[scarletthj]

Hello i am having difficulty with the oxidation numbers for the following equation.

4KNO3 + 7C + S ------> 3CO2 + 3CO + 2N2 + K2CO3 + K2S

I think that the oxidation numbers may be ~

4KNO3 = if i start with the oxygens i get -2 x 3 = -6, K = 0, N = 0, so is the oxidation number -6?

7C = 0

S = 0

3CO2 = If i start with the oxygens i get -2 x 2 = -4, and C = 0 x 3, so is the oxidation number just -4?

3CO = C = 0, O = -2, so is the oxidation number here just -2 (carbon 0 x 3 = 0)?

2N2 = 0

K2CO3 = K = 0, C = 0 O x 3 = -6, so just -6 for this part?

K2S = K = 0, S = 0, so oxidation number is 0?


Thanks for your help.

[Borek]

Oxidation numbers are assigned to ATOMS, not molecules.

http://www.chembuddy.com/?left=balancing-stoichiometry&right=oxidation-numbers-method

[scarletthj]

I have looked again the question and think the oxidation numbers are ~

            +1 +5 -6
4KNO3 = 4K N O3

7C = 0

S = 0
                -4         
3CO2 = 3C O2

          +2 0
3CO = 3C O

2N2 = 0
             -2 +4 -6
K2CO3 = K2 C  O3
       
         -2  -2
K2S = K2  S

Are any of the above correct?

I have always struggled with oxidation numbers.

Any help would be much appreciated.

[BluRay]

Maybe one answer is not enough for you, so I say it too: oxidation numbers are assigned to *atoms*, not molecules. So the o.n. of oxygen is -2, not -6.
In the reaction you have written, you have:

atom                                 o.n.
S in K₂S                             -2
C in CO                              +2
C in CO₂ or in K₂CO₃           +4
N in KNO₃                          +5
K in all those compounds     +1   
O in all those compounds     -2     

Furthermore, all elements have o.n. = 0.           
Of course, if you have more atoms in a compound, you multiply its o.n. by the number of atoms to get the overall formal charge.