Determining the Identity of an Unknown Metal Salt Containing Bromide Ion

[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?

 

[phinds]

Glad to hear you got it figured out. Too bad you didn't post your corrected work so others could learn from it.

 

[Borek]

.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

Nope - don't round intermediate results that you use for further calculations. 0.003 and 0.00324 (which has a correct number of significant digits) differ by over 7%.

17% Metal

As 48% was wrong this is wrong too.

Mole ratio of M to Br is 1:2 so:

moles M = x
(.003 mol Br)

2x = .003
x = .002 moles M

0.003 was wrong, 0.002 is not much better.

.002 mol M = 17g / x

17g and 17% are not the same thing, you meant 17% of 0.500 grams (initial mass sample). That was a good thinking, but even after using 17% of 0.500g as a metal mass you won't get the right answer, as both 0.002 moles and 17% are wrong due to the rounding errors.

Note, that while finding the identity of the metal is possible (Ca), the hydrate is not stoichiometric - there is no integer n which produces results described in the question. It is not a hydrate, more like just a wet salt. Quite wet.