To calculate the mass of the sample after all the water is removed, we first need to determine the mass of the anhydrous sodium carbonate (Na₂CO₃) remaining after the water of hydration is lost.
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Calculate the molar mass of Na₂CO₃·10H₂O:
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Molar mass of Na₂CO₃:
- Sodium (Na): \( 2 \times 22.99 , \text{g/mol} = 45.98 , \text{g/mol} \)
- Carbon (C): \( 1 \times 12.01 , \text{g/mol} = 12.01 , \text{g/mol} \)
- Oxygen (O): \( 3 \times 16.00 , \text{g/mol} = 48.00 , \text{g/mol} \)
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Total for Na₂CO₃: \[ 45.98 + 12.01 + 48.00 = 105.99 , \text{g/mol} \]
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Molar mass of 10H₂O:
- Hydrogen (H): \( 2 \times 10 \times 1.01 , \text{g/mol} = 20.20 , \text{g/mol} \)
- Oxygen (O): \( 10 \times 16.00 , \text{g/mol} = 160.00 , \text{g/mol} \)
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Total for 10H₂O: \[ 20.20 + 160.00 = 180.20 , \text{g/mol} \]
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Total molar mass of Na₂CO₃·10H₂O: \[ 105.99 + 180.20 = 286.19 , \text{g/mol} \]
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Determine the mass contribution of the anhydrous Na₂CO₃:
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The mass of Na₂CO₃ in the hydrate can be calculated using the ratio of the molar masses: \[ \text{mass of Na₂CO₃} = \left( \frac{105.99 , \text{g/mol}}{286.19 , \text{g/mol}} \right) \times 60.0 , \text{g} \]
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Calculate this value: \[ \text{mass of Na₂CO₃} = \left( \frac{105.99}{286.19} \right) \times 60.0 \approx 22.18 , \text{g} \]
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Therefore, after heating and removing all water from the original 60.0-gram sample of Na₂CO₃·10H₂O, the mass of the remaining sample (anhydrous Na₂CO₃) is approximately 22.18 grams.