StepWorking
01Given
Mass of organic compound w=0.5 g, mass of AgBr formed m=0.40 g, atomic mass of Ag=108 u, atomic mass of Br=80 u.
02Strategise
In Carius method, all bromine in the organic compound is converted into AgBr. Molecular mass of AgBr=108+80=188 g/mol. Mass of bromine in m g of AgBr is 18880×m. Percentage of Br=Mass of compoundMass of Br×100=188×w80×m×100.
03Execute
Percentage of Br=188×0.580×0.40×100=9432×100≈34.04%. Rounding to nearest integer gives 34%.
✓Verify
32/94≈16/47≈0.3404. Nearest integer is 34.