To solve the problem step by step, we will follow the logical sequence outlined in the video transcript.
### Step 1: Calculate the moles of the compound
Given:
- Volume of the solution = 50 mL = 0.050 L
- Molarity (M) = 0.2 M
To find the moles of the compound, we use the formula:
\[
\text{Moles} = \text{Volume (L)} \times \text{Molarity (M)}
\]
\[
\text{Moles} = 0.050 \, \text{L} \times 0.2 \, \text{mol/L} = 0.010 \, \text{mol} = 10 \, \text{mmol}
\]
### Step 2: Calculate the moles of AgCl formed
Given:
- Mass of AgCl = 1.435 g
- Molar mass of AgCl = 143.5 g/mol
To find the moles of AgCl, we use the formula:
\[
\text{Moles of AgCl} = \frac{\text{Mass of AgCl}}{\text{Molar mass of AgCl}}
\]
\[
\text{Moles of AgCl} = \frac{1.435 \, \text{g}}{143.5 \, \text{g/mol}} \approx 0.010 \, \text{mol} = 10 \, \text{mmol}
\]
### Step 3: Relate moles of AgCl to moles of chloride ions
From the reaction, we know that:
- Each mole of AgCl corresponds to one mole of Cl⁻ ions.
Thus, the moles of chloride ions (Cl⁻) provided by the compound is also:
\[
\text{Moles of Cl⁻} = \text{Moles of AgCl} = 10 \, \text{mmol}
\]
### Step 4: Determine the formula of the compound
The empirical formula of the compound is given as CoCl₃·4NH₃. We know that:
- The empirical formula contains 3 moles of Cl⁻ ions.
Since we have found that the compound provides 10 mmol of Cl⁻ ions, we can determine the number of empirical formula units in the compound:
\[
\text{Moles of Cl⁻ from empirical formula} = 3 \, \text{(from CoCl₃)}
\]
To find the number of empirical units (n):
\[
n = \frac{\text{Total moles of Cl⁻}}{\text{Moles of Cl⁻ from empirical formula}} = \frac{10 \, \text{mmol}}{3 \, \text{mmol}} \approx 3.33
\]
This indicates that the compound contains approximately 3 times the empirical formula.
### Conclusion
Thus, the molecular formula of the compound is:
\[
\text{Molecular formula} = \text{(CoCl}_3\cdot4\text{NH}_3)_3 = \text{Co}_3\text{Cl}_9\cdot12\text{NH}_3
\]
### Final Answer
The formula of the compound is \( \text{Co}_3\text{Cl}_9\cdot12\text{NH}_3 \).
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