To determine the empirical formula of the original compound Co\(x\)Cl\(y\)·\(z\)H₂O, we need to calculate the moles of cobalt (Co), chlorine (Cl), and water (H₂O) based on the mass changes observed in the experiment.
Step 1: Calculate the mass of water lost
- Initial mass of salt: 3.865 g
- Final mass after heating: 2.108 g
Mass of water lost (H₂O): \[ \text{Mass of H}_2\text{O} = \text{Initial mass} - \text{Final mass} = 3.865, \text{g} - 2.108, \text{g} = 1.757, \text{g} \]
Step 2: Calculate moles of water lost
Molar mass of H₂O:
- H: 1.01 g/mol × 2 = 2.02 g/mol
- O: 16.00 g/mol \[ \text{Molar mass of H}_2\text{O} = 2.02 + 16.00 = 18.02, \text{g/mol} \] Moles of H₂O: \[ \text{Moles of H}_2\text{O} = \frac{1.757, \text{g}}{18.02, \text{g/mol}} \approx 0.0974, \text{mol} \]
Step 3: Analyze the reaction with Aluminum
The aluminum reacts with cobalt ions to form a cobalt precipitate. The mass of the blue precipitate formed is given as 0.957 g. We have to determine how many moles of Co are in this blue precipitate.
Assuming the precipitate is cobalt(II) hydroxide Co(OH)₂, the molar mass calculation is as follows:
Molar mass of Co(OH)₂:
- Co: 58.93 g/mol
- O: 16.00 g/mol × 2 = 32.00 g/mol
- H: 1.01 g/mol × 2 = 2.02 g/mol \[ \text{Molar mass of Co(OH)}_2 = 58.93 + 32.00 + 2.02 \approx 92.95, \text{g/mol} \] Moles of Co(OH)₂: \[ \text{Moles of Co(OH)}_2 = \frac{0.957, \text{g}}{92.95, \text{g/mol}} \approx 0.0103, \text{mol} \]
Since the cobalt(II) hydroxide is formed from cobalt ions, this means there are also approximately 0.0103 moles of Co in the original salt.
Step 4: Determine the moles of cobalt
Moles of Co: \[ \text{Moles of Co} = 0.0103, \text{mol} \]
Step 5: Calculate moles of chlorine
When cobalt chloride (CoCl₂) dissociates:
- CoCl₂ → Co\(^{2+}\) + 2 Cl\(^{-}\)
This means for every mole of Co, there will be 2 moles of Cl. Therefore: Moles of Cl: \[ \text{Moles of Cl} = 2 \times 0.0103, \text{mol} = 0.0206, \text{mol} \]
Step 6: Calculate the ratio and write the empirical formula
Now we have the approximate moles of the components:
- Co: 0.0103 mol
- Cl: 0.0206 mol
- H₂O: 0.0974 mol
Next, we need to express these in a simple whole-number ratio. To do this, we divide by the smallest number of moles (for Co):
Dividing by Co's moles (0.0103): \[ \text{Ratio of Co: } \frac{0.0103}{0.0103} \approx 1 \] \[ \text{Ratio of Cl: } \frac{0.0206}{0.0103} \approx 2 \] \[ \text{Ratio of H₂O: } \frac{0.0974}{0.0103} \approx 9.45 \approx 9 \text{ (rounding to nearest whole number)} \]
Thus:
- Co = 1
- Cl = 2
- H₂O = 9
Final Empirical Formula
The empirical formula of the original hydrate is: \[ \text{CoCl}_2 \cdot 9\text{H}_2\text{O} \]
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