Hello, Abdelmounaim, I think you can figure this out.  First, waters of hydration will leave and so you are left with Cu(NO3)2.  If your final solid is CuO, that leaves 2 N and 5O to account for.

After you try to figure out, check this site - it only differs by the number of waters of hydration.   http://www.nhehs.org.uk/chemistry/_Copper_nitrate_Heating.html

Provided there is no possibility of reduction CuO is it

Dear Abdelmounaim

The temperature causes leaving waters of hydration only and please mention the  temperature as it may be not CuO at all. 

Please read the following papers:

1.Seung Kon Ryu, Woon Kyu Lee and Soo Jin Park, Thermal Decomposition of Hydrated Copper Nitrate [Cu(NO3)2·3H2O] on Activated Carbon Fibers,Carbon Science,  Vol. 5, No. 4 December 2004 pp. 180-185.

2. I.V. Morozov, K.O. Znamenkov, Yu.M. Korenev, Thermal decomposition of Cu(NO3)2·3H2O at reduced pressures, Thermochimica Acta 403 (2003) 173–179.

Best regards

P. Sagan

The temperature causes leaving H2O only for this reaction but if you have

[Cu(CO3)2·3H2O] can find CuO and CO2