Question

A sample of a substance with the empirical formula XCl3 weighs 0.5183 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCI by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to he 1.4071 g. The chemical reaction is XCl3 +3 AgNO3 3AgCI + X(NO3)3 (a) Calculate the formula mass of XCl3. Formula mass XCI3- g mol (b) Calculate the atomic mass of X Atomic mass X- g mol

Answer 1

Mass of XCl_3 0.5183 g mass of total Age 1.4071 g Reaction: - Xcl_3 + 3 AgNO_3 rightarrow 3 AgCl + X((NO_3)_3 Molar mass of Agel 143.32 h /mol So mol of Age mass/molar mass 1.4071 g/143.32 g/mol 9.82 times 10^-3 mol Now: - 3 mol of age from 1 mol of XCl_3 So i mol of Age from 1/3 mol of XCl_3 therefore 9.82 times 10^-3 mol age 1/3 times 9.82 times 10^-3 3.27 times 10^-3 mol XCl_3 mass of 3.27 times 10^-3 mol XCl_3 0.5183 g So mass of 1 mol XCl_3 0.5183/3.27 times 10^-3 158.50 g/mol fromula mass of Xcl_3 158.50 g/mol Atomic mass of X (formula mass of XCl_3 - 3X(Atomic mass of) (158.50) - (3 times 35.5) 52.0 Atomic mass of X = 52.0 g/mol.