Calculating the Percentage of Carbon by Mass in Glucose

Key Points

• Molecular Formula of Glucose: C₆H₁₂O₆
• Objective: Calculate the percentage of carbon by mass in a sample of pure glucose
• Atomic Masses Provided:
  • Carbon (C): 12.01 g/mol
  • Hydrogen (H): 1.008 g/mol
  • Oxygen (O): 16.00 g/mol

Steps to Solve

1. Assumptions:

   • Assume we have 1 mole of glucose.

2. Moles of Carbon in Glucose:

   • Each glucose molecule has 6 carbon atoms.
   • Therefore, 1 mole of glucose = 6 moles of carbon.

3. Calculating Mass of Carbon:

   • Mass of 6 moles of carbon = 6 moles × 12.01 g/mol = 72.06 grams.

4. Calculating Molar Mass of Glucose:

   • Glucose molecular composition: 6 carbons, 12 hydrogens, 6 oxygens.
   • Formula for molar mass:
     • Mass of Carbons = 6 × 12.01 g = 72.06 g
     • Mass of Hydrogens = 12 × 1.008 g = 12.096 g
     • Mass of Oxygens = 6 × 16.00 g = 96.00 g
   • Total Molar Mass of Glucose = 72.06 g + 12.096 g + 96.00 g = 180.156 grams.

5. Percentage Calculation:

   [ \text{Percentage of Carbon by Mass} = \frac{\text{Mass of Carbon}}{\text{Mass of Glucose}} × 100 % ]

   • Substituting values: [ \text{Percentage of Carbon by Mass} = \frac{72.06}{180.156} × 100 % ]
   • Calculate and round to significant figures: [ \text{Percentage of Carbon by Mass} ≈ 0.4000 × 100 = 40.00 %]

Conclusion

• The percentage of carbon by mass in glucose, when rounded to four significant figures, is 40.00%.