[ae86]

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 ?

[Hunter2]

Yes it is

https://en.m.wikipedia.org/wiki/Ceric_ammonium_nitrate

[ae86]

Yes but why didn't I get m = 6 ? The compound is not synthesized by myself it is from sigma aldrich.

[Borek]

No idea, but at least I can confirm 6.85 looks like a correct output from the data given.

[Orcio_87]

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.

[Borek]

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.

[ae86]

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.

[Borek]

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?