Question

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 \(0.105 \mathrm{~g}\) of compound \(\mathrm{A}\) was strongly heated in \(\mathrm{ex}\) cess \(\mathrm{O}_{2}, 0.0203 \mathrm{~g} \mathrm{CrO}_{3}\) was formed. ii. In a second experiment it took \(32.93 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) to titrate completely the \(\mathrm{NH}_{3}\) present in \(0.341 \mathrm{~g}\) compound \(\mathrm{A}\). 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 \(\mathrm{A}\) was dissolved in \(10.00 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\left(K_{\mathrm{f}}=\right.\) \(\left.1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\right)\) What is the formula of the compound? What is the structure of the complex ion present? (Hints: \(\mathrm{Cr}^{3+}\) is expected to be sixcoordinate, with \(\mathrm{NH}_{3}\) and possibly \(\mathrm{I}^{-}\) as ligands. The \(\mathrm{I}^{-}\) ions will be the counterions if needed.)

Short answer

The formula of compound A is Cr(NH3)16I10, and the structure of the complex ion present is [Cr(NH3)6]3+ with NH3 and I- as counterions. The compound can be represented as Cr(NH3)6I3·10NH3I7.

Step-by-step solution

Determine the moles of CrO3 formed

We are given that 0.0203 g of CrO3 was formed after heating 0.105 g of compound A. To find the moles of CrO3, we'll first need the molar mass of CrO3: Molar mass of CrO3 = 52.00 g/mol (Cr) + 3 * 16.00 g/mol (O) = 100.00 g/mol. Now we can find the moles of CrO3: Moles of CrO3 = mass / molar mass = 0.0203 g / 100.00 g/mol ≈ 2.03 x 10^(-4) mol.

Determine the moles of Cr in compound A

We know that the amount of Cr is conserved in this reaction, so the moles of Cr in compound A must be equal to the moles of CrO3 formed: Moles of Cr = 2.03 x 10^(-4) mol

Determine the moles of NH3 in compound A

We are given that 32.93 mL of 0.100 M HCl was required to titrate all the NH3 in 0.341 g of compound A. Since the ratio of NH3 to HCl is 1:1 in this reaction, we can find the moles of NH3 as follows: Moles of NH3 = V(HCl) * C(HCl) = 0.03293 L * 0.100 mol/L = 3.293 x 10^(-3) mol NH3

Determine the moles of iodine in compound A

We are given that compound A contains 73.53% iodine by mass. From this information, we can find the mass of iodine in 0.341 g of compound A: Mass of iodine = 0.341 g * 73.53% = 0.2505 g Now, we'll find the moles of iodine: Moles of iodine = mass / molar mass = 0.2505 g / (127 g/mol) ≈ 1.97 x 10^(-3) mol

Determine the empirical formula of compound A

To find the empirical formula, we need to determine the mole ratio of Cr, NH3, and I in compound A. First, we need to normalize the number of moles of each component by the smallest number of moles: Mole ratio of Cr, NH3, and I = 2.03 x 10^(-4) mol : 3.293 x 10^(-3) mol : 1.97 x 10^(-3) mol Mole ratio = 1 : 16.23 : 9.7 Since this ratio isn't a whole number, let's round it to the nearest whole number to get the empirical formula: Empirical formula = Cr(NH3)16I10

