Chemical Forums

Please login or register.

Login with username, password and session length

Sponsored links

Pages: [1]   Go Down

Author Topic: Problem of the week - 04/03/2013  (Read 5370 times)

0 Members and 1 Guest are viewing this topic.

Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Mole Snacks: +1539/-389
  • Offline Offline
  • Gender: Male
  • Posts: 23384
  • I am known to be occasionally wrong.
    • Chembuddy
Problem of the week - 04/03/2013
« on: March 04, 2013, 08:39:08 AM »

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.
Logged
Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Mole Snacks: +1539/-389
  • Offline Offline
  • Gender: Male
  • Posts: 23384
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Problem of the week - 04/03/2013
« Reply #1 on: March 07, 2013, 09:06:43 AM »

No takers?
Logged
Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Corribus

  • Chemist
  • Sr. Member
  • *
  • Mole Snacks: +356/-20
  • Offline Offline
  • Gender: Male
  • Posts: 2261
Re: Problem of the week - 04/03/2013
« Reply #2 on: March 07, 2013, 04:32:29 PM »

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 CC 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 ED 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+ + 2H2:rarrow: CT3OH + H3O+

Ok.. how’d I do?
Logged
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Mole Snacks: +1539/-389
  • Offline Offline
  • Gender: Male
  • Posts: 23384
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Problem of the week - 04/03/2013
« Reply #3 on: March 07, 2013, 10:01:39 PM »

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, I found it accidentally last week while doing some database cleaning.
« Last Edit: March 08, 2013, 12:53:38 AM by Borek »
Logged
Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Corribus

  • Chemist
  • Sr. Member
  • *
  • Mole Snacks: +356/-20
  • Offline Offline
  • Gender: Male
  • Posts: 2261
Re: Problem of the week - 04/03/2013
« Reply #4 on: March 08, 2013, 04:35:13 AM »

Great, spent all that time and all I had to do was a forum search. :)
Logged
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Mole Snacks: +1539/-389
  • Offline Offline
  • Gender: Male
  • Posts: 23384
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Problem of the week - 04/03/2013
« Reply #5 on: March 08, 2013, 04:53:41 AM »

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.
Logged
Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Big-Daddy

  • Sr. Member
  • *****
  • Mole Snacks: +28/-94
  • Offline Offline
  • Posts: 1177
Re: Problem of the week - 04/03/2013
« Reply #6 on: March 08, 2013, 06:45:49 AM »

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.
Logged

Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Mole Snacks: +1539/-389
  • Offline Offline
  • Gender: Male
  • Posts: 23384
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Problem of the week - 04/03/2013
« Reply #7 on: March 08, 2013, 07:32:20 AM »

Note: it is not just a Corribus notation, H, D and T are pretty much established as symbols for protium, deuterium and tritium.
Logged
Chembuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info, pH-meter.info

Corribus

  • Chemist
  • Sr. Member
  • *
  • Mole Snacks: +356/-20
  • Offline Offline
  • Gender: Male
  • Posts: 2261
Re: Problem of the week - 04/03/2013
« Reply #8 on: March 08, 2013, 07:47:41 AM »

IUPAC discourages it, but it's a heck of a lot easier than writing all the subscripts and superscripts.
Logged
What men are poets who can speak of Jupiter if he were like a man, but if he is an immense spinning sphere of methane and ammonia must be silent?  - Richard P. Feynman

Big-Daddy

  • Sr. Member
  • *****
  • Mole Snacks: +28/-94
  • Offline Offline
  • Posts: 1177
Re: Problem of the week - 04/03/2013
« Reply #9 on: March 08, 2013, 08:37:19 AM »

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.
Logged
Pages: [1]   Go Up
 

Mitch Andre Garcia's Chemical Forums 2003-Present.

Page created in 0.093 seconds with 24 queries.