for the percentage composition of Carbon (C), Hydrogen (H), and Oxygen (O) in cholesterol (C27H46O).

Question

ScanSolve AI Tutor · Accepted Answer

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: $\text{C}=12$ , $\text{H}=1$ , $\text{O}=16$ .

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:
$M = (27 \times 12) + (46 \times 1) + (1 \times 16)$ $M = 324 + 46 + 16$ $M = 386 \text{ g/mol}$
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: $1 \times 16 = 16 \text{ g}$
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.

For the percentage composition of Carbon (C), Hydrogen (H), and Oxygen (O) in cholesterol (C27H46O).

Answer

Step-by-step solution

Calculate the molar mass of $\text{C}_{27}\text{H}_{46}\text{O}$ .

Calculate the mass contribution of each element.

Calculate the percentage of each element.

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