- #1
SyntheticVisions
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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?
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?
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