Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on March 04, 2013, 02:39:08 PM
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A radioactive gas A is known to react with copper(II) oxide at 500oC. One of the products of this reaction, B, at the same temperature, reacts with compound M (which contains 75% by mass of Al) to yield a gas C with molar mass equal to 24, and a compound containing 53% by mass Al. Radioactive decay of one of the atoms of compound C leads to its decomposition giving some charged particles D with molar mass equal to 21, and helium atoms 3He. Contact of gas C with water vapors gives the molecules of an alcohol E with molar mass of 38, and hydronium ions.
What compounds are represented by letters A, B, C, D, E, and M.
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No takers?
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I feel bad - this problem is all lonely and neglected, so I'll take a stab at it. I’m not sure about some of my answers, because nuclear radiolysis reactions are not really my area of expertise, but what they heck.
I will describe my thought process, then provide my answers. Let’s see how close I got.
The most obvious place for me to start was actually molecule C. C is radioactive like A, but here we’re given two important clues: one is the low molecular weights involved and the other is that the product of the radioactive decay is a light helium, 3He. There really are two major possibilities here for a radioactive isotope with low molecular weight: tritium and carbon-14. Both undergo beta decay, but only tritium gives up a light helium as well. Therefore C and A, I conclude, both involve tritium, henceforth abbreviated as T.
Moving to A. There are a couple of common gases that have tritium in them that might react with cupric oxide: di-tritium, tritiated ammonia, and tritiated methane. NT3 and T2 both react with cupric oxide to give water, copper metal, and – in the case of ammonia – nitrogen gas. Tritiated methane produces tritiated water and CO2. Not much help at the moment, since the products are somewhat similar.
Ok, I file that in the back of my mind and take a look at the reaction in the middle, the one with aluminum. The compound with 53% by mass of aluminum is clearly aluminum oxide, Al2O3. Well, it’s the first possibility I checked since it’s the most obvious. Sadly, and most unfairly, I get no credit for this achievement because it’s not one of my unknowns. But it does tell me that mystery compound C must contain my tritium source. I know C must also react with water to give an alcohol (E), so therefore I conclude that C must have some carbon in it, and playing around a little with the weights to get MW = 24 gives me a convenient tritiated methane, CT4, for compound C. Given that this was also one of my possibilities for A, and thinking it’d be rather redundant to have the same compound for A and C, I eliminate CT4 from my possibilities for A.
My original thought for M was a tritiated version of aluminum trihidride – very reactive and, by coincidence, conveniently having almost exactly 75% by weight of aluminum. That coincidence threw me off for quite a while, because that meant B had to be my carbon source, which made no sense after I eliminated CT4 from possibilities for A. Banged my head against the wall for awhile at that one, because I was dead set on AlT3, with the nice 75% and all, but no, M just had to have my carbon source. No other way. Trimethyl aluminum was my first thought here but it’s not 75% aluminum by weight - crap. Almost emailed Borek and asserted that there was a typo, 75% couldn't be right. LOL at my ego, thinking a typo is more likely than that I was wrong. Well, I’ll be honest here and admit that at this point I just started to look around on Wikipedia for aluminum compounds that contain carbon. Came up with aluminum carbide – 75% aluminum by weight and, what do you know, reacts with water to give methane and alumina. Bingo. Now does that make me a cheat, or just resourceful? I'll let the chemistry gods decide.
Anyway, that means M is aluminum carbide, Al4C3, and that means B is tritiated water, T2O. Front half of the problem done.
Back half: I thought originally this would be easy until I realized I know squat about self-radiolysis reactions. I knew C was CT4. Decay of one of the tritiums to the light helium knocks 3 g/mol off of CT4’s weight, which gives D, conveniently at molar mass of 21. D must therefore be CT3- or CT3+. I put aside for the moment which of these two possibilities it is and move to the last reaction: Reaction of D with water gives an alcohol E with molar mass of 38 and some hydronium, H3O+. Well, methanol is the obvious choice for the alcohol, and tritiated methanol CT3OH has a mass of 38, so I think this is E. D can’t be a carbanion and produce hydronium – if D was a carbanion and it reacted with water, I surmise it would give a hydroxyl and produce CT3H, another tritiated methane. No go. Therefore D must be the short-lived methenium, CT3+, which should nicely rip off a hydroxyl from water and produce a bunch of extra protons.
So in conclusion:
A = T2
B = T2O
C = CT4
D = CT3+
E = CT3OH
M = Al4C3
Here are the balanced equations:
CT4 (g) + CuO(s) :rarrow: Cu (S) + T2O (l)
6 T2O (l) + Al4C3 (s) :rarrow: 2 Al2O3 (s) + 3 CT4(g)
CT4 :rarrow: CT3+ + 3He + e-
CT3+ + 2H2O :rarrow: CT3OH + H3O+
Ok.. how’d I do?
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A = T2
B = T2O
C = CT4
D = CT3+
E = CT3OH
M = Al4C3
Hard to deny, that's the correct answer :)
Sadly, we have at least three Chemistry Olympiad wannabes and none of them attempted the problem >:(
Edit: forgot to mention, this question was already posted at CF several years ago (http://www.chemicalforums.com/index.php?topic=9083.0), I found it accidentally last week while doing some database cleaning.
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Great, spent all that time and all I had to do was a forum search. :)
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No, the old thread was soft deleted, I found it while checking what should be deleted permanently. I have undeleted it around the time I posted the link.
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Sorry Borek!
I got as far as C is CT4 (to use Corribus' notation), D is CT3+ and E is CT3OH.
However, without looking at Corribus' answer I was unable to work out A, B or M.
Clearly I have a long way to go in these types of problems.
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Note: it is not just a Corribus notation, H, D and T are pretty much established as symbols for protium, deuterium and tritium.
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IUPAC discourages it, but it's a heck of a lot easier than writing all the subscripts and superscripts.
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Note: it is not just a Corribus notation, H, D and T are pretty much established as symbols for protium, deuterium and tritium.
I've always written the odd isotopes out (e.g. 14C, 2H etc.)but I have seen D for deuterium many times - rarely encountered 3H i.e. T so I thought it was worth mentioning.