Question

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

When 0.105g of compound A was strongly heated in excess O₂, 0.0203g of CrO₃was formed.

In a second experiment, it took 32.93 mL of 0.100M HCl to titrate completely the NH₃present IN 0.341g of compound A

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

The freezing point of water was lowered by 0.64 ^oC when 0.601g of compound A was dissolved in 10.00g of H₂O.

(K_f= 1.86 pC kg/mol)

What is the formula of the compound? What is the structure of the complex ion present? [Hints: Cr³⁺is expected to be six-coordinate with NH₃and (possibly) I^- acting as ligands. The I^- ions will be the counter ions if needed.]

Short answer

The empirical formula of the compound is$\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{I}}_3$

Step-by-step solution

Part i)

First, calculate the mass of Cr present.

$0.203\mathrm{g}{\mathrm{CrO}}_3\times \frac{52.00\mathrm{g}\mathrm{Cr}}{100.0\mathrm{g}{\mathrm{CrO}}_3}=0.0106\mathrm{Cr}$

Part ii)

Next, calculate the mass % of Cr

$\frac{0.0106\mathrm{g}}{0.105\mathrm{g}}\times 100=10.1\%\mathrm{Cr}$

Part iii)

Next calculate the amount of NH₃present:

$32.93\mathrm{mL}\mathrm{HCl}\times \frac{0.100\mathrm{mmol}\mathrm{HCl}}{\mathrm{ml}}\times \frac{1\mathrm{mmol}{\mathrm{NH}}_3}{1\mathrm{mmol}\mathrm{HCl}}=56.1\%$

Next, calculate the mass % of ammonia

$\frac{56.1\mathrm{mg}}{341\mathrm{mg}}\times 100=16.5\%$

Part iv)

Next, combine the mass % values to check if other elements are needed

74.53% +16.5% +10.1% = 100%

No other elements present.

Part v)

Next, assume you have 100.00g of compound

$$\begin{gathered} 10.1\mathrm{g}\mathrm{Cr}\times \frac{1\mathrm{mol}}{52.00\mathrm{g}}=0.194\mathrm{mol} \\ 16.5\mathrm{g}{\mathrm{NH}}_3\times \frac{1\mathrm{mol}}{17.03\mathrm{g}}=0.969\mathrm{mol} \\ 74.54\mathrm{g}\mathrm{I}\times \frac{1\mathrm{mol}}{126.9\mathrm{g}}=0.5794\mathrm{mol} \end{gathered}$$

Divide each molar amount by the lowest molar amount present.

$$\begin{gathered} \frac{0.194}{0.194}=1 \\ \frac{0.969}{0.194}=4.99 \\ \frac{0.5794}{0.194}=2.99 \end{gathered}$$

This means that $\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_5{\mathrm{I}}_3$is the empirical formula. Cr(III) forms octahedral complexes. This means that compound A is made of the octahedral${\left[\mathrm{Cr}\left({\mathrm{NH}}_3\right)\mathrm{I}\right]}^{2+}$ complex ion and two I- as counter ions. The formula is then$\left[\mathrm{Cr}{\left({\mathrm{NH}}_3\right)}_6\mathrm{I}\right]{\mathrm{I}}_2$