Question

In the Carius method of estimation of halogen, 0.4g of an organic compound gave 0.188g of AgBr. What is the percentage of bromine in the organic compound? (The atomic mass of Ag = 108 g mol\(^{-1}\) & Br = 80 g mol\(^{-1}\))

• A. \(20%\)
• B. \(10%\)
• C. \(15%\)
• D. \(25%\)
• E. \(30%\)

Hint:

In the Carius method, to find the halogen percentage, use the formula:

\[ \% X = \left( \frac{\text{Mass of halogen in AgX}}{\text{Mass of organic compound}} \right) \times 100 \]

where \( \text{AgX} \) is the corresponding silver halide.

Answer 1

The Correct Option is A

Step 1: Determine the mass fraction of bromine in AgBr Molecular mass of silver bromide (\({AgBr}\)): \[ M_{{AgBr}} = 108 + 80 = 188 { g/mol} \] Mass fraction of bromine in AgBr: \[ \frac{{Mass of Br}}{{Mass of AgBr}} = \frac{80}{188} \]
Step 2: Calculate the mass of bromine in the given sample Given mass of AgBr = 0.188 g Mass of bromine in AgBr: \[ {Mass of Br} = \frac{80}{188} \times 0.188 \] Computing the value: \[ {Mass of Br} = \frac{80 \times 0.188}{188} = 0.08 { g} \]
Step 3: Calculate the percentage of bromine Given mass of organic compound = 0.4 g Percentage of bromine:

• Given mass of organic compound = 0.4 g
• Percentage of bromine: \[ \% \text{Br} = \left( \frac{0.08}{0.4} \right) \times 100 \]
• Simplifying: \[ \% \text{Br} = 20\% \]