Log in

View Full Version : A difficult chemistry question


TheMatrix
December 4th, 2012, 02:23 AM
Hey guys

I have a lab report that I(and some others, but I'll probably end up doing everything as usual) have to do. Basically, it's about finding a mystery carbonate.
Now while the teacher explained it in class, it makes no sense to me, and I haven't been able to find anything on the internet, either.

I was given 13.81g of "XCO3", where X is the unknown metal. It was narrowed down to the alkali and alkali earth metals, though. And obviously it can't be radioactive.
I was to react that with 2MHCl, which I used 100mL of. Long story short, I needed to find the amount of escaped carbon dioxide, which came out to be 13.99 grams. Some of it may have stayed in the flask, though.

But now I need to figure out what "X" is. Here is the chemical equation I was given and used:
XCO3 + 2MHCl --> CO2 + H2O + XCl2
Of course you have to change that in a simple way for elements not in the first column.

Can anybody explain it to me? I've been thinking about this for days, and nobody around me at school can explain it, either.

Thanks!

WoShiDavid95
December 7th, 2012, 01:26 AM
Amt of CO2: 13.99/[12 + (16 X 2)] = 0.318 mol (3 s.f.)
By mole ratio, amt of XCO3 = 0.318 mol
Mr of XCO3: 13.81/0.318 = 43.4
Mr of X: 43.4 - 12 - (16 x 3) = -16.6

Which is impossible... you have to know the exact mass of CO2 collected, in your case most of it has probably escaped and now the answer cannot be determined.

TheMatrix
December 7th, 2012, 01:41 AM
Amt of CO2: 13.99/[12 + (16 X 2)] = 0.318 mol (3 s.f.)
By mole ratio, amt of XCO3 = 0.318 mol
Mr of XCO3: 13.81/0.318 = 43.4
Mr of X: 43.4 - 12 - (16 x 3) = -16.6

Which is impossible... you have to know the exact mass of CO2 collected, in your case most of it has probably escaped and now the answer cannot be determined.

That's what I came out to, as well. I think I get the concept now, after reading through the text again and trying some other examples.

For the report, I just changed the numbers to make it work. :P
(hopefully the instructor won't look too closely at it and say "that's impossible")