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