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Chemistry Forums for Students => Inorganic Chemistry Forum => Topic started by: ae86 on May 10, 2022, 10:54:09 AM

Title: thermal analysis cerium nitrate ammonium
Post by: ae86 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 ?
Title: Re: thermal analysis cerium nitrate ammonium
Post by: Hunter2 on May 10, 2022, 11:51:45 AM
Yes it is

https://en.m.wikipedia.org/wiki/Ceric_ammonium_nitrate
Title: Re: thermal analysis cerium nitrate ammonium
Post by: ae86 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.
Title: Re: thermal analysis cerium nitrate ammonium
Post by: Borek 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.
Title: Re: thermal analysis cerium nitrate ammonium
Post by: Orcio_87 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.
Title: Re: thermal analysis cerium nitrate ammonium
Post by: Borek 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.
Title: Re: thermal analysis cerium nitrate ammonium
Post by: ae86 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.
Title: Re: thermal analysis cerium nitrate ammonium
Post by: Borek 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?