Chapter 13, Problem 20PE

(a)

Interpretation Introduction

Interpretation:

Mass of water in $125\text{ g}{\text{AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}$ has to be calculated.

Concept Introduction:

Mass percent is one of the commonly used concentration terms to determine concentration of any species. The expression for mass percent of any species present in sample is as follows:

  $\text{Mass percent}=\left(\frac{\text{Mass of species}}{\text{Mass of sample}}\right)\left(100\text{ \%}\right)$

Answer to Problem 20PE

Mass of water in $125\text{ g}{\text{AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}$ is $55.98\text{ g}$.

Explanation of Solution

Molecular mass of ${\text{AlCl}}_3$ is $133.34\text{ g/mol}$ and ${\text{H}}_{\text{2}}\text{O}$ is $18.02\text{ g/mol}$. Expression to calculate total mass of water is calculated as follows:

  $\text{Total mass of water}=\left(\text{Number of molecules of water}\right)\left(\text{Mass of water}\right)$        (1)

Substitute 6 for number of molecules of water and $18.02\text{ g/mol}$ for mass of water in equation (1).

  $\begin{array}{c}\text{Total mass of water}=\left(\text{6}\right)\left(18.02\text{ g/mol}\right)\\ =108.12\text{ g/mol}\end{array}$

The formula to calculate mass percent of water is as follows:

  $\text{Mass percent of water}=\left(\frac{{\text{Total mass of H}}_{\text{2}}\text{O}}{\left({\text{molecular mass of AlCl}}_3\right)\text{+}\left({\text{total mass of H}}_{\text{2}}\text{O}\right)}\right)\left(\text{100 \%}\right)$        (2)

Substitute $108.12\text{ g/mol}$ for total mass of ${\text{H}}_{\text{2}}\text{O}$ and $133.34\text{ g/mol}$ for molecular mass of ${\text{AlCl}}_3$ in equation (2).

  $\begin{array}{c}\text{Mass percent of water}=\left(\frac{108.12\text{ g/mol}}{\left(133.34\text{ g/mol}\right)\text{+}\left(108.12\text{ g/mol}\right)}\right)\left(\text{100 \%}\right)\\ =\left(\frac{108.12\text{ g/mol}}{241.46\text{ g/mol}}\right)\left(\text{100 \%}\right)\\ =44.78\text{ \%}\end{array}$

Hence, mass percent of water in ${\text{AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}$ is $44.78\text{ \%}$.

The formula to calculate mass of water is as follows:

  $\text{Mass of water}=\frac{\left({\text{Mass of AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}\right)\left(\text{Mass percent of water}\right)}{\text{100 \%}}$        (3)

Substitute $125\text{ g}$ for mass of ${\text{AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}$ and $44.78\text{ \%}$ for mass percent of water in equation (3).

  $\begin{array}{c}\text{Mass of water}=\frac{\left(125\text{ g}\right)\left(44.78\text{ \%}\right)}{\text{100 \%}}\\ =55.98\text{ g}\end{array}$

Hence, mass of water in $125\text{ g}$ ${\text{AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}$ is $55.98\text{ g}$.

(b)

Interpretation Introduction

Interpretation:

Mass of anhydrous compound of ${\text{AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}$ has to be calculated.

Concept Introduction:

Compound that does not contain any water molecule is termed as anhydrous compound. Mass of anhydrous compound is calculated as follows:

  $\text{Mass of anhydrous compound}=\left(\text{Mass of hydrated compond}\right)-\left(\text{Mass of water}\right)$

Answer to Problem 20PE

Mass of anhydrous compound is $\text{69}\text{.02 g}$.

Explanation of Solution

The expression used to calculate mass of anhydrous compound is as follows:

  $\text{Mass of anhydrous salt}=\left({\text{Mass of AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}\right)-\left(\text{Mass of water}\right)$        (4)

Substitute $125\text{ g}$ for ${\text{AlCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}$ and $55.98\text{ g}$ for $\text{mass of water}$.

  $\begin{array}{c}\text{Mass of anhydrous salt}=\text{125 g}-55.98\text{ g}\\ =\text{69}\text{.02 g}\end{array}$

Hence, mass of anhydrous compound is $\text{69}\text{.02 g}$.