Okay, Candido. Let's tackle the first question from PART I. The question asks for the percentage composition of Carbon (C), Hydrogen (H), and Oxygen (O) in cholesterol ($\text{C}_{27}\text{H}_{46}\text{O}$). Given atomic masses: $\text{C}=12$, $\text{H}=1$, $\text{O}=16$. Step 1: Calculate the molar mass of $\text{C}_{27}\text{H}_{46}\text{O}$. The molecule contains: • 27 atoms of Carbon (C) • 46 atoms of Hydrogen (H) • 1 atom of Oxygen (O) Molar mass: $$ M = (27 \times 12) + (46 \times 1) + (1 \times 16) $$ $$ M = 324 + 46 + 16 $$ $$ M = 386 \text{ g/mol} $$ Step 2: Calculate the mass contribution of each element. • Mass of C: $27 \times 12 = 324 \text{ g}$ • Mass of H: $46 \times 1 = 46 \text{ g}$ • Mass of O: $1 \times 16 = 16 \text{ g}$ Step 3: Calculate the percentage of each element. Percentage of an element = $\frac{\text{mass of element}}{\text{molar mass of compound}} \times 100\%$ Percentage of C: $$ \% \text{C} = \frac{324 \text{ g}}{386 \text{ g}} \times 100\% $$ $$ \% \text{C} \approx 83.94\% $$ Percentage of H: $$ \% \text{H} = \frac{46 \text{ g}}{386 \text{ g}} \times 100\% $$ $$ \% \text{H} \approx 11.92\% $$ Percentage of O: $$ \% \text{O} = \frac{16 \text{ g}}{386 \text{ g}} \times 100\% $$ $$ \% \text{O} \approx 4.14\% $$ Rounding these percentages to the nearest whole number: • C: $83.94\% \approx 84\%$ • H: $11.92\% \approx 12\%$ • O: $4.14\% \approx 4\%$ Comparing these values with the given options: A 27, 46, 1 B 54, 23, 23 C 64, 28, 8 D 84, 12, 4 The calculated percentages match option D. The correct option is D. The percentages of C, H, and O are approximately 84%, 12%, and 4%, respectively. That's 2 down. 3 left today — send the next one.