To determine the formula for the hydrate of cerium(III) iodine (CeI₃), we first need to find the number of water molecules (H₂O) per formula unit of CeI₃.
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Calculate the molar masses:
- Molar mass of Ce (Cerium) = 140.12 g/mol
- Molar mass of I (Iodine) = 126.90 g/mol
- Molar mass of H (Hydrogen) = 1.01 g/mol
- Molar mass of O (Oxygen) = 16.00 g/mol
Molar mass of CeI₃: \[ Molar\ mass\ of\ CeI₃ = 140.12 + (3 \times 126.90) = 140.12 + 380.70 = 520.82\ g/mol \]
Molar mass of H₂O: \[ Molar\ mass\ of\ H₂O = (2 \times 1.01) + 16.00 = 2.02 + 16.00 = 18.02\ g/mol \]
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Percentage composition gives the mass in a 100 g sample:
- Mass of CeI₃ in the hydrate = 76.3 g
- Mass of H₂O in the hydrate = 23.7 g
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Convert these masses to moles:
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Moles of CeI₃: \[ \text{Moles of CeI₃} = \frac{76.3\ g}{520.82\ g/mol} \approx 0.146\ mol \]
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Moles of H₂O: \[ \text{Moles of H₂O} = \frac{23.7\ g}{18.02\ g/mol} \approx 1.315\ mol \]
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Find the ratio of moles:
- To find the ratio of moles of H₂O to moles of CeI₃: \[ \text{Ratio} = \frac{1.315\ mol\ H₂O}{0.146\ mol\ CeI₃} \approx 9 \]
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Writing the formula of the hydrate: Thus, the formula for the hydrate can be written as: \[ \text{CeI}_3 \cdot 9\text{H}_2\text{O} \]
The formula for the hydrate of cerium(III) iodine is CeI₃ · 9H₂O.