A mono-iodo organic compound (containing exactly one iodine atom per molecule) is analyzed via the Carius method. It is found that 0.408 g of the compound yields 0.470 g of silver iodide ( AgI ) precipitate. The molar mass of the organic compound is: (Given molar masses: Ag = 108 g mol ⁻¹ , I = 127 g mol ⁻¹ )

Step-by-step solution

Molar mass of AgI = 108 + 127 = 235 g mol ⁻¹ Moles of AgI precipitate formed = 0.470 235 = 0.002 mol Since the organic compound is mono-iodo, one molecule of the compound contains exactly one iodine atom. Therefore, 1 mole of the organic compound will produce 1 mole of AgI . Moles of the organic compound taken = Moles of AgI = 0.002 mol We are given that the mass of the organic compound is 0.408 g . Molar mass of the compound = Mass Moles = 0.408 0.002 = 204 g mol ⁻¹ Answer: 204 g mol ⁻¹