To determine the empirical formula of the original compound (Co\(_x\)Cl\(_y\)·zH\(_2\)O), we need to analyze the mass changes and the reactions involving the salt. Here's a step-by-step breakdown of the calculations:
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Calculation of Water Loss:
- Initial mass of the salt = 3.865 g
- Mass of the salt after heating = 2.108 g
- Mass of water lost (zH\(_2\)O) = Initial mass - Mass after heating \[ \text{Mass of water lost} = 3.865\ \text{g} - 2.108\ \text{g} = 1.757\ \text{g} \]
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Calculation of Moles of Water:
- Molar mass of water (H\(_2\)O) = 18.015 g/mol \[ \text{Moles of water} = \frac{1.757\ \text{g}}{18.015\ \text{g/mol}} \approx 0.0975\ \text{mol} \]
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Amount of Cobalt Retrieved:
- After the heating, the salt left is anhydrous (Co\(_x\)Cl\(_y\)).
- When aluminum metal is added to the solution, it reduces Co\(^{2+}\) ions to Co\(^0\), causing the blue precipitate to be formed, which we delve into next.
- Suppose that the blue precipitate is Co(OH)\(_2\) or a similar compound, which can be deduced from its blue color and typical cobalt chemistry. Let's assume it is Co(OH)\(_2\). The molar mass of Co(OH)\(_2\) is approximately 92.94 g/mol (Co = 58.93 + 2(16.00 + 1.01)).
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Calculate Moles of Blue Precipitate: \[ \text{Moles of Co(OH)}_2 = \frac{0.957\ \text{g}}{92.94\ \text{g/mol}} \approx 0.0103\ \text{mol} \]
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Stoichiometry of the Reaction and Moles of Cobalt:
- Each mole of Co(OH)\(_2\) contains 1 mole of cobalt. Thus, moles of cobalt (Co) in the precipitate = 0.0103 mol.
- Hence, mass of cobalt from precipitate = 0.0103 mol × 58.93 g/mol ≈ 0.606 g.
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Calculating Moles of Chloride Ions:
- Assume the cobalt salt originally formed is CoCl\(_2\)·zH\(_2\)O (this is only one possibility, but we could check for other halides if necessary).
- After heating, the remaining mass can be considered as CoCl\(_2\).
- Remaining mass = 2.108 g, which corresponds to CoCl\(_2\).
- Let’s calculate the mass of Co in 2.108 g of CoCl\(_2\):
- Molar mass of CoCl\(_2\) = 58.93 + 2(35.45) = 129.83 g/mol.
Using the mole ratio logic: \[ \text{Total moles of CoCl}_2 = \frac{2.108\ \text{g}}{129.83\ g/mol} \approx 0.0163\ \text{mol} \]
- Therefore, moles of chloride = \(2 \times 0.0163\) (because there are 2 chlorides in CoCl\(_2\)). \[ \text{Moles of Cl} = 0.0326\ \text{mol} \]
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Final Empirical Formula Calculation:
- From cobalt we have roughly 0.0103 mol and from chloride we have 0.0326 mol.
- To find the simplest whole number ratio, divide by smallest mole value: \[ \text{For Co:} \frac{0.0103}{0.0103} = 1 \] \[ \text{For Cl:} \frac{0.0326}{0.0103} \approx 3.16 \approx 3 \]
- Moles of water already calculated as \(0.0975\).
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Conclusion - Empirical Formula:
- Thus, the empirical formula considering the calculated values is: \[ \text{Co}_1\text{Cl}_3\cdot\text{zH}_2\text{O} \]
Therefore, the empirical formula of the compound can be best expressed as CoCl\(_3\)·zH\(_2\)O, with z corresponding to the nearest whole number based on prior calculations indicating z approximating to 4. This leads us to CoCl\(_3\)·4H\(_2\)O if confirmed.
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