Problem 118 - Chemistry The Molecular Science

Question

Analysis of a coordination compound gives these results: \(22.0 \%\) Co, \(31.4 \% \mathrm{~N}, 6.78 \% \mathrm{H},\) and \(39.8 \% \mathrm{Cl} .\) One mole of the compound dissociates in water to form 4 mol ions. (a) Determine the formula of the compound. (b) Write the equation for its dissociation in water.

Step-by-Step Solution

Answer

(a) The formula is \([Co(NH_3)_6]Cl_3\). (b) The dissociation equation is \([Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-\).

Step 1: Convert Percentage to Masses

Firstly, assume 100g of the compound, which allows us to directly convert percentages to grams. Thus, the compound contains: - 22.0g of Co - 31.4g of N - 6.78g of H - 39.8g of Cl.

Step 2: Convert Masses to Moles

Calculate the number of moles of each element using their atomic masses:- Moles of Co: \(\frac{22.0}{58.93} \approx 0.373 \text{ moles}\)- Moles of N: \(\frac{31.4}{14.01} \approx 2.24 \text{ moles}\)- Moles of H: \(\frac{6.78}{1.008} \approx 6.73 \text{ moles}\)- Moles of Cl: \(\frac{39.8}{35.45} \approx 1.12 \text{ moles}\)

Step 3: Determine Mole Ratio

Divide the moles of each element by the smallest number of moles (0.373 for Co) to find the simplest whole number ratio:- Co: \(\frac{0.373}{0.373} = 1\)- N: \(\frac{2.24}{0.373} \approx 6\)- H: \(\frac{6.73}{0.373} \approx 18\)- Cl: \(\frac{1.12}{0.373} \approx 3\)Thus, the empirical formula is \(\text{CoN}_6\text{H}_{18}\text{Cl}_3\).

Step 4: Account for Coordination and Charge

To determine the molecular formula, consider that the compound forms 4 ions in solution. A possible formula respecting the stoichiometry and charge balance is \([Co(NH_3)_6]Cl_3\). This complex dissociates into \([Co(NH_3)_6]^{3+}\) and \(3 \text{Cl}^-\) ions, totaling 4 ions.

Step 5: Write the Dissociation Equation

The dissociation of the compound in water is represented by:\[[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^-\]

Key Concepts

Empirical Formula

The empirical formula of a compound gives the simplest whole-number ratio of atoms of each element present. For the given coordination compound, the first step is to convert the provided percentages into masses by assuming a 100 g sample of the compound. This assumption makes it easy to equate the percentages directly to grams, which simplifies the process of calculation.
Next, you need to determine the number of moles of each element by dividing the mass of each element by its atomic mass:

• For Cobalt (Co): 22.0 g divided by its atomic mass of 58.93 gives approximately 0.373 moles.
• For Nitrogen (N): 31.4 g divided by 14.01 gives about 2.24 moles.
• For Hydrogen (H): 6.78 g divided by 1.008 gives about 6.73 moles.
• For Chlorine (Cl): 39.8 g divided by 35.45 gives about 1.12 moles.

The next step is to find the simplest whole-number ratio of these moles by dividing by the smallest number (in this case, 0.373 for Co).
This results in the empirical formula for the compound as \(\text{CoN}_6\text{H}_{18}\text{Cl}_3\). This empirical formula shows the simplest ratio, but it does not indicate the actual numbers of atoms in a molecule, which is especially important in coordination compounds.

Mole Ratio

Understanding the mole ratio is crucial when determining the empirical formula, and it gives insight into the proportion of each element in the compound. To accomplish this, we compute the ratio by dividing the number of moles of each element by the smallest mole value obtained.
Here, the smallest mole value is 0.373 moles, which corresponds to Cobalt (Co). Here's how it aligns:

• Cobalt (Co): The ratio is exactly 1 (\(\frac{0.373}{0.373} = 1\)).
• Nitrogen (N): The ratio is approximately 6 (\(\frac{2.24}{0.373} \approx 6\)).
• Hydrogen (H): The ratio is about 18 (\(\frac{6.73}{0.373} \approx 18\)).
• Chlorine (Cl): The ratio turns out to be around 3 (\(\frac{1.12}{0.373} \approx 3\)).

These mole ratios offer a clear picture of the elemental composition compared to one another within the compound, resulting in the empirical formula \(\text{CoN}_6\text{H}_{18}\text{Cl}_3\). This is a critical step in establishing both the empirical and molecular formulas of coordination compounds.

Dissociation Equation

Writing a dissociation equation reveals how a compound breaks down in solution. In the case of the coordination compound, we need to determine how the compound separates into individual ions when it is dissolved.
Considering that 1 mole of the compound produces 4 moles of ions in solution, we account for both the central metal complex and the counter ions. In coordination chemistry, the molecular formula must reflect the charge balance between the cation and anion components.
The compound \([Co(NH_3)_6]Cl_3\) dissociates in water according to the equation:\[[Co(NH_3)_6]Cl_3 \rightarrow [Co(NH_3)_6]^{3+} + 3Cl^- \]This equation shows that the complex dissociates into a cation \([Co(NH_3)_6]^{3+}\) and three chloride anions \(Cl^-\). Each component contributes to the total of 4 ions in solution, maintaining charge balance and illustrating the compound's behavior in water.

Ions in Solution

Coordination compounds often dissociate into ions when dissolved, revealing their structural and chemical attributes. The exercise given describes a compound dissociating into four ions in solution, demonstrating the coordination sphere and ions distinct from the central metallic complex.
When \([Co(NH_3)_6]Cl_3\) dissolves, it dissociates into one complex cation and three anions. The cation, \([Co(NH_3)_6]^{3+}\), consists of a central Co atom surrounded by six ammonia ligands, forming a stable coordination entity.
The chloride ions \( (Cl^-) \) are the counter ions that balance the charge of the cation. Understanding the species formed in solution is crucial for insight into the coordination compound's properties, such as solubility and reactivity. Knowing the behavior of these ions helps in the application of coordination chemistry, including catalysis and biochemical functions.