Chapter 21, Problem 107IP

Ammonia and potassium iodide solutions are added to an aqueous solution of Cr(NO₃)₃. A solid is isolated (compound A), and the following data are collected:

i. When 0.105 g of compound A was strongly heated in excess 0 2, 0.0203 g CrO₃ was formed.

ii. In a second experiment it took 32.93 mL of 0.100 M HCI to titrate completely the NH₃ present in 0.341 g compound A.

iii. Compound A was found to contain 73.53% iodine by mass.

iv. The freezing point of water was lowered by 0.64°C when 0.601 g compound A was dissolved in 10.00 g H₂O (K_f =1.86°C·kg/mol).

What is the formula of the compound? What is the structure of the complex ion present? (Hints: Cr³⁺ is expected to be sixcoordinate, with NH₃ and possibly I⁻ as ligands. The I⁻ ions will be the counterions if needed.)

Interpretation Introduction

Interpretation: The formation of compound A by the addition of ammonia and potassium iodide solutions to an aqueous solution of $\text{Cr}{\left({\text{NO}}_{\text{3}}\right)}_{\text{3}}$ is given. The formula of the compound and the structure of complex ion present on the basis of given information is to be stated.

Concept introduction: Electronic configuration is used to describe the distribution of the electrons in the orbitals of an atom. Structure of an atom can be defined by its electronic configuration. It can also be used to denote an atom which is ionized to a cation or anion formed by the loss or gain of electrons in their respective orbitals.

To determine: The formula of the compound A and the structure of complex ion present on the basis of given information.

Answer to Problem 107IP

Answer

The formula of the compound is $\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{I}\right]{\text{I}}_2$ and the complex ion is octahedral ${\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{I}\right]}^{2+}$.

Explanation of Solution

Explanation

The mass percent of $\text{Cr}$ is $\underset{\_}{10.1\%Cr}$.

The molar mass of $\text{Cr}$ is $51.99\text{g}$.

The molar mass of ${\text{CrO}}_3$ is $99.99\text{g}$.

In $99.99\text{g}$ ${\text{CrO}}_3$, the mass of $\text{Cr}$ present is $51.99\text{g}$.

In $0.0203\text{g}$ ${\text{CrO}}_3$, the mass of $\text{Cr}$ present $\begin{array}{c}=\frac{51.99\times 0.0203}{99.99}\\ =0.0106\text{g}\text{Cr}\end{array}$

Mass percent of $\text{Cr}$ is calculated by the formula,

$\text{Mass}\text{percent}=\frac{\text{Mass}\text{of}\text{Cr}}{\text{Mass}\text{of}\text{compound}\text{A}}\times 100\%$

The mass of compound A is $\text{0}\text{.105}\text{g}$.

The mass of $\text{Cr}$ is $0.0106\text{g}\text{Cr}$.

Substitute the value of mass of compound A and $\text{Mass}\text{of}\text{Cr}$ in the above formula.

$\begin{array}{c}\text{Mass}\text{percent}=\frac{0.0106\text{g}}{\text{0}\text{.105}\text{g}}\times 100\%\\ =\underset{\_}{10.1\%Cr}\end{array}$

The mass percent of ${\text{NH}}_{\text{3}}$ is $\underset{\_}{16.5\%N{H}_3}$.

One $\text{mL}$ $\text{HCl}$ equals to $\text{0}\text{.100}\text{HCl}$.

Therefore, $\text{32}\text{.93}\text{mL}\text{HCl}$ is equal to $\text{32}\text{.93}\times \text{0}\text{.100}\text{mL}\text{HCl}$

Molar mass of ${\text{NH}}_3$ is $\text{17}\text{.04}\text{g}$.

One $\text{mL}$ of $\text{HCl}$ titrate $17.04\text{mg}{\text{NH}}_3$.

Therefore, the mass of ${\text{NH}}_3$ titrated by $\text{32}\text{.93}\times \text{0}\text{.100}\text{mL}\text{HCl}$ is calculated as,

$17.04\times \text{32}\text{.93}\times 0.100=56.1\text{g}{\text{NH}}_3$

Mass percent of ${\text{NH}}_3$ is calculated by the formula,

$\text{Mass}\text{percent}=\frac{\text{Mass}\text{of}{\text{NH}}_3}{\text{Mass}\text{of}\text{compound}\text{A}}\times 100\%$

The mass of compound A is this case is $\text{0}\text{.341}\text{g}$.

The mass of ${\text{NH}}_3$ is $56.1\text{g}$.

Substitute the value of mass of compound A and mass of ${\text{NH}}_3$ in the above formula.

$\begin{array}{c}\text{Mass}\text{percent}=\frac{56.1\text{g}}{\text{0}\text{.341}\text{g}}\times 100\%\\ =\underset{\_}{16.5\%N{H}_3}\end{array}$

Except chromium, ammonia and iodine, no other element is needed in compound A.

