Nevermind, I found my mistake.. I've been trying to figure this problem out for some time now, and I can't seem to get it right: You are given an unknown hydrated metal salt containing bromide ion, [tex]MBr_2 \bullet nH_2O[/tex] (M and n being variables). You dissolve 0.500g of this salt in water and add excess silver nitrate solution, [tex]AgNO_3[/tex], to precipitate the bromide ion as insoluble silver bromide, [tex]AgBr[/tex]. After filtering, washing, drying, and weighing, the [tex]AgBr[/tex] is found to weigh .609g. What is the weight percent bromide in this metal salt? I tried doing the following: .609g AgBr / 187.77amu AgBr = .003 mol Br (.003 mol)(79.90 amu Br) = .240g Br .240 g / .500 g = And got: 48.0% [tex]Br[/tex] The question goes on to say: A second 0.500g sample is dehydrated to remove water of hydration. After drying, the sample is found to weigh 0.325g. What is the weight percent in the metal salt? I did the following: (.500g - .325g) / .500 g = And got: 35.0% [tex]H_2O[/tex] And then it says: The metal cation has a valence of two. What is the atomic weight of the metal? What is the identity of the metal? I did this: 17% Metal Mole ratio of M to Br is 1:2 so: moles M = x (.003 mol Br) 2x = .003 x = .002 moles M AW Metal = x .002 mol M = 17g / x .002x = 17 x = 8500 See there is where it all blows up in my face. Came up with an atomic weight for the metal of 8500. Can someone tell me where I went wrong?