Chapter 13, Problem 23PE

Interpretation Introduction

Interpretation:

Empirical formula of hydrated iron chloride has to be determined.

Concept Introduction:

The steps required for calculation of empirical formula is as follows:

1 Convert the data provided of each element in grams if not given in grams. If data is provided in percentage than total mass of compound is taken as $100\text{ g}$.

2 Mass of each element is changed into moles by division of given mass of each element with its atomic mass respectively.

3 The value of mole of each element is divided by the smallest mole value and round off the resultant value to positive integer.

4 If resultant value is in fraction then multiply with smallest number to make it integer.

5 The final value of mole is used as a subscript for empirical formula.

Explanation of Solution

The hydrated iron chloride contains $20.66\text{ g Fe}$, $39.35\text{ g Cl}$ and $39.99{\text{ g H}}_2\text{O}$. Therefore the empirical formula is calculated as follows:

Step 1: The percentage $"\%"$ is replaced by $"\text{g"}$. Thus, $100\text{ g}$ hydrated iron chloride contains $20.66\text{ g Fe}$, $39.35\text{ g Cl}$ and $39.99{\text{ g H}}_2\text{O}$.

Step 2: The expression to calculate moles of elements is as follows:

  $\text{ Moles}=\frac{\text{given mass}}{\text{molar mass}}$

The formula to calculate moles of $\text{ Fe}$ is as follows:

  $\text{ Moles}\left(\text{Fe}\right)=\frac{\text{mass of Fe}}{\text{atomic mass of Fe}}$        (1)

Substitute $20.66\text{ g}$ for mass of $\text{ Fe}$ and $55.85\text{ g/mol}$ for molar mass of $\text{ Fe}$ in equation (1).

  $\begin{array}{c}\text{ Moles}\left(\text{Fe}\right)=\frac{20.66\text{ g}}{\text{55}\text{.85 g/mol}}\\ =0.37\text{ mol}\end{array}$

The formula to calculate moles of $\text{Cl}$ is as follows:

  $\text{ Moles}\left(\text{Cl}\right)=\frac{\text{mass of Cl}}{\text{atomic mass of Cl}}$        (2)

Substitute $39.35\text{ g}$ for mass of $\text{Cl}$ and $35.5\text{ g/mol}$ for molar mass of $\text{Cl}$ in equation (2).

  $\begin{array}{c}\text{ Moles}\left(\text{Cl}\right)=\frac{39.35\text{ g}}{\text{35}\text{.5 g/mol}}\\ =1.11\text{ mol}\end{array}$

The formula to calculate moles of ${\text{H}}_2\text{O}$ is as follows:

  $\text{ Moles}\left({\text{H}}_2\text{O}\right)=\frac{{\text{mass of H}}_2\text{O}}{{\text{molar mass of H}}_2\text{O}}$        (3)

Substitute $39.99\text{ g}$ for mass of ${\text{H}}_2\text{O}$ and $18.02\text{ g/mol}$ for molar mass of ${\text{H}}_2\text{O}$ in equation (3).

  $\begin{array}{c}\text{ Moles}\left({\text{H}}_2\text{O}\right)=\frac{39.99\text{ g}}{18.02\text{ g/mol}}\\ =2.22\text{ mol}\end{array}$

Step 3: The positive integer for $\text{ Fe}$ is calculated as follows:

  $\text{Subscript for Fe}=\frac{\text{moles of Fe }}{\text{smallest mole value}}$        (4)

Substitute $0.37\text{ mol}$ for moles of $\text{ Fe}$ and $0.37\text{ mol}$ for smallest mole value in equation (4).

  $\begin{array}{c}\text{Subscript for Fe}=\frac{0.37\text{ mol}}{0.37\text{ mol}}\\ =1.0\end{array}$

The positive integer for $\text{Cl}$ is calculated as follows:

  $\text{Subscript for Cl}=\frac{\text{moles of Cl }}{\text{smallest mole value}}$        (5)

Substitute $1.11\text{ mol}$ for moles of $\text{Cl}$ and $0.37\text{ mol}$ for smallest mole value in equation (5).

  $\begin{array}{c}\text{Subscript for Cl}=\frac{1.11\text{ mol }}{0.37\text{ mol}}\\ \simeq 3\end{array}$

The positive integer for ${\text{H}}_2\text{O}$ is calculated as follows:

  ${\text{Subscript for H}}_2\text{O}=\frac{{\text{moles of H}}_2\text{O }}{\text{smallest mole value}}$        (6)

Substitute $2.22\text{ mol}$ for moles of ${\text{H}}_2\text{O}$ and $0.37\text{ mol}$ for smallest mole value in equation (6).

  $\begin{array}{c}{\text{Subscript for H}}_2\text{O}=\frac{2.22\text{ mol }}{0.37\text{ mol}}\\ \simeq 6.0\end{array}$

Therefore, the empirical formula for hydrated iron chloride is ${\text{FeCl}}_3\cdot 6{\text{H}}_{\text{2}}\text{O}$.