Question

Calculate percent composition for the compound C₁₂H₂₂O₁₁

Answer 1

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.

1. **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:

\[
\text{Total molar mass} = 144.12 \, \text{g/mol} + 22.176 \, \text{g/mol} + 176.00 \, \text{g/mol} = 342.296 \, \text{g/mol}
\]

2. **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.

- Percent composition of Carbon (C):
\[
\frac{144.12 \, \text{g/mol}}{342.296 \, \text{g/mol}} \times 100 = 42.13\%
\]

- Percent composition of Hydrogen (H):
\[
\frac{22.176 \, \text{g/mol}}{342.296 \, \text{g/mol}} \times 100 = 6.46\%
\]

- Percent composition of Oxygen (O):
\[
\frac{176.00 \, \text{g/mol}}{342.296 \, \text{g/mol}} \times 100 = 51.41\%
\]

3. **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%