Question
When a sample of hydrate is heated, water of hydration is lost.
If you start with 60.0-gram sample of Na₂CO₃⦁10H₂O, what will be the mass of the sample (in grams) after all of the water is removed by heating?
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If you start with 60.0-gram sample of Na₂CO₃⦁10H₂O, what will be the mass of the sample (in grams) after all of the water is removed by heating?
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Answers
GPT-4o mini
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.
1. **Calculate the molar mass of Na₂CO₃·10H₂O**:
- 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): \( 3 \times 16.00 \, \text{g/mol} = 48.00 \, \text{g/mol} \)
- Total for Na₂CO₃:
\[
45.98 + 12.01 + 48.00 = 105.99 \, \text{g/mol}
\]
- Molar mass of 10H₂O:
- Hydrogen (H): \( 2 \times 10 \times 1.01 \, \text{g/mol} = 20.20 \, \text{g/mol} \)
- Oxygen (O): \( 10 \times 16.00 \, \text{g/mol} = 160.00 \, \text{g/mol} \)
- Total for 10H₂O:
\[
20.20 + 160.00 = 180.20 \, \text{g/mol}
\]
- Total molar mass of Na₂CO₃·10H₂O:
\[
105.99 + 180.20 = 286.19 \, \text{g/mol}
\]
2. **Determine the mass contribution of the anhydrous Na₂CO₃**:
- The mass of Na₂CO₃ in the hydrate can be calculated using the ratio of the molar masses:
\[
\text{mass of Na₂CO₃} = \left( \frac{105.99 \, \text{g/mol}}{286.19 \, \text{g/mol}} \right) \times 60.0 \, \text{g}
\]
- Calculate this value:
\[
\text{mass of Na₂CO₃} = \left( \frac{105.99}{286.19} \right) \times 60.0 \approx 22.18 \, \text{g}
\]
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**.
1. **Calculate the molar mass of Na₂CO₃·10H₂O**:
- 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): \( 3 \times 16.00 \, \text{g/mol} = 48.00 \, \text{g/mol} \)
- Total for Na₂CO₃:
\[
45.98 + 12.01 + 48.00 = 105.99 \, \text{g/mol}
\]
- Molar mass of 10H₂O:
- Hydrogen (H): \( 2 \times 10 \times 1.01 \, \text{g/mol} = 20.20 \, \text{g/mol} \)
- Oxygen (O): \( 10 \times 16.00 \, \text{g/mol} = 160.00 \, \text{g/mol} \)
- Total for 10H₂O:
\[
20.20 + 160.00 = 180.20 \, \text{g/mol}
\]
- Total molar mass of Na₂CO₃·10H₂O:
\[
105.99 + 180.20 = 286.19 \, \text{g/mol}
\]
2. **Determine the mass contribution of the anhydrous Na₂CO₃**:
- The mass of Na₂CO₃ in the hydrate can be calculated using the ratio of the molar masses:
\[
\text{mass of Na₂CO₃} = \left( \frac{105.99 \, \text{g/mol}}{286.19 \, \text{g/mol}} \right) \times 60.0 \, \text{g}
\]
- Calculate this value:
\[
\text{mass of Na₂CO₃} = \left( \frac{105.99}{286.19} \right) \times 60.0 \approx 22.18 \, \text{g}
\]
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**.