Chapter 13, Problem 24PE

Interpretation Introduction

Interpretation:

Empirical formula of hydrated nickel 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 nickel chloride contain $24.69\text{ \% Ni}$, $29.83\text{ \% Cl}$ and $45.48{\text{ \% H}}_2\text{O}$. Therefore the empirical formula is calculated as follows:

Step 1: The percentage $"\%"$ is replaced by $"\text{g"}$. Thus, $100\text{ g}$ hydrated iron chloride contain $24.69\text{ g Ni}$, $29.83\text{ g Cl}$ and $45.48{\text{ g H}}_2\text{O}$.

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

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

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

  $\text{ Moles}\left(\text{Ni}\right)=\frac{\text{mass of Ni}}{\text{molar mass of Ni}}$        (1)

Substitute $24.69\text{ g}$ for mass of $\text{Ni}$ and $58.69\text{ g/mol}$ for molar mass of $\text{Ni}$ in equation (1).

  $\begin{array}{c}\text{ Moles}\left(\text{Ni}\right)=\frac{24.69\text{ g}}{58.69\text{ g/mol}}\\ =\text{ 0}\text{.42 mol}\end{array}$

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

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

Substitute $29.83\text{ g}$ for mass of $\text{Cl}$ and $35.5\text{ g/mol}$ for molar mass of $\text{Cl}$ in equation (2).

  $\begin{array}{c}\text{ Moles}\left(\text{Cl}\right)=\frac{29.83\text{ g}}{\text{35}\text{.5 g/mol}}\\ =0.84\text{ mol}\end{array}$

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 $45.48\text{ g}$ 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).

  $\begin{array}{c}\text{ Moles}\left({\text{H}}_2\text{O}\right)=\frac{45.48\text{ g}}{18.02\text{ g/mol}}\\ =2.52\text{ mol}\end{array}$

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

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

Substitute $\text{0}\text{.42 mol}$ for moles of $\text{Ni}$ and $\text{0}\text{.42 mol}$ for smallest mole value in equation (4).

  $\begin{array}{c}\text{Subscript for Ni}=\frac{0.42\text{ mol }}{0.42\text{ mol}}\\ \simeq 1.0\end{array}$

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 $0.84\text{ mol}$ for moles of $\text{Cl}$ and $\text{0}\text{.42 mol}$ for smallest mole value in equation (5).

  $\begin{array}{c}\text{Subscript for Cl}=\frac{0.84\text{ mol }}{0.42\text{ mol}}\\ \simeq 2.0\end{array}$

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 $2.52\text{ mol}$ for moles of ${\text{H}}_2\text{O}$ and $\text{0}\text{.42 mol}$ for smallest mole value in equation (6).

  $\begin{array}{c}{\text{Subscript for H}}_2\text{O}=\frac{2.52\text{ mol }}{\text{0}\text{.42 mol}}\\ \simeq 6.0\end{array}$

Therefore, the empirical formula for hydrated iron chloride is ${\text{NiCl}}_2\cdot 6{\text{H}}_{\text{2}}\text{O}$.