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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Problem of the Week Archive => Topic started by: Borek on April 29, 2013, 05:20:52 PM

Title: Problem of the week - 29/04/2013
Post by: Borek on April 29, 2013, 05:20:52 PM
1.000 g of a roasted salt decomposes producing three oxides - 0.878 g of a solid one, 0.0354 g of a liquid one and 0.0866 g of a gaseous one (at STP).

Determine the salt identity.
Title: Re: Problem of the week - 29/04/2013
Post by: Rutherford on April 30, 2013, 01:17:36 PM
I got Pb(OH)2·PbCO3 or (PbOH)2CO3, not quite sure  :-\.
Title: Re: Problem of the week - 29/04/2013
Post by: Big-Daddy on April 30, 2013, 01:41:04 PM
(PbOH)2CO3,

Surely this can't exist ... That would mean the (PbOH)+ ion exists which I doubt.
Title: Re: Problem of the week - 29/04/2013
Post by: Rutherford on April 30, 2013, 02:35:45 PM
Okay, then maybe the first one.
Title: Re: Problem of the week - 29/04/2013
Post by: Big-Daddy on April 30, 2013, 02:38:44 PM
Borek when you say oxides do you just mean the salt contains oxygen in some form, or specifically the O2- ion? (Or can you not tell us?)
Title: Re: Problem of the week - 29/04/2013
Post by: Borek on April 30, 2013, 03:28:08 PM
Borek when you say oxides do you just mean the salt contains oxygen in some form, or specifically the O2- ion? (Or can you not tell us?)

No idea what you are asking about. Have you seen gaseous oxide containing O2-?
Title: Re: Problem of the week - 29/04/2013
Post by: Big-Daddy on April 30, 2013, 04:27:16 PM
Borek when you say oxides do you just mean the salt contains oxygen in some form, or specifically the O2- ion? (Or can you not tell us?)

No idea what you are asking about. Have you seen gaseous oxide containing O2-?

No, but then Raderford is suggesting a carbonate, which I would not normally class as an oxide. So by oxide you mean any salt containing O somewhere?
Title: Re: Problem of the week - 29/04/2013
Post by: Borek on April 30, 2013, 04:42:38 PM
No, but then Raderford is suggesting a carbonate, which I would not normally class as an oxide. So by oxide you mean any salt containing O somewhere?

Have you read the question?
Title: Re: Problem of the week - 29/04/2013
Post by: Big-Daddy on April 30, 2013, 05:20:20 PM
No, but then Raderford is suggesting a carbonate, which I would not normally class as an oxide. So by oxide you mean any salt containing O somewhere?

Have you read the question?

 :-[ That was stupid of me.
Title: Re: Problem of the week - 29/04/2013
Post by: delta609 on May 01, 2013, 08:17:58 PM
TlHCO3
Title: Re: Problem of the week - 29/04/2013
Post by: delta609 on May 01, 2013, 08:18:29 PM
 I didn't get the exact amounts you have posted, but very close. 

2TlHCO3  :rarrow: Tl2CO3 + CO2 + H2O
Tl

Tl2CO3 = .883g
CO2 = .083g
H2O = .034g
Title: Re: Problem of the week - 29/04/2013
Post by: Borek on May 02, 2013, 03:00:08 AM
Tl2CO3 is not an oxide.
Title: Re: Problem of the week - 29/04/2013
Post by: delta609 on May 02, 2013, 01:12:43 PM
Pb2(OH)2CO3

Pb2(OH)2CO3  :rarrow: 2PbO + CO2 + H2O

numbers are spot on this time.
Title: Re: Problem of the week - 29/04/2013
Post by: Borek on May 02, 2013, 02:29:04 PM
Raderford was first:

I got Pb(OH)2·PbCO3 or (PbOH)2CO3, not quite sure  :-\.

No matter how you write it, this is the intended answer.

And I should slap myself again, as I have not checked details. Problem was used in a chemistry competition, and after solving and checking that the numbers are OK I assumed the compound is just a basic lead carbonate. Turns out it is not - so called white lead has a slightly different formula (2PbCO3·Pb(OH)2). It doesn't mean Pb(OH)2·PbCO3 doesn't exist, as there are several similar compounds, but I prefer problems with the obviously existing real solutions.
Title: Re: Problem of the week - 29/04/2013
Post by: blaisem on May 03, 2013, 08:33:25 AM
Would anyone mind providing a link or a keyword for google where I can read how a problem like this is approached/solved?

Thank you
Title: Re: Problem of the week - 29/04/2013
Post by: Rutherford on May 03, 2013, 09:35:06 AM
It requires a lot of practice and trial and error. I usually encounter problems when solving tasks like this one.
Here you had a liquid oxide. The most famous liquid oxide is water, so you assume that it is water, you calculate its quantity and compare, e.g. with the gaseous oxide. If they got formed in 1:1 ratio, then the gaseous oxide has a molar mass of 44g/mol which corresponds to few gasses, but a good guess would be carbon dioxide (because when acidic or basic carbonate salts decompose, they can form solid oxide, water and carbon dioxide). You do the same for the solid oxide. Assume 1:1 ratio, and you will get a very big molar mass that doesn't correspond to any simple oxide I can think of. Then try the 2:1 ratio, and you will end with a molar mass of 223g/mol. Now you assume that there is one oxygen atom in the oxide, so the other element that composites the oxide has a molar mass of 207g/mol (Pb). Now you need to pack these oxides into one compound. As you see, many assumptions have to be made.