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Unknown metal determination

  1. Feb 10, 2004 #1
    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?
     
    Last edited: Feb 10, 2004
  2. jcsd
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