Question

Determine the percent composition by mass of oxygen in the compound C6H12O6

Answer 1

To determine the percent composition by mass of oxygen in the compound C₆H₁₂O₆ (glucose), we can use the following steps:

1. **Calculate the molar mass of C₆H₁₂O₆**:
- Carbon (C): 6 atoms × 12.01 g/mol = 72.06 g/mol
- Hydrogen (H): 12 atoms × 1.008 g/mol = 12.096 g/mol
- Oxygen (O): 6 atoms × 16.00 g/mol = 96.00 g/mol

Now, we sum these values to obtain the total molar mass of C₆H₁₂O₆:

\[
\text{Molar mass of C₆H₁₂O₆} = 72.06 \, \text{g/mol} + 12.096 \, \text{g/mol} + 96.00 \, \text{g/mol} = 180.156 \, \text{g/mol}
\]

2. **Calculate the mass percent of oxygen**:
- The total mass of oxygen in one mole of C₆H₁₂O₆ is 96.00 g (from the calculation above).
- The mass percent of oxygen is calculated using the formula:

\[
\text{Percent composition of oxygen} = \left( \frac{\text{Mass of oxygen}}{\text{Molar mass of the compound}} \right) \times 100\%
\]

Substituting the values:

\[
\text{Percent composition of oxygen} = \left( \frac{96.00 \, \text{g}}{180.156 \, \text{g/mol}} \right) \times 100\% \approx 53.29\%
\]

Thus, the percent composition by mass of oxygen in glucose (C₆H₁₂O₆) is approximately **53.29%**.