# Decomposition products of hexaurea-chromic-chloride-trihydrate

I am sitting over a TG diagram of the decomposition steps of hexaurea-chromic-chloride-trihydrate ($\ce{[Cr(CO(NH2)2)6]Cl3 \cdot 3 H2O}$) and cannot really make sense of the products it might decompose into.

It is clear that water leaves first (-3.5%), and the next step might be (with a wide error margin) fumes of $\ce{CO2}$, $\ce{H2O}$ and $\ce{2 NH3}$. What is left is then $\ce{[Cr(CO(NH2)2)5]Cl3}$.

I have not found any resources on what might happen further - the four urea ligands could decompose into $$\ce{4 CO} + \ce{5 NH3} + \ce{3/2 N2} + \ce{1/2 H}$$ and I am left with $\ce{CrCl3}$ and urea, but those steps show up last (17.9% + 17.3% + 5.4% = 40.6%), leaving the 15.5% in the middle unexplained.

What other decomposition could take place after water and two urea ligands are gone? Is there a way to figure that out other than by guessing or experiment?

1. Decomposition of $\ce{3 H2O}$ and $\ce{2 CO(NH2)2}$: $\ce{[Cr(CO(NH2)2)6]Cl3$\cdot$3 H2O -> [Cr(CO(NH2)2)4]Cl3 + 3 H2O \uparrow + 2 CO(NH2)2 \uparrow}$
2. Decomposition of $\ce{2 CO(NH2)2}$ and $\ce{2 HCl}$: $\ce{[Cr(CO(NH2)2)4]Cl3 -> CrH6C2N4ClO2 + 2 CO(NH2)2 \uparrow + 2 HCl \uparrow}$
3. Decomposition of $\ce{2 CO2}$, $\ce{N2}$, $\ce{3 H2}$ and $\ce{1/2 Cl2}$ and oxidation of $\ce{Cr}$ to form $\ce{CrN2}$: $\ce{CrH6C2N4ClO2 -> CrN2 + 2 CO2 \uparrow + N2 \uparrow + 3 H2 \uparrow + 1/2 Cl2 \uparrow}$
Given the temperature range and pressure (probably standard pressure), the formation of $\ce{CrO2}$ is unlikely.