An organic compound is known to contain exactly two bromine atoms per molecule. When 0.404 g of this compound is subjected to the Carius method, it yields 0.752 g of silver bromide precipitate. The molar mass of the organic compound is ________ g/mol . (Given: Atomic masses of Ag = 108 u , Br = 80 u )

Step-by-step solution

Molar mass of AgBr = 108 + 80 = 188 g/mol Moles of AgBr formed = 0.752 188 = 0.004 mol Since all the bromine in the precipitate comes from the organic compound, the moles of bromine atoms = 0.004 mol . The organic compound contains exactly 2 bromine atoms per molecule. Therefore, the moles of the organic compound = 0.004 2 = 0.002 mol Molar mass of the organic compound = Mass Moles = 0.404 0.002 = 202 g/mol Answer: 202