2.0 g of a bromo hydrocarbon (X) was subjected to Carius analysis, gave 3.36 g of AgBr . The percentage of carbon in the compound (X) is 26.7 % . Total number of carbon atoms in the empirical formula for compound (X) is _____. (Given molar mass in g mol ⁻¹ H: 1 , C: 12 , Br: 80 , Ag: 108 )

Step-by-step solution

Mass of AgBr formed = 3.36 g Molar mass of AgBr = 108 + 80 = 188 g mol ⁻¹ Moles of AgBr formed = 3.36 188 0.01787 mol Since 1 mole of AgBr contains 1 mole of Br , moles of Br in the compound = 0.01787 mol. Mass of Br = 0.01787 80 = 1.43 g Percentage of Br in compound (X) = 1.43 2.0 100 = 71.5 % Given percentage of C = 26.7 % Percentage of H = 100 % - (71.5 % + 26.7 %) = 1.8 % Now, we find the molar ratio of the elements in the compound: Moles of C = 26.7 12 = 2.225 Moles of H = 1.8 1 = 1.8 Moles of Br = 71.5 80 = 0