Ok, today I did the salt analysis of hydrated manganese ethanoate.

One of the tests included the decomposition of the solid by heating, firstly gentle heating then strong heating.

I got water of crystallisation on gentle heating and the solid melted to a plastic like liquid and of off-white colour. Just wanted to confirm the equation of the reaction:

(CH_3COO)_2Mn.4H2O (s)\frac{\triangle}{\rightarrow}Mn(OH)_2(aq) + 2(CH_3COOH)(g)

Then on strong heating, I got a black deposit, and acetone smell. Here I don't know the reaction. What are the products besides propanone (acetone). Is it Mn(OH)2 or Mn(OH)3 or another compound? I'll balance afterwards.

(CH_3COO)_2Mn(s) \frac{\triangle}{\rightarrow} Mn(OH)_2(s) + (CH_3)_2CO(g)

Thar one seems no good... :(

Although I've done a lot of organometallic chemistry, I can't say I've ever done anything like this.

The first equation looks correct except there's no water on the right (except the "(aq)" -- I'm not sure that's correct), but I'm pretty sure you understand that. You are driving off water so I'd put it on the right and remove the "(aq)".

Manganese usually forms Manganese (II) compounds. Depending on how hot it got, it is probably MnO (loses 1 more mole of water).

Mn(OH)_2 = MnO + H_2O

It's probably a stepwise process. First heating to drive off ethanol and form Manganese (II) hydroxide, then strong heating to drive off more water, and form MnO, and also decompose any additional organic material in the residue.

I am puzzled. Acetone formation from Ethanoate requires oxidation. That implies some reduction process. Are you forming manganese metal? Probably not. Maybe it's a disproportionation reaction of the organic, itself. You must be forming carbon (I guess it does turn black, as you mentioneed.) The reduction of carbon supports the oxidation of ethanoate (ethanol) to acetone.

Hey! I stated I'm doing salt analysis!! There were other test which showed me that the salt was (CH3COO)2Mn and the teacher told us that the salt has 4 moles of water of crystallisation for each mole of (CH3COO)2Mn.

I also know that MnO is white... or is does it become black upon strong heating?

(CH3COO)2Mn is the starting material, right? A salt of acetic acid. I probably thought incorrectly when I said ethanol. I got confused (old folks do that). Why did you call it ethanoate instead of acetate?

You undoubtedly are driving off acetic acid. Depending on how hot you heat it, you will form Mn(OH)2 or MnO. After heating hot, it undoubtedly goes to MnO.

The black undoubtedly comes from carbon. On heating to high temperatures, the residual salt will decompose further. (CH3COO)2Mn definitely can't stand up under high heat.

(CH_3COO)_2Mn \frac{\triangle}{\rightarrow} C + MnO + (CH_3)_2CO

Ok, I now get to this after some reflexion... I'll balance if it's good...

The problem is that this type of reaction usually isn't very clean. You get other garbage in addition to what you've written. Unless you do a complete analysis (mass spectrometer?), it's difficult to balance.

Yep, I've tried it. There always is 2 more O on the LHS... I'm thinking perhaps there's an CO2 produced... do you think so?

I can almost guarantee that there would be some CO2 produced.

Ok thanks! I'll try it!