Predicting the product of a combustion reaction with CH5NO

__CH5NO + __O2 --> ?
I'm having trouble trying to determine the oxidation numbers in CH5NO. I can't seem to find a way to make the overall charge neutral. Does it matter? Could it be a typo? My worksheet has (N-->N+5) above the N in CH5NO, if that helps.

This is as far as I've managed to get
__CH5NO + __O2 --> __H2O + __CO2 + __N?

Help?

The compound that is being combusted is could be either O-methylhydroxylamine (CH3ONH2) or N-methylhydroxylamine (HONH(CH3), both of which are CH5NO. The products you've indicated are, water, carbon dioxide and nitrogen, but nitrogen is diatomic. However, it is unlikely that either compound would combust to nitrogen, and much more likely that you'd get an oxide of nitrogen, probably NO2. Here's the equation I'd use:

4 CH5NO + 11 O2 -------------> 4 CO2 + 10 H2O + 4 NO2

Now if your teacher is forcing you to use nitrogen as one of the products, even though I'm certain that it isn't one, use this equation 4 CH5NO + 7 O2 -------------> 4 CO2 + 10 H2O + 2 N2

H2NCH2OH ie, amino methanol
H @ +1
O @ -2
C @ + 4
N @ -3
N is only @+5 in nitro [-NO3] compounds, which you do not have here.

__CH5NO + __O2 --> __H2O + __CO2 + __N? ==== That's good; now balance it....
Start with 2 CH5NO so you get 1 N2 [no fractions] [also integral H2O]
Then see how many N2, CO2, and H2O you can get
2 CH5NO ---> 5 H2O + 2 CO2 + N2
OK, how many O2 needed to do that ?
2 CH5NO + 3.5 O2 ---> 5 H2O + 2 CO2 + N2
Oh well, double the whole equation to get rid of the fraction

