formula mass and atomic mass

Question

A sample of a substance with the empirical formula XCl₂ weighs 0.4530 g. When it is dissolved in water and all its chlorine is converted to insoluble AgCl by addition of an excess of silver nitrate, the mass of the resulting AgCl is found to be 0.9656 g. The chemical reaction is

XCl₂ + 2 AgNO₃2AgCl + X(NO₃)₂

(a) Calculate the formula mass of XCl₂.

Formula mass XCl₂ =  g mol^-1

(b) Calculate the atomic mass of X.

Atomic mass X =  g mol^-1

Answer 1

moles AgCl formed = 0.9656 g AgCl x 1 mol AgCl / 143.3 g = 6.74x10^-3 moles AgCl

moles of XCl₂ present = 6.74x10^-3 moles AgCl x 1 mol XCl2 / 2 mols AgCl = 3.369x10^-4 moles XCl₂

a) Formula mass XCl₂ = grams / moles = 0.4530 g / 3.369x10^-4 moles = 134.45 g/mol

b) Since the atomic mass of Cl = 35.45 g/mol and we have two of them, we can find atomic mass of X...

Atomic mass X = 134.45 - (2x35.45) = 134.45 - 70.90 = 63.55 g / mole

Extra credit: The unknown X is Cu and the unknown compound is CuCl₂