Determine the molecular formula of compound A from the freezing point depression

We are given that dissolving 0.601 g of compound A in 10.00 g water results in a freezing point depression of 0.64°C. Using the freezing point depression formula: ΔT = Kf * molality * i Where ΔT is the freezing point depression, Kf is the cryoscopic constant for water, molality is the molality of the solution, and i is the van't Hoff factor, which is equal to the number of ions generated from compound A in the solution. We're given that Kf = 1.86°C kg/mol. First, we need to find the molality: Molality = ΔT / (Kf * i) = 0.64°C / (1.86°C kg/mol * 1) ≈ 0.34 mol/kg Now, we can find the moles of compound A in 0.601 g: Moles of compound A = mass / molar mass We know that the empirical formula is Cr(NH3)16I10. Let's find the molar mass of this empirical formula: Molar mass of Cr(NH3)16I10 = 52.00 g/mol (Cr) + 16 * 17.03 g/mol (NH3) + 10 * 126.9 g/mol (I) = 1671.48 g/mol Now we find the moles of compound A: Moles of compound A = 0.601 g / 1671.48 g/mol ≈ 3.60 x 10^(-4) mol Comparing the moles of Cr in compound A (2.03 x 10^(-4) mol), we can now determine the molecular formula: Cr(NH3)_16I_10 has a 1:1 ratio between Cr and the entire formula-unit, and thus, the molecular formula is equal to the empirical formula: Molecular formula = Cr(NH3)16I10

Determine the structure of the complex ion present

From the molecular formula, we know that the compound A consists of Cr(III) complex with NH3 ligands and I- counterions. It is mentioned that Cr(III) is expected to be six-coordinate. Since there are 16 ammonia molecules, we can propose that: 1. Cr is coordinated with 6 NH3 ligands 2. The remaining 10 NH3 ligands and 10 I- ions act as counterions Thus, the complex ion present in compound A is [Cr(NH3)6]3+ which forms the compound Cr(NH3)6I3·10NH3I7 (Compound A).

Key concepts

Molar Mass Calculation

To understand coordination compounds, calculating the molar mass is crucial. Molar mass is the sum of the atomic masses of all atoms in a molecule. For compound A, it includes calculating each element in the given formula. Let's say we derived the empirical formula: \( \text{Cr(NH}_3\text{)}_6\text{I}_3 \). Each component contributes to the total weight because:

• Cr has an atomic mass of 52.00 g/mol.
• NH3 comprises nitrogen (14.01 g/mol) and hydrogen (3 x 1.01 g/mol), totaling 17.03 g/mol.
• Iodine weighs 126.90 g/mol.

Sum these values, multiplying by the number of each type of atom, to find the total molar mass of compound A.

Empirical Formula Determination

Finding an empirical formula involves determining the simplest ratio of atoms in a compound. Here's the process for compound A:

• Measure the moles of each element: Cr, NH3, and I.
• For Cr, it came from the CrO3 data; for NH3, it was from the titration; for I, from mass percentage.
• Use the smallest mole value to normalize the others into a ratio.
• If the numbers aren't whole, multiply each by a common factor to make integers, crafting the simplest formula.

This process yields the basic composition pattern like \( \text{Cr(NH}_3\text{)}_x\text{I}_y \).

Molecular Formula

The molecular formula reveals the actual number of atoms per molecule, typically a multiple of the empirical formula. From freezing point depression data, you can determine the exact molar mass. Compare this with the empirical mass:

• Use given molar mass to verify the number of empirical units in the molecular form.
• In compound A, the inverse freezing point depression calculation helped to confirm the formula as \( \text{Cr(NH}_3\text{)}_{16}\text{I}_{10} \).

This process ensures the detailed molecular structure is accurate and consistent with observed data.

Freezing Point Depression

A drop in a solvent’s freezing point occurs when a solute is added. This property can help determine molar mass of a solute through the calculation:

• Use \( \Delta T = K_f \times m \times i \), where \( K_f \) is the cryoscopic constant, \( m \) is molality, and \( i \) is the van’t Hoff factor.
• By rearranging, you find the molality from known values then use it to find moles of the compound.
• From the number of moles, calculate molar mass by relating it to the mass of solute used.

This aids in correlating to the compound’s formula, verifying it with known data.

Ligands

Ligands are ions or molecules that bond to a metal atom, forming a coordination complex. In compound A, ammonia (\( \text{NH}_3 \)) acts as a primary ligand:

• Ligands donate electrons to the central metal ion, here \( \text{Cr}^{3+} \).
• This donation forms a stable structure such as \( [\text{Cr(NH}_3\text{)}_6]^{3+} \).
• Iodide ions also play a role but may act as counterions to balance charges outside the coordination sphere.

Understanding ligands and their functions is vital to grasp the nature and properties of coordination compounds.