Calculate the percent of carbon, hydrogen and oxygen in ethanol (`C_(2)H_(5)OH`)

Answer

To calculate the percent composition of carbon, hydrogen, and oxygen in ethanol (C₂H₅OH), we will follow these steps:

### Step 1: Determine the molecular mass of ethanol (C₂H₅OH).
- **Carbon (C)**: There are 2 carbon atoms, and the atomic mass of carbon is approximately 12 g/mol.
\[ \text{Mass of Carbon} = 2 \times 12 = 24 \text{ g/mol} \]

- **Hydrogen (H)**: There are 6 hydrogen atoms (5 from C₂H₅ and 1 from OH), and the atomic mass of hydrogen is approximately 1 g/mol.
\[ \text{Mass of Hydrogen} = 6 \times 1 = 6 \text{ g/mol} \]

- **Oxygen (O)**: There is 1 oxygen atom, and the atomic mass of oxygen is approximately 16 g/mol.
\[ \text{Mass of Oxygen} = 1 \times 16 = 16 \text{ g/mol} \]

Now, we can calculate the total molecular mass of ethanol:
\[ \text{Molecular mass of C₂H₅OH} = \text{Mass of Carbon} + \text{Mass of Hydrogen} + \text{Mass of Oxygen} = 24 + 6 + 16 = 46 \text{ g/mol} \]

### Step 2: Calculate the percent composition of each element.

#### Percent of Carbon:
\[ \text{Percent of Carbon} = \left( \frac{\text{Mass of Carbon}}{\text{Total Mass}} \right) \times 100 = \left( \frac{24}{46} \right) \times 100 \approx 52.17\% \]

#### Percent of Hydrogen:
\[ \text{Percent of Hydrogen} = \left( \frac{\text{Mass of Hydrogen}}{\text{Total Mass}} \right) \times 100 = \left( \frac{6}{46} \right) \times 100 \approx 13.04\% \]

#### Percent of Oxygen:
\[ \text{Percent of Oxygen} = \left( \frac{\text{Mass of Oxygen}}{\text{Total Mass}} \right) \times 100 = \left( \frac{16}{46} \right) \times 100 \approx 34.78\% \]

### Final Results:
- Percent of Carbon in ethanol: **52.17%**
- Percent of Hydrogen in ethanol: **13.04%**
- Percent of Oxygen in ethanol: **34.78%**

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