# Unknown metal determination

1. Feb 10, 2004

### SyntheticVisions

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, $$MBr_2 \bullet nH_2O$$ (M and n being variables).

You dissolve 0.500g of this salt in water and add excess silver nitrate solution, $$AgNO_3$$, to precipitate the bromide ion as insoluble silver bromide, $$AgBr$$. After filtering, washing, drying, and weighing, the $$AgBr$$ 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% $$Br$$

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% $$H_2O$$

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