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thermal analysis cerium nitrate ammonium

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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:

--- 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 ?
--- End quote ---
You have two peaks on the graph, not just one. So it is not only one reaction that you are looking for.

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