Nevermind, I found my mistake..(adsbygoogle = window.adsbygoogle || []).push({});

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?

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Unknown metal determination

Can you offer guidance or do you also need help?

Draft saved
Draft deleted

**Physics Forums | Science Articles, Homework Help, Discussion**