0.301 g of an organic compound gave 0.282 g of silver bromide by a halogen estimation method. Find the percentage of bromine in the compound.

Answer

To find the percentage of bromine in the organic compound, we can follow these steps:

### Step 1: Determine the moles of silver bromide (AgBr) produced.
The molar mass of silver bromide (AgBr) can be calculated as follows:
- Molar mass of Ag = 107.87 g/mol
- Molar mass of Br = 79.90 g/mol
- Molar mass of AgBr = 107.87 g/mol + 79.90 g/mol = 187.77 g/mol

Now, we can calculate the moles of AgBr produced:
\[ \text{Moles of AgBr} = \frac{\text{mass of AgBr}}{\text{molar mass of AgBr}} = \frac{0.282 \, \text{g}}{187.77 \, \text{g/mol}} \approx 0.00150 \, \text{mol} \]

### Step 2: Determine the moles of bromine (Br) in the organic compound.
From the reaction, we know that 1 mole of AgBr contains 1 mole of Br. Therefore, the moles of bromine in the compound is the same as the moles of AgBr produced:
\[ \text{Moles of Br} = 0.00150 \, \text{mol} \]

### Step 3: Calculate the mass of bromine in the organic compound.
Now, we can calculate the mass of bromine using its molar mass:
\[ \text{Mass of Br} = \text{moles of Br} \times \text{molar mass of Br} = 0.00150 \, \text{mol} \times 79.90 \, \text{g/mol} \approx 0.11985 \, \text{g} \]

### Step 4: Calculate the percentage of bromine in the organic compound.
To find the percentage of bromine in the organic compound, we use the formula:
\[ \text{Percentage of Br} = \left( \frac{\text{mass of Br}}{\text{mass of organic compound}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage of Br} = \left( \frac{0.11985 \, \text{g}}{0.301 \, \text{g}} \right) \times 100 \approx 39.83\% \]

### Final Answer:
The percentage of bromine in the organic compound is approximately **39.83%**.

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