NoSpoon
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Spinel (unknown cation)
well, I have a Spinell that is made of 43.9%oxygen, 18.5%aluminum and 37.6%unknown cations(mass-percent)
so how do I calculate? I could:
43.9 grams O / 16 grams mole^-1 = 2.78
18.5 grams Al / 27 grams mole^-1=0.68
Al : O = 1 : 4
since a spinel needs a 3:4 cation : anion ratio I still need two times 0.68 mol cation which would be the 37.6% unknown stuff ... but that would mean
it has the molar mass of 27. which would be aluminum again?! Al304 doesn't exist as far as I know ^^
and anyway aluminum can't be only one out of three cations since its oxidationstate is 3+ and we need 8+ (because of the four oxidate anions O2-) the
remaining two cations needed to be +2.5 ... I hope somebody understands what I mean ... :-)
so how do I solve this problem?
thx 
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Eclectic
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Here is a starting point:
http://www.galleries.com/minerals/oxides/spinel.htm
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NoSpoon
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thx for your help,
I could say the 37.6 grams must contain another 0.68 moles of Al(since a spinel needs two trivalent ions?), meaning I have 19.1 grams unknown left,
which has to have the molar mass of 19.1/0.68 but that would be 28.. which might bei Si...
SiAl2O4 doesn't exist either... as it seems I'm going the wrong way...
"The general formula of the Spinel Group is AB2O4. The A represents a divalent metal ion such as Magnesium, Iron, Nickel, Manganese and/or Zinc. The
quad valent lead ion can also occupy this site. The B represents trivalent metal ions such as Aluminum, Iron, Chromium and/or Manganese, Titanium may
also occupy this site with a +4 charge and lead at +2 can occupy this site" is the starting point in here? But how do I add +5 to the charge of
Aluminum +3 to get to +8 balancing out the -8 of the oxide ions..
pls help me again
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DeAdFX
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Two alumumins 4 oxygen 1 other metal like Mg Zn etc... I think thats why you are having trouble?
2Al 4O ratio???
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NoSpoon
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yeah, but then try to calculate .. I get SiAl2O4 and that doesn't exist does it?
do you know what I mean? =)
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12AX7
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Al2O3 + SiO2 = Al2SiO5 is kyanite.
Si(II) is unstable in general. It's probably not the right size for a spinel anyway.
I didn't check your math but Mg is close to your result at 24.3 and fits the spinel structure quite well.
Tim
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NoSpoon
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unfortunately not close enough ... i tried ... an I'm still trying
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12AX7
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| Quote: | Originally posted by NoSpoon
well, I have a Spinell that is made of 43.9%oxygen, 18.5%aluminum and 37.6%unknown cations(mass-percent)
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Well let's see...(tired as I am, I feel like stabbing this one now)
We know it is of the form RAl2O4, with R being a divalent (Mg, Fe(II), Mn(II), Zn, Ni(II), Co(II) I think, and possibly Cr(II) ). That means 4 * 16 =
64 atw oxygen, 2*27 = 54 aluminum, and an unknown amount of R. R + Al + O = 100%, so R + 54 + 64 = x --> R + 118 = x.
We are given R = 37.6%, Al = 18.5% and O = 43.9%. This means Al2O4 = 62.4% = 118atw. Thus, total atw = 189.1 and R = 71.1. This has an error of 1%
at most, with the starting values in error by at most 1%. Assuming the data is correct, it must be either a mixture of cations, or Pr(IV), which has
the right atomic weight to valence ratio and an even valence, but may not fit the spinel structure.
Um...okay...
Tim
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chemoleo
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I'm not quite sure what you guys did (not having read thru all) but.....
It's 43.9 % Oxygen, [O]
Thus, per 100 grams, there is
43.9/16 = 2.74 moles of oxygen, [O].
Per 100 grams of spinel there is
18.5/27 = 0.685 moles of Al.
2.74 / (2*0.685) = 2, so the ration of Al vs O = 2, which is what we want.
As a result, there should be 0.685 moles of metal X, which makes up the mass of 37.6 g. Thus, per mole of metal X, this would weigh 37.6/0.685 = 54.88
g/mole.
Looking at the periodic table, manganese has an atomic mass of 54.91 g/mole, and it stably occurs in the Mn2+ state.
So your spinel has the formula MnAl2O4.
[Edited on 12-12-2005 by chemoleo]
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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NoSpoon
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| Quote: |
2.74 / (2*0.685) = 2, so the ration of Al vs O = 2, which is what we want.
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but why do you multiply it by two? we only do have a 0.685 moles of Al, ( 18.5 grams) how come you can just multiply it by 2 ? I already tried this
version, but I thought it's not allowed to just mulitply it by two ^^ cause it would have to consist of 18.5 % *2 Al ... ?!
i'm confused .,,,
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chemoleo
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Yes you are correct, I got ahead of myself.
I don't have a solution to your question  .
At this point I'd rather question the questioneer than try to answer the question - there is no obvious solution to it. You seem to have twice as
much oxygen as you should have, there's nothing you can do about it (2.74/4(O) = 0.68; 0.68/2(Al)=0.34; 0.68/0.34 = 2:1 O vs Al). If it's i.e. a
double salt, i.e. a mixture of i.e. Cr2O4 and Al2O4, with differing divalent cations (where Cr is part of cations X, together with the divalent
cations), you get a large number of permutations. Not exactly an obvious solution. Besides this, at least with Cr, it would be too heavy.
The only thing that baffles me is that it fits so nicely with Mn, or the half mass of Mn, which is Al 
Surely that can't be a coincidence?
Never Stop to Begin, and Never Begin to Stop...
Tolerance is good. But not with the intolerant! (Wilhelm Busch)
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NoSpoon
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yeah I'm sure unfortunately... otherwise I would't have asked for help ....
but thx anyway, your attempts gave me some selfconfidence... I was already crying ^^ not beeing able to solve that(litte stinky piece of ....)
problem...
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