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Author Topic: Problem of the week - 04/02/2013  (Read 5340 times)

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Borek

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Problem of the week - 04/02/2013
« on: February 03, 2013, 11:04:14 PM »

This one was posted some time ago by Raderford, and - as Arckon mentioned - it is POTW type of the question. So, why not try it as such?  ;)

A white solid substance X underwent a series of tests in which samples of X were calcined in a flow of various gases and the resultant solid residues were weighed. The experimental data are summarized in the attached table. In all tests the mixture on exit contained, aside from the initial gases, the same unknown gas Y. In test No. 5, a reddish-brown substance Z condensed on the colder parts of the apparatus. Identify the lettered compounds.

Note: I am not 100% sure the reactions will work exactly as expected, but I don't think there is anything wrong in expecting they will.
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Borek

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Re: Problem of the week - 04/02/2013
« Reply #1 on: February 18, 2013, 12:18:38 AM »

Oh come on, it is not that hard.

I must say I was surprised by the red compound - I mean, it is a known fact that some similar compounds have quite a covalent character (which makes them surprisingly volatile), but I didn't know about this particular one.

Initially I started looking for these volatile compounds and tried to see chemistry of which one will fit the data, but it didn't work. So I took the opposite approach - I started guessing possible formulae to find the molar mass. And the first answer I came to was to my surprise the correct one, even if my first thought was "nah, impossible".
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Rutherford

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Re: Problem of the week - 04/02/2013
« Reply #2 on: February 24, 2013, 07:48:07 AM »

Should I write it so we move on or wait? I would really like to see a new problem of the week.
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Re: Problem of the week - 04/02/2013
« Reply #3 on: February 24, 2013, 08:35:46 AM »

Fire away if you know the answer.
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Rutherford

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Re: Problem of the week - 04/02/2013
« Reply #4 on: February 24, 2013, 08:42:49 AM »

X=FeCO3;
Y=CO2;
Z=FeCl3.
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Re: Problem of the week - 04/02/2013
« Reply #5 on: February 24, 2013, 11:04:48 PM »

Care to elaborate how you solved it?
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Rutherford

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Re: Problem of the week - 04/02/2013
« Reply #6 on: February 25, 2013, 01:19:27 AM »

According to experiment 5, X is a carbonate because it is completely dissolved in HCl and it releases a gas. The cation is a transition metal, because it can change oxidation states easily (that's why Cl2 is used). The metal chloride is volatile (Z).

According to experiment 4, another metal chloride is produced (the oxidation state of the metal is lower now that in Z). Then it can be written:
M(CO3)x/2 :rarrow: MClx
1.095(M+30x)=M+35.5x
M=27.9x, for x=2, M=55.8g/mol which would fit for iron. X is FeCO3, Z is a chloride of iron where iron's oxidation state is bigger than in FeCl2, so it is FeCl3. CO2 is released in experiment 5 and 4, so it should be Y.
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Re: Problem of the week - 04/02/2013
« Reply #7 on: February 25, 2013, 01:55:51 AM »

According to experiment 5, X is a carbonate because it is completely dissolved in HCl and it releases a gas.

That's not a correct reasoning IMHO. First of all, we don't dissolve X - it reacts in a solid phase. Fact that it releases a gas in the reaction with HCl is also not necessarily pointing to carbonate - after all it releases a gas in ALL experiments, so HCl presence is not decisive.

But I agree carbonate is a good guess.

Quote
The cation is a transition metal, because it can change oxidation states easily (that's why Cl2 is used). The metal chloride is volatile (Z).

Right.

Quote
According to experiment 4, another metal chloride is produced (the oxidation state of the metal is lower now that in Z). Then it can be written:
M(CO3)x/2 :rarrow: MClx
1.095(M+30x)=M+35.5x
M=27.9x, for x=2, M=55.8g/mol which would fit for iron.

Right.

Quote
X is FeCO3, Z is a chloride of iron where iron's oxidation state is bigger than in FeCl2, so it is FeCl3. CO2 is released in experiment 5 and 4, so it should be Y.

CO2 is released in ALL experiments, so your argument about experiments 4 & 5 is off.

But yes, your approach - with the assumptions about changing oxidation state and using it to calculate possible values of the molar mass - is the same one I used to solve the problem.
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Rutherford

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Re: Problem of the week - 04/02/2013
« Reply #8 on: February 25, 2013, 02:14:45 AM »

"CO2 is released in ALL experiments, so your argument about experiments 4 & 5 is off."

I didn't mean that. What I wanted to say is, if CO2 is released in both exp. 4 and 5, and it is the only gaseous product of exp. 4, it has to be Y.

For the carbonate, you are right. It is probably more a guess than a good reasoning. Sulfide could fit, too and maybe some other anion. The calculation would show which one is correct.
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