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
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, 3
He. 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
, 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
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, Al2
. 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
, 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, Al4
, and that means B is tritiated water, T2
O. 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
. 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-
. 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, H3
. Well, methanol is the obvious choice for the alcohol, and tritiated methanol CT3
OH has a mass of 38, so I think this is E
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 CT3
H, 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
Here are the balanced equations:
(g) + CuO(s)
Cu (S) + T2
O (l) + Al4
(s) + 3 CT4
He + e-
OH + H3
Ok.. how’d I do?