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Topic: Problem of the week - 05/03/2012  (Read 14645 times)

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Offline Borek

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Problem of the week - 05/03/2012
« on: March 05, 2012, 08:39:39 AM »
1.4511 g sample of a yellow, anhydrous ionic compound A, was roasted in the air stream, leaving a black residue. All gaseous products of the roasting were absorbed in the KOH pellets, increasing their mass by 1.4960 g. The black residue was dissolved in an acid yielding a green solution that was electrolysed on a platinum electrode, increasing its mass by almost exactly half a gram. Product of the electrolysis was removed from the electrode by a stream of the carbon monoxide, yielding exactly 1.4511 g of a volatile, highly toxic liquid B.

1. Using information from the question give formulas of both compounds A and B. If you are not sure about the exact formula, give the most probable one.
2. Give names of both compounds (don't worry if you don't know them outright – you can use any sources you want).
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Offline stewie griffin

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Re: Problem of the week - 05/03/2012
« Reply #1 on: March 05, 2012, 10:15:18 PM »
I haven't done any math with the numbers yet, but from the descriptive chemistry provided I believe compound A is NiCl2 (nickel chloride anhydrous) and compound B is Ni(CO)4 (nickel tetracarbonyl).

Offline Borek

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Re: Problem of the week - 05/03/2012
« Reply #2 on: March 06, 2012, 02:55:54 AM »
Check if they survive the math then.
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Offline stewie griffin

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Re: Problem of the week - 05/03/2012
« Reply #3 on: March 06, 2012, 01:07:47 PM »
I can't seem to get my numbers to work out right from the get go.
NiCl2= 129.6g/mol and would produce 2 moles of HCl during roasting.
1.4511 g of NiCl2 x 129.6 g/mol x 2 mol HCl/1 mol x 36 g/ mol HCl = .817g of HCl.
However the problem suggests we need to be able to generate 1.4960g of gas so NiCl2 would fall short.
NiBr2 is also yellow but would only provide 1.07g of HBr.
I'll have to think more...

Offline stewie griffin

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Re: Problem of the week - 05/03/2012
« Reply #4 on: March 06, 2012, 06:54:31 PM »
Well I'm stuck so I'll throw out some of my thoughts and see if others can point out my errors, or if they can use my thoughts to solve it. I can't seem to figure out what A is but I still think B is Ni(CO)4

Since the moles of whatever metal is in this ionic compound doesn't change, and the weight of A used eventually provides the same weight of B, then shouldn't A and B have the same molecular weight? By my assumptions, A should have a molecular weight of 170 since Ni(CO)4= 170 g/mol. Furthermore, the anion portion of A must have a molecular weight of 112 since 170 - 58 (since I assume the metal is nickel) = 112. I can't think of any combination of anions or polyatomic anions which would equal 112 though.

The fact that the weight of the KOH pellets increases by more than the amount of A used implies that the anion of A is being oxidized by O2 during roasting. Otherwise, how could we add more weight to the KOH than we started with? So perhaps the anion has a sulfur in it, which would be oxidized to sulfate or sulfite?

Offline Borek

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Re: Problem of the week - 05/03/2012
« Reply #5 on: March 07, 2012, 04:13:50 AM »
Plenty of good and correct ideas in what you posted. Don't be surprised by the fact identity of A is not obvious - I was surprised to learn about the compound.

As Sherlock Holmes once said - "When you have eliminated the impossible, whatever remains, however improbable, must be the truth"  :P
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Offline stewie griffin

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Re: Problem of the week - 05/03/2012
« Reply #6 on: March 07, 2012, 08:49:13 AM »
Nickel thiosulfate has a molecular weight of 170 and some sulfurs which could get oxidized in the roasting process. What exactly thiosulfate gets oxidized to I'm not sure.

Offline stewie griffin

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Re: Problem of the week - 05/03/2012
« Reply #7 on: March 08, 2012, 03:32:30 PM »
Well I imagine thiosulfate would be oxidized to sulfate/sulfuric acid, but the math doesn't hold up with that either.
I think the problem of the week has bested me this time  :)

Offline Borek

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Re: Problem of the week - 05/03/2012
« Reply #8 on: March 08, 2012, 06:08:36 PM »
Hint: both A and B have exactly the same mass.
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Offline stewie griffin

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Re: Problem of the week - 05/03/2012
« Reply #9 on: March 08, 2012, 06:46:20 PM »
Borek, after your Sherlock quote I was quite tempted to say that A = B = NiCO4 but I don't really consider nickel tetracarbonyl to be an ionic compound as described in the problem. Perhaps it's just semantics.

Well, after your last comment, I'll go ahead and throw it out there.... A=B= NiCO4.

Offline stewie griffin

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Re: Problem of the week - 05/03/2012
« Reply #10 on: March 08, 2012, 07:59:33 PM »
Aha!
Roasting NiCO4 would result in 4 moles of CO2.
1.4511 / 170.7 x 4 x 44 = 1.496. So the math would work out as well.
 

Offline Borek

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Re: Problem of the week - 05/03/2012
« Reply #11 on: March 09, 2012, 03:38:49 AM »
You are just a step away from an interesting discovery.
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Offline Dan

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Re: Problem of the week - 05/03/2012
« Reply #12 on: March 09, 2012, 04:14:05 AM »
Well, after your last comment, I'll go ahead and throw it out there.... A=B= NiCO4.

A is a yellow solid, B is liquid. I agree that the molecular formulas must be the same though...

Ni(CO)4 and Nickel but-2-ynedioate?
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Offline Borek

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Re: Problem of the week - 05/03/2012
« Reply #13 on: March 09, 2012, 04:41:48 AM »
Nickel but-2-ynedioate?

That's not the compound I had on mind, but the answer you gave shows the question has a potential of being even more interesting. Apparently it was not precise enough.

But with some more effort you can probably square this question up ;)
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Offline Wastrel

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Re: Problem of the week - 05/03/2012
« Reply #14 on: March 09, 2012, 06:18:44 AM »
There are potentially a very large number of candidates for A any number of which could be yellow.  There is an old compound that could fit neetly if it is yellow, particulally with yellow being the colour of honey, but I can see from the last hint that probably isn't the one you were thinking of.

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