Ammonia and potassium iodide solutions are added to an aqueous solution of \(\mathrm{Cr}\left(\mathrm{NO}_{3}\right)_{3} .\) 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 $\mathrm{O}_{2}, 0.0203 \mathrm{g} \mathrm{CrO}_{3}$ was formed. ii. In a second experiment it took 32.93 \(\mathrm{mL}\) of 0.100$M \mathrm{HCl}\( to titrate completely the \)\mathrm{NH}_{3}\( present in \)0.341 \mathrm{g} \mathrm{com}-$ pound A. iii. Compound A was found to contain 73.53\(\%\) iodine by mass. iv. The freezing point of water was lowered by \(0.64^{\circ} \mathrm{C}\) when 0.601 \(\mathrm{g}\) compound A was dissolved in 10.00 $\mathrm{g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\( \)1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol} )$ What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be six-coordinate, with \(\mathrm{NH}_{3}\) and possibly I- - as ligands. The I- ions will be the counterions if needed.)

Step by step solution

01

Determine the moles of CrO₃ formed

We are given that 0.105 g of compound A was strongly heated in excess O₂, and 0.0203 g CrO₃ was formed. To find the moles of CrO₃, we use the molar mass of CrO₃ (151.99 g/mol): Moles of CrO₃ = mass / molar mass Moles of CrO₃ = 0.0203 g / 151.99 g/mol = 1.33 x 10⁻⁴ moles

02

Calculate the moles of Cr present in compound A

Since only one Chromium atom is present in CrO₃, there is an equal number of moles of Chromium in compound A: Moles of Cr = 1.33 x 10⁻⁴ moles

03

Calculate the moles of NH₃ present in compound A

It took 32.93 mL of 0.100 M HCl to titrate completely the NH₃ present in 0.341 g compound A. We can calculate the moles of NH₃ in compound A by using the volume and concentration of HCl: Moles of NH₃ = 32.93 mL * 0.100 mol/L = 3.293 x 10⁻³ moles

04

Calculate the moles of I present in compound A

Compound A was found to contain 73.53 % iodine by mass. We can use this percentage to calculate the mass of iodine in 0.105 g of compound A: Mass of I = 0.105 g * 0.7353 = 0.0772 g To find the moles of I, we must use the molar mass of I (126.90 g/mol): Moles of I = 0.0772 g / 126.90 g/mol = 6.09 x 10⁻⁴ moles

05

Determine the molecular formula of compound A

To determine the molecular formula of compound A, we need to find the ratio of moles of Cr, NH₃, and I in the compound. To simplify, we will divide each mole value by the smallest value (1.33 x 10⁻⁴ moles Cr): - Cr: (1.33 x 10⁻⁴) / (1.33 x 10⁻⁴) = 1 - NH₃: (3.293 x 10⁻³) / (1.33 x 10⁻⁴) = 24.74 ≈ 25 - I: (6.09 x 10⁻⁴) / (1.33 x 10⁻⁴) = 4.58 ≈ 5 Therefore, the molecular formula of compound A is approximately Cr(NH₃)₂₅I₅.

06

Determine the structure of the complex ion present

We are given a hint that Cr³⁺ is expected to be six-coordinate, with NH₃ and possibly I⁻ as ligands. The I⁻ ions will be the counterions if needed. Since there are a large number of ammonia ligands (25 of them) in the molecular formula, it seems likely that the complex ion present in compound A contains Cr³⁺ surrounded by six ammonia ligands, forming [Cr(NH₃)₆]³⁺. The I⁻ ions will provide charge balance, and the resulting formula would be [Cr(NH₃)₆]I₃, but this formula doesn't conform fully to the stoichiometry found in step 5. The specific arrangement of ammonia ligands and iodide ions in compound A may not be easily determined, but we can infer that the complex ion in compound A is [Cr(NH₃)₆]³⁺ or a similar complex ion containing chromium with six-coordinate ammonia ligands.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!