To calculate the percent composition of the compound \( \text{C}{12}\text{H}{22}\text{O}_{11} \), we first need to find its molar mass.
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Calculate the molar mass:
- Carbon (C): \( 12.01 , \text{g/mol} \)
- Hydrogen (H): \( 1.008 , \text{g/mol} \)
- Oxygen (O): \( 16.00 , \text{g/mol} \)
Now, we calculate the contribution of each element in the compound:
\[ \text{Molar mass of C}{12} = 12 , \text{C} \times 12.01 , \text{g/mol} = 144.12 , \text{g/mol} \] \[ \text{Molar mass of H}{22} = 22 , \text{H} \times 1.008 , \text{g/mol} = 22.176 , \text{g/mol} \] \[ \text{Molar mass of O}_{11} = 11 , \text{O} \times 16.00 , \text{g/mol} = 176.00 , \text{g/mol} \]
Now, we sum these values to get the total molar mass of the compound:
\[ \text{Total molar mass} = 144.12 , \text{g/mol} + 22.176 , \text{g/mol} + 176.00 , \text{g/mol} = 342.296 , \text{g/mol} \]
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Calculate percent composition:
Now, we can find the percent composition by dividing the mass contribution of each element by the total molar mass and multiplying by 100.
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Percent composition of Carbon (C): \[ \frac{144.12 , \text{g/mol}}{342.296 , \text{g/mol}} \times 100 = 42.13% \]
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Percent composition of Hydrogen (H): \[ \frac{22.176 , \text{g/mol}}{342.296 , \text{g/mol}} \times 100 = 6.46% \]
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Percent composition of Oxygen (O): \[ \frac{176.00 , \text{g/mol}}{342.296 , \text{g/mol}} \times 100 = 51.41% \]
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Final results:
- Percent composition of C: \( 42.13% \)
- Percent composition of H: \( 6.46% \)
- Percent composition of O: \( 51.41% \)
So, the percent composition of \( \text{C}{12}\text{H}{22}\text{O}_{11} \) is approximately:
- C: 42.13%
- H: 6.46%
- O: 51.41%