Answer :
Sure, let's fill in the table step by step for the given molecule [tex]\(C_{27} H_{45} O\)[/tex].
### Step-by-Step Solution
1. Determine the number of atoms and their atomic masses:
- Carbon (C):
- Number of Carbon atoms = 27
- Atomic mass of Carbon (C) = 12.01 amu
- Hydrogen (H):
- Number of Hydrogen atoms = 45
- Atomic mass of Hydrogen (H) = 1.008 amu
- Oxygen (O):
- Number of Oxygen atoms = 1
- Atomic mass of Oxygen (O) = 16.00 amu
2. Calculate the mass contributed by each type of atom:
- Mass of Carbon (C) atoms:
[tex]\[ \text{C\_mass} = 27 \times 12.01 = 324.27 \text{ amu} \][/tex]
- Mass of Hydrogen (H) atoms:
[tex]\[ \text{H\_mass} = 45 \times 1.008 = 45.36 \text{ amu} \][/tex]
- Mass of Oxygen (O) atoms:
[tex]\[ \text{O\_mass} = 1 \times 16.00 = 16.00 \text{ amu} \][/tex]
3. Calculate the total formula mass of the molecule [tex]\(C_{27} H_{45} O\)[/tex]:
[tex]\[ \text{formula\_mass} = \text{C\_mass} + \text{H\_mass} + \text{O\_mass} = 324.27 + 45.36 + 16.00 = 385.63 \text{ amu} \][/tex]
4. Calculate the composition (mass contribution) of each type of atom as a percentage of the formula mass:
- Percentage of Carbon (C):
[tex]\[ \text{percent\_C} = \left( \frac{\text{C\_mass}}{\text{formula\_mass}} \right) \times 100 = \left( \frac{324.27}{385.63} \right) \times 100 \approx 84.09\% \][/tex]
- Percentage of Hydrogen (H):
[tex]\[ \text{percent\_H} = \left( \frac{\text{H\_mass}}{\text{formula\_mass}} \right) \times 100 = \left( \frac{45.36}{385.63} \right) \times 100 \approx 11.76\% \][/tex]
- Percentage of Oxygen (O):
[tex]\[ \text{percent\_O} = \left( \frac{\text{O\_mass}}{\text{formula\_mass}} \right) \times 100 = \left( \frac{16.00}{385.63} \right) \times 100 \approx 4.15\% \][/tex]
### Completed Table
Here is the completed table for cholesterol [tex]\(C_{27} H_{45} O\)[/tex]:
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline \text{Atom} & \text{Number} & \text{Atomic mass (amu)} & \begin{tabular}{c} \text{Mass (Number} $\times$ \text{atomic} \\ \text{mass)} \end{tabular} & \begin{tabular}{c} \text{Composition} \\ \text{(Mass} $\div$ \text{formula mass} $\times$ 100) \end{tabular} \\ \hline C & 27 & 12.01 & 324.27 & 84.09\% \\ \hline H & 45 & 1.008 & 45.36 & 11.76\% \\ \hline O & 1 & 16.00 & 16.00 & 4.15\% \\ \hline \end{tabular} \][/tex]
### Formula Mass of [tex]\(C_{27} H_{45} O\)[/tex]:
[tex]\[ 385.63 \text{ amu} \][/tex]
This breakdown provides a detailed understanding of the composition of the cholesterol molecule in terms of its constituent elements.
### Step-by-Step Solution
1. Determine the number of atoms and their atomic masses:
- Carbon (C):
- Number of Carbon atoms = 27
- Atomic mass of Carbon (C) = 12.01 amu
- Hydrogen (H):
- Number of Hydrogen atoms = 45
- Atomic mass of Hydrogen (H) = 1.008 amu
- Oxygen (O):
- Number of Oxygen atoms = 1
- Atomic mass of Oxygen (O) = 16.00 amu
2. Calculate the mass contributed by each type of atom:
- Mass of Carbon (C) atoms:
[tex]\[ \text{C\_mass} = 27 \times 12.01 = 324.27 \text{ amu} \][/tex]
- Mass of Hydrogen (H) atoms:
[tex]\[ \text{H\_mass} = 45 \times 1.008 = 45.36 \text{ amu} \][/tex]
- Mass of Oxygen (O) atoms:
[tex]\[ \text{O\_mass} = 1 \times 16.00 = 16.00 \text{ amu} \][/tex]
3. Calculate the total formula mass of the molecule [tex]\(C_{27} H_{45} O\)[/tex]:
[tex]\[ \text{formula\_mass} = \text{C\_mass} + \text{H\_mass} + \text{O\_mass} = 324.27 + 45.36 + 16.00 = 385.63 \text{ amu} \][/tex]
4. Calculate the composition (mass contribution) of each type of atom as a percentage of the formula mass:
- Percentage of Carbon (C):
[tex]\[ \text{percent\_C} = \left( \frac{\text{C\_mass}}{\text{formula\_mass}} \right) \times 100 = \left( \frac{324.27}{385.63} \right) \times 100 \approx 84.09\% \][/tex]
- Percentage of Hydrogen (H):
[tex]\[ \text{percent\_H} = \left( \frac{\text{H\_mass}}{\text{formula\_mass}} \right) \times 100 = \left( \frac{45.36}{385.63} \right) \times 100 \approx 11.76\% \][/tex]
- Percentage of Oxygen (O):
[tex]\[ \text{percent\_O} = \left( \frac{\text{O\_mass}}{\text{formula\_mass}} \right) \times 100 = \left( \frac{16.00}{385.63} \right) \times 100 \approx 4.15\% \][/tex]
### Completed Table
Here is the completed table for cholesterol [tex]\(C_{27} H_{45} O\)[/tex]:
[tex]\[ \begin{tabular}{|c|c|c|c|c|} \hline \text{Atom} & \text{Number} & \text{Atomic mass (amu)} & \begin{tabular}{c} \text{Mass (Number} $\times$ \text{atomic} \\ \text{mass)} \end{tabular} & \begin{tabular}{c} \text{Composition} \\ \text{(Mass} $\div$ \text{formula mass} $\times$ 100) \end{tabular} \\ \hline C & 27 & 12.01 & 324.27 & 84.09\% \\ \hline H & 45 & 1.008 & 45.36 & 11.76\% \\ \hline O & 1 & 16.00 & 16.00 & 4.15\% \\ \hline \end{tabular} \][/tex]
### Formula Mass of [tex]\(C_{27} H_{45} O\)[/tex]:
[tex]\[ 385.63 \text{ amu} \][/tex]
This breakdown provides a detailed understanding of the composition of the cholesterol molecule in terms of its constituent elements.
