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Topic: thermal analysis cerium nitrate ammonium  (Read 594 times)

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Offline ae86

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thermal analysis cerium nitrate ammonium
« on: May 10, 2022, 10:54:09 AM »
Hello,

I have an assignment to do with the the thermal analysis cerium nitrate ammonium. In fact I need to deduct m. The chemical formula is (NH4)2Ce(NO3)m. All I know is that at the end I have only CeO2. So I did:

molecular mass of (NH4)2Ce(NO3)m: M(salt)= 2*(14+4)+140,116+m(14+3*16) = (176,116+62m) g/mol
molecular mass of CeO2: M(1)=140,116+2*16 = 172,116 g/mol
weight loss : WL = 29,78+18,20+23,38 = 71,36 % (see the picture)

WL = (M(salt)-M(1))/M(salt)
     = (176,116+62m-172,116)/(176,116+62m)

     0,7136(176,116+62m)=4+62m
     125,69+44,25m=4+62m
     121,69=17,75m
     m=121,69/17,75
     m=6,85

The problem (if I didn't do any mistakes) is that on the market the cerium nitrate ammonium has that chemical formula (NH4)2Ce(NO3)6 (m=6)

Please can anybody help me ?

Offline Hunter2

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Offline ae86

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Re: thermal analysis cerium nitrate ammonium
« Reply #2 on: May 10, 2022, 12:13:20 PM »
Yes but why didn't I get m = 6 ? The compound is not synthesized by myself it is from sigma aldrich.

Offline Borek

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Re: thermal analysis cerium nitrate ammonium
« Reply #3 on: May 10, 2022, 12:41:11 PM »
No idea, but at least I can confirm 6.85 looks like a correct output from the data given.
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Offline Orcio_87

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Re: thermal analysis cerium nitrate ammonium
« Reply #4 on: May 10, 2022, 04:05:52 PM »
Quote
The problem (if I didn't do any mistakes) is that on the market the cerium nitrate ammonium has that chemical formula (NH4)2Ce(NO3)6 (m=6)

Please can anybody help me ?
You have two peaks on the graph, not just one. So it is not only one reaction that you are looking for.

Offline Borek

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Re: thermal analysis cerium nitrate ammonium
« Reply #5 on: May 10, 2022, 05:19:18 PM »
You have two peaks on the graph, not just one. So it is not only one reaction that you are looking for.

Perhaps I am missing something, but for the composition it should be enough to compare initial state ((NH4)2Ce(NO3)m) with the final one (CeO2), path nor intermediates don't matter.
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Offline ae86

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Re: thermal analysis cerium nitrate ammonium
« Reply #6 on: May 11, 2022, 07:35:03 AM »
You don't have to take note of the intermediates to calculate the number of molecule of nitrate.
Well if we add to  ((NH4)2Ce(NO3)m) 3 molecules of H20, we have approximately the mass loss at the end of the decomposition. I don't know if this salt could be under the crystallized form.

Offline Borek

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Re: thermal analysis cerium nitrate ammonium
« Reply #7 on: May 11, 2022, 12:01:56 PM »
Crystal water can be always a factor, no doubt about it. Thing is, in the case of hydrates such water is typically clearly stated in the formula.

Loosing this water should be visible as a step on the curve. Does the water mass fit the first mass loss step?
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