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 \(\mathrm{A}\) ), and the following data are collected: i. When \(0.105 \mathrm{~g}\) of compound \(\mathrm{A}\) 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 \(0.100 \mathrm{M} \mathrm{HCl}\) 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 \(0.601\) g compound \(A\) was dissolved in \(10.00 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) \(\left(K_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \cdot \mathrm{kg} / \mathrm{mol}\right)\) 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)6I3, and its structure consists of a complex ion [Cr(NH3)6]^3+ with three I- counterions.

Step-by-step solution

Calculate moles of CrO3

We're given that \(0.105 \mathrm{~g}\) of compound A produced \(\displaystyle 0.0203 \mathrm{~g}\) of \(\displaystyle \mathrm{CrO_{3}}\) upon heating in excess O2. Firstly, determine the moles of CrO3 produced: Molar mass of CrO3 = 52.00 (Cr) + 3(16.00) (O) = 100.00 g/mol Moles of CrO3 = (0.0203 g CrO3) / (100.00 g/mol) = 2.03 × 10^{-4} moles CrO3 Since there is one Cr atom in each CrO3, we have 2.03 × 10^{-4} moles of Cr in compound A.

Calculate moles of NH3

We are also told that it took \(\displaystyle 32.93 \mathrm{~mL}\) \(\displaystyle 0.100 \mathrm{M}\) HCl to titrate the NH3 in \(\displaystyle 0.341 \mathrm{~g}\) of compound A. Convert the titration data into moles of NH3: Moles H+ = Moles NH3 = (0.100 mol/L) × (32.93 × 10^{-3} L) = 3.293 × 10^{-3} moles

Calculate moles of I

We are given that compound A contains \(\displaystyle 73.53 \%\) iodine by mass. Determine the number of moles of I from this percentage: Mass I = 0.7353 * 0.341 g = 0.2508 g I Molar mass I = 126.90 g/mol Moles of I = (0.2508 g I) / (126.90 g/mol) = 1.974 × 10^{-3} moles I

Determine the ratio of Cr:NH3:I in compound A

We now have the moles of each element in the compound A. We can calculate the molar ratios by dividing each value by the smallest value: Molar ratio of Cr : 2.03 × 10^{-4} / 2.03 × 10^{-4} = 1.00 Molar ratio of NH3 : 3.293 × 10^{-3} / 2.03 × 10^{-4} = 16.2 Molar ratio of I : 1.974 × 10^{-3} / 2.03 × 10^{-4} = 9.7 Round the ratios to get whole numbers: CrNH3xIx = CrNH3(16)3 = Cr(NH3)6I3

Deduce the structure of the complex ion

The empirical formula of compound A is Cr(NH3)6I3. The hint tells us that Cr^3+ is expected to be six-coordinate, with NH3 and possibly I- as ligands. The I- ions will be the counterions if needed. Based on the empirical formula, we can propose the structure of the complex ion as [Cr(NH3)6]^3+, with three I- counterions. Thus, the formula of compound A is: Cr(NH3)6I3

Key concepts

Molar Mass Calculation

Understanding the molar mass calculation is key to working with chemical substances, as it enables us to convert mass into moles and vice versa. The molar mass is the mass of one mole of a substance, and it’s numerically equal to the substance’s atomic or molecular weight in grams per mole. For practical calculations, finding the molar mass involves adding up the atomic weights of each element in a compound.

For instance, in the step-by-step solution above, the molar mass of chromium trioxide (CrO3) is calculated by the sum of the atomic weights of one chromium atom (52.00 g/mol) and three oxygen atoms (16.00 g/mol each), totaling 100.00 g/mol. This fundamental step is crucial for later steps, where the moles of CrO3 are calculated by dividing the mass of CrO3 obtained from the experiment by the molar mass previously determined.

It's important to remember that the correctness of further calculations depends on accurate molar mass calculation, as it establishes the foundation for stoichiometry, which is the study of the quantitative relationships between the amounts of reactants and products in a chemical reaction.

Titration

Titration is an analytical technique used in chemistry to determine the concentration of a reactant in a solution. This process typically involves adding a solution of known concentration (the titrant) to a solution of unknown concentration until the reaction reaches an endpoint, which is often indicated by a color change due to an indicator or by reaching a pH that corresponds to the completion of the reaction.

In our example, titration was used to find out the amount of ammonia (NH3) in compound A. The known titrant in this case was hydrochloric acid (HCl), which reacts with NH3 to form ammonium chloride (NH4Cl). Since the concentration of HCl and the volume used are known, we can calculate the moles of HCl, which in a one-to-one reaction, are equal to the moles of NH3 present in the sample. This principle is essential in the field of analytical chemistry and has a diverse range of applications, from environmental monitoring to medical diagnostics.

Colligative Properties

Colligative properties are characteristics of solutions that depend on the number of solute particles present, regardless of their identity. In the context of the exercise, the colligative property in question is the freezing point depression, which is the decrease in the freezing point of a solvent due to the presence of a solute.

The relationship between the freezing point depression and the number of solute particles can be represented by the formula \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the freezing point depression, \( i \) is the van't Hoff factor (which represents the number of particles the compound dissociates into), \( K_f \) is the freezing point depression constant for the solvent, and \( m \) is the molality of the solution.

In the given problem, the mass of compound A and the lowering of the freezing point of water are used to determine the molality and, ultimately, the molar mass of compound A. This property is vital to understand not just in chemistry but also in various real-world applications, such as anti-freeze solutions in cars and the salting of roads during winter to prevent ice formation.