g}/\text{mol}}\\ & =& 0.50537\text{mol}\end{array}$

Therefore, the number of moles of ${\text{MnCl}}_2$ in $100.0\text{ g}$ hydrate is $0.50537\text{mol}$.

Conclusion

The percent by mass of water in the ${\text{MnCl}}_{\text{2}}$ is calculated as $63.6017\%$. The mass of ${\text{MnCl}}_{\text{2}}$ in $100.0\text{ g}$ hydrate is calculated as $63.6017\text{g}$. The number of moles of ${\text{MnCl}}_2$ in $100.0\text{ g}$ hydrate is calculated as $0.50537\text{mol}$.

Interpretation Introduction

(g)

Interpretation:

The number of moles of water present per mole ${\text{MnCl}}_2$ is to be calculated.

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 1ASA

The number of moles of water present per mole ${\text{MnCl}}_2$ is $4$.

Explanation of Solution

The number of moles of ${\text{MnCl}}_2$ in $100.0\text{ g}$ hydrate is $0.50537\text{mol}$.

The number of moles of water in $100.0\text{ g}$ hydrate is $2.02033\text{mol}$.

The number of moles of water present per mole ${\text{MnCl}}_2$ can be calculated by the ratio of the number of moles of water and ${\text{MnCl}}_2$ as shown below.

$\text{Ratio}=\frac{\text{Number of moles of water}}{{\text{Number of moles of MnCl}}_2}$

Substitute the value of the number of moles of water and the number of moles of ${\text{MnCl}}_2$ in the above equation.

$\begin{array}{rll}\text{Ratio}& =& \frac{2.02033\text{mol}}{0.50537\text{mol}}\\ & \approx & 4\end{array}$

Therefore, the number of moles of water present per mole ${\text{MnCl}}_2$ is $4$.

Conclusion

The number of moles of water present per mole ${\text{MnCl}}_2$ is calculated as $4$.

Interpretation Introduction

(h)

Interpretation:

The formula of hydrate of ${\text{MnCl}}_2$ 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 1ASA

The formula of hydrate is ${\text{MnCl}}_2\cdot 4{\text{H}}_{\text{2}}\text{O}$.

Explanation of Solution

The given sample is a pink manganese $\left(\text{II}\right)$ chloride hydrate.

The number of moles of water present per mole ${\text{MnCl}}_2$ in the hydrate is $4$. Four molecules of water will be attached to one ${\text{MnCl}}_2$ unit.

Therefore, the formula of hydrate is ${\text{MnCl}}_2\cdot 4{\text{H}}_{\text{2}}\text{O}$.

Conclusion

The formula of hydrate is ${\text{MnCl}}_2\cdot 4{\text{H}}_{\text{2}}\text{O}$.