Given

The mass percent of iodine is $73.53\%$.

To check whether any other element is needed or not, mass percent of $\text{Cr}$, ${\text{NH}}_3$ and iodine are added shown in the formula given below.

$\text{Mass}\text{percent}\text{of}\text{Cr}+\text{Mass}\text{percent}\text{of}{\text{NH}}_{\text{3}}+\text{Mass}\text{percent}\text{of}\text{I}$

Substitute the values of mass percent in the above formula.

$74.53\%+16.5\%+10.1\%=100\%$

Therefore, no other element is required in the compound A.

The number of moles of $\text{Cr}$, ${\text{NH}}_3$ and iodine is calculated below.

The compound A is assumed to have mass $100\text{g}$.

The number of moles is calculated by the formula,

$\text{Number}\text{of}\text{moles}=\frac{\text{Given}\text{mass}}{\text{Molar}\text{mass}}$

One mole of $\text{Cr}$ have mass $\text{51}\text{.99}\text{g}$.

Therefore, number of moles of $\text{Cr}$ having mass $10.1\text{g}$ $\begin{array}{l}=\frac{10.1}{52}\\ =0.194\text{mol}\end{array}$

One mole of ${\text{NH}}_3$ have mass $17.03\text{g}$.

Therefore, number of moles of ${\text{NH}}_3$ having mass $16.5\text{g}$ $\begin{array}{l}=\frac{16.5}{17.03}\\ =0.969\text{mol}\end{array}$

One mole of iodine have mass $\text{126}\text{.9}\text{g}$.

Therefore, number of moles of iodine having mass $73.53\text{g}$ $\begin{array}{l}=\frac{73.53}{126.9}\\ =0.5794\text{mol}\end{array}$

The formula of the compound is $\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{I}\right]{\text{I}}_2$ and the complex ion is octahedral ${\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{I}\right]}^{2+}$.

Each molar amount is divided by lowest molar amount for calculating the empirical formula.

The lowest molar amount is $0.194\text{mol}$.

The number of $\text{Cr}$ atoms is calculated as,

$\text{Number}\text{of}\text{Cr}\text{atoms}=\frac{\text{Moles}\text{of}\text{Cr}}{\text{Lowest}\text{molar}\text{amount}}$

Number of moles of $\text{Cr}$ is $0.194$. Substitute the value of moles of $\text{Cr}$ and lowest molar amount in the above formula.

$\begin{array}{c}\text{Number}\text{of}\text{Cr}\text{atoms}=\frac{\text{Moles}\text{of}\text{Cr}}{\text{Lowest}\text{molar}\text{amount}}\\ \frac{0.194}{0.194}=1\end{array}$

The number of ${\text{NH}}_3$ ligands is calculated by the formula,

$\text{Number}\text{of}{\text{NH}}_{\text{3}}\text{ligands}=\frac{\text{Moles}\text{of}{\text{NH}}_3}{\text{Lowest}\text{molar}\text{amount}}$

Number of moles of ${\text{NH}}_3$ is $0.969$. Substitute the value of moles of ${\text{NH}}_3$ and lowest molar amount in the above formula.

$\begin{array}{c}\text{Number}\text{of}{\text{NH}}_{\text{3}}\text{ligands}=\frac{\text{Moles}\text{of}{\text{NH}}_3}{\text{Lowest}\text{molar}\text{amount}}\\ \frac{0.969}{0.194}=4.99\\ \approx 5\end{array}$

The number of iodine ligands is calculated by the formula,

$\text{Number}\text{of}\text{iodine}\text{ligands}=\frac{\text{Moles}\text{of}\text{iodine}}{\text{Lowest}\text{molar}\text{amount}}$

Number of moles of iodine is $0.5794$. Substitute the value of moles of iodine and lowest molar amount in the above formula.

$\begin{array}{c}\text{Number}\text{of}\text{iodine}\text{ligands}=\frac{\text{Moles}\text{of}\text{iodine}}{\text{Lowest}\text{molar}\text{amount}}\\ \frac{0.5794}{0.194}=2.99\\ \approx 3\end{array}$

This indicates that the empirical formula is $\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}{\text{I}}_{\text{3}}$. This means that ${\text{Cr}}^{\text{3+}}$ is expected to be six coordinated with ${\text{NH}}_3$ and ${\text{I}}^{-}$ as ligands. Therefore the compound A is composed of ${\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{I}\right]}^{2+}$ complex ion and iodine ions acts as counter ions. Therefore, the formula of compound is, $\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{I}\right]{\text{I}}_2$.

Conclusion

Conclusion

The formula of the compound is $\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{I}\right]{\text{I}}_2$ and the complex ion is octahedral ${\left[\text{Cr}{\left({\text{NH}}_{\text{3}}\right)}_{\text{5}}\text{I}\right]}^{2+}$.