Question

In Carius method of estimation of halogen, 0.25 g of an organic compound gave 0.15 g of silver bromide (AgBr). The percentage of Bromine in the organic compound is ________________ × 10^-1% (Nearest integer).
(Given: Molar mass of Ag is 108 and Br is 80 g mol^-1)

Hint:

In Carius method, the amount of halogen can be calculated by the formation of silver halide, and the mass of the halogen can be derived using stoichiometry. Always use the molar mass of the halide to find the exact amount of halogen present.

Answer 1

The Carius method is used to determine the amount of halogen in an organic compound by converting the halogen into a silver halide (in this case, AgBr for bromine). To find the percentage of bromine in the compound, we follow these steps:

Step 1: Calculate the moles of AgBr formed

We are given the mass of AgBr formed, which is 0.15 g. The molar mass of AgBr is the sum of the atomic masses of silver (Ag) and bromine (Br):

Molar mass of AgBr = 108 + 80 = 188 g/mol

Now, calculate the moles of AgBr formed using the formula:

Moles of AgBr = (Mass of AgBr) / (Molar mass of AgBr) = 0.15 g / 188 g/mol = 0.00079787 mol

Step 2: Moles of bromine (Br) in the compound

From the reaction between the organic compound and silver nitrate, we know that 1 mole of AgBr contains 1 mole of bromine. So, the moles of bromine (Br) will be the same as the moles of AgBr:

Moles of Br = 0.00079787 mol

Step 3: Calculate the mass of bromine (Br) in the organic compound

The atomic mass of bromine (Br) is given as 80 g/mol. So, the mass of bromine in the organic compound is:

Mass of Br = (Moles of Br) × (Atomic mass of Br) = 0.00079787 mol × 80 g/mol = 0.06383 g

Step 4: Calculate the percentage of bromine in the organic compound

We are given that the mass of the organic compound is 0.25 g. The percentage of bromine in the compound is calculated as:

Percentage of Br = (Mass of Br / Mass of the organic compound) × 100 = (0.06383 g / 0.25 g) × 100 = 25.532%

Step 5: Convert the percentage to the required format

The final answer should be in the format × 10^-1%. Therefore, we need to multiply the result by 10 and round to the nearest integer:

25.532 × 10 = 255.32 ≈ 255

Conclusion

The percentage of bromine in the organic compound is 255 × 10^-1%.

Answer 2

Step 1: Understand the relationship between the mass of AgBr and the amount of bromine in the organic compound.
In Carius method, bromine from the organic compound reacts with silver (Ag) to form silver bromide (AgBr). The mass of AgBr formed is related to the mass of bromine in the organic compound.
The molecular mass of AgBr is the sum of the atomic masses of silver (Ag) and bromine (Br): \[ \text{Molar mass of AgBr} = 108 \, \text{g/mol (Ag)} + 80 \, \text{g/mol (Br)} = 188 \, \text{g/mol}. \] The mass of bromine can be calculated using the ratio of the molar masses of bromine and silver bromide.

Step 2: Calculate the moles of AgBr.
The mass of AgBr obtained is 0.15 g. To calculate the moles of AgBr: \[ \text{Moles of AgBr} = \frac{0.15 \, \text{g}}{188 \, \text{g/mol}} = 0.0007979 \, \text{mol}. \]

Step 3: Calculate the moles of Bromine.
Since the molar ratio of AgBr to bromine is 1:1, the moles of bromine will be the same as the moles of AgBr: \[ \text{Moles of Br} = 0.0007979 \, \text{mol}. \]

Step 4: Calculate the mass of Bromine.
The molar mass of bromine is 80 g/mol, so the mass of bromine is: \[ \text{Mass of Br} = 0.0007979 \, \text{mol} \times 80 \, \text{g/mol} = 0.063832 \, \text{g}. \]

Step 5: Calculate the percentage of Bromine in the organic compound.
The mass of the organic compound is 0.25 g, so the percentage of bromine in the compound is: \[ \text{Percentage of Br} = \frac{0.063832 \, \text{g}}{0.25 \, \text{g}} \times 100 = 25.53\%. \]

Step 6: Convert the answer to the required form.
The required answer is in the form of \( \text{Percentage of Br} \times 10^{-1} \): \[ 25.53\% \times 10^{-1} = 255 \, (\text{nearest integer}). \]

Final Answer:
\[ \boxed{255}. \]