Step 1: Write the molecular formula.
The compound is C57H110O6. We need to find the mass percentage of carbon in this molecule.
Step 2: Find the molar mass of each element's contribution.
Carbon (C): atomic mass = 12 g/mol, so 57 carbon atoms contribute \(57 \times 12 = 684\) g/mol.
Hydrogen (H): atomic mass = 1 g/mol, so 110 hydrogen atoms contribute \(110 \times 1 = 110\) g/mol.
Oxygen (O): atomic mass = 16 g/mol, so 6 oxygen atoms contribute \(6 \times 16 = 96\) g/mol.
Step 3: Calculate the total molar mass.
\[ M = 684 + 110 + 96 = 890 \text{ g/mol} \]
Step 4: Apply the mass percentage formula.
The formula for mass percentage of an element is: \[ \% \text{ of element} = \frac{\text{mass contributed by element}}{\text{total molar mass}} \times 100 \]
Step 5: Calculate mass % of carbon.
\[ \% C = \frac{684}{890} \times 100 \]
Dividing: \(684 / 890 = 0.7685...\)
So \(\% C = 76.85\%\)
Step 6: Identify the correct answer.
The mass percentage of carbon in C57H110O6 is approximately 76.85%. This compound is a triglyceride (a fat molecule), and fats are indeed carbon-rich, which makes this answer physically reasonable.
\[ \boxed{76.85\%} \]