Glucose has the formula C6H12O6. What is the percent composition of oxygen in glucose to three significant digits?
The correct answer and explanation is :
To calculate the percent composition of oxygen in glucose (C₆H₁₂O₆), we need to determine the mass of oxygen in one mole of glucose and then divide it by the molar mass of glucose. After that, we multiply by 100 to find the percentage of oxygen in glucose.
Step 1: Calculate the Molar Mass of Glucose (C₆H₁₂O₆)
The molar mass of glucose is the sum of the atomic masses of its constituent elements:
- Carbon (C): 12.01 g/mol
- Hydrogen (H): 1.008 g/mol
- Oxygen (O): 16.00 g/mol
Now, let’s calculate the total molar mass of glucose:
[
\text{Molar mass of glucose} = (6 \times 12.01) + (12 \times 1.008) + (6 \times 16.00)
]
[
\text{Molar mass of glucose} = 72.06 + 12.096 + 96.00 = 180.156 \, \text{g/mol}
]
Step 2: Calculate the Mass of Oxygen in One Mole of Glucose
There are 6 oxygen atoms in one molecule of glucose (C₆H₁₂O₆), and each oxygen atom has a molar mass of 16.00 g/mol. Therefore, the total mass of oxygen in one mole of glucose is:
[
\text{Mass of oxygen in glucose} = 6 \times 16.00 = 96.00 \, \text{g/mol}
]
Step 3: Calculate the Percent Composition of Oxygen in Glucose
To find the percent composition of oxygen, we use the formula:
[
\text{Percent composition of oxygen} = \left( \frac{\text{Mass of oxygen}}{\text{Molar mass of glucose}} \right) \times 100
]
[
\text{Percent composition of oxygen} = \left( \frac{96.00}{180.156} \right) \times 100 \approx 53.34\%
]
Final Answer:
The percent composition of oxygen in glucose is 53.3% (rounded to three significant digits).
Explanation:
The percent composition is a way of expressing how much of a compound’s mass is contributed by each element. By calculating the molar mass of glucose and the mass of oxygen within that, we can determine that approximately 53.34% of glucose’s mass comes from oxygen atoms. This type of calculation is fundamental in chemistry, especially in stoichiometry and molecular composition analysis.