In `CH_(3)CH_(2)Br,%` of Br is

Options

A. 80
B. 75
C. 70
D. 7

Explanation

To find the percentage of bromine in ethyl bromide (C2H5Br), we will follow these steps:

### Step 1: Determine the atomic mass of bromine (Br)
The atomic mass of bromine is approximately **80 g/mol**.

### Step 2: Calculate the molecular mass of ethyl bromide (C2H5Br)
- **Carbon (C)**: There are 2 carbon atoms, and the atomic mass of carbon is **12 g/mol**.
- Total mass from carbon = 2 × 12 = **24 g/mol**

- **Hydrogen (H)**: There are 5 hydrogen atoms, and the atomic mass of hydrogen is **1 g/mol**.
- Total mass from hydrogen = 5 × 1 = **5 g/mol**

- **Bromine (Br)**: The mass from bromine is **80 g/mol**.

Now, we can calculate the total molecular mass of C2H5Br:
\[ \text{Molecular mass of C2H5Br} = \text{mass of C} + \text{mass of H} + \text{mass of Br} \]
\[ = 24 + 5 + 80 = 109 \text{ g/mol} \]

### Step 3: Calculate the percentage of bromine in C2H5Br
The formula to calculate the percentage of bromine is:
\[ \text{Percentage of Br} = \left( \frac{\text{mass of Br}}{\text{molecular mass of C2H5Br}} \right) \times 100 \]

Substituting the values:
\[ \text{Percentage of Br} = \left( \frac{80}{109} \right) \times 100 \]

Calculating this gives:
\[ \text{Percentage of Br} \approx 73.39\% \]

### Step 4: Round the percentage to the nearest whole number
Since we need to approximate the percentage:
\[ 73.39\% \approx 74\% \]

### Final Answer
Thus, the percentage of bromine in ethyl bromide (C2H5Br) is approximately **74%**.

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