To find out how much water of hydration will be lost when heating a sample of sodium carbonate decahydrate (Na₂CO₃·10H₂O), we first need to determine the molar mass of the compound and the water it contains.
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Calculate the molar mass of Na₂CO₃·10H₂O:
- Molar mass of Na (Sodium) = 22.99 g/mol
- Molar mass of C (Carbon) = 12.01 g/mol
- Molar mass of O (Oxygen) = 16.00 g/mol
- Molar mass of H₂O (Water) = 18.02 g/mol
Now, calculate the total for Na₂CO₃ and 10H₂O:
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Molar mass of Na₂CO₃ = 2(22.99) + 12.01 + 3(16.00) = 45.98 + 12.01 + 48.00 = 105.99 g/mol
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Molar mass of 10H₂O = 10 × 18.02 = 180.20 g/mol
Therefore, the total molar mass of Na₂CO₃·10H₂O is: \[ 105.99 + 180.20 = 286.19 \text{ g/mol} \]
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Calculate the mass fraction of water in Na₂CO₃·10H₂O: The mass of water in one mole of Na₂CO₃·10H₂O is 180.20 g (from the 10H₂O part).
To determine the fraction of the hydrate that is water: \[ \text{Mass fraction of water} = \frac{180.20 \text{ g}}{286.19 \text{ g}} \approx 0.629 \]
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Determine the amount of water lost from 60.0 g of Na₂CO₃·10H₂O: Multiply the mass fraction of water by the mass of the hydrate sample: \[ \text{Mass of water lost} = 60.0 \text{ g} \times 0.629 \approx 37.74 \text{ g} \]
Thus, when heating the 60.0-gram sample of Na₂CO₃·10H₂O, approximately 37.74 grams of water will be lost.