[Rutherford]

A 1.000 g sample of the nitrate of a certain metal was thermally decomposed at 220 °C to give 0.4858 g of a solid residue. What nitrate was used in the experiment?

I did it on trial and error.
-Firstly assumed that a nitrite and oxygen are made;
-then the metal, oxygen and nitrogen(IV)-oxide;
-and then the oxide and nitrogen(IV)-oxide.
Here I got the answer Mn (55g/mol) to be the metal. The procedure I wrote takes a lot of time. Can I conclude somehow what could be made?

[sankalpmittal]

A 1.000 g sample of the nitrate of a certain metal was thermally decomposed at 220 °C to give 0.4858 g of a solid residue. What nitrate was used in the experiment?

I did it on trial and error.
-Firstly assumed that a nitrite and oxygen are made;
-then the metal, oxygen and nitrogen(IV)-oxide;
-and then the oxide and nitrogen(IV)-oxide.
Here I got the answer Mn (55g/mol) to be the metal. The procedure I wrote takes a lot of time. Can I conclude somehow what could be made?

Not a good method. Let the metal be M. Then write the equation for
M_x(NO₃)_y------->

Then metal oxide can be treated as residue , and POAC (principle of atomic conservation) or mole method can be applied.

[Rutherford]

Why only the metal oxide? A solid residue could also be the nitrite of the metal or the metal itself? In every case I did it similar to the way you posted: M(NO₃)_x ... I want to know if there is a way to solve this problem quicker than just assuming and repeating.

[sankalpmittal]

Why only the metal oxide? A solid residue could also be the nitrite of the metal or the metal itself? In every case I did it similar to the way you posted: M(NO₃)_x ... I want to know if there is a way to solve this problem quicker than just assuming and repeating.

Ok , I see. Since you do not know what type of element (i.e. group number etc...) you can not be sure of. For eg. Na etc give nitrite in addition to metal oxide as residue. Assuming and repeating brute force sort of method is only possible (which you tried) , if its a stoichiometric problem.

[Rutherford]

Okay then  . I am not happy with it being the only way but I have to accept it. Thanks for the *delete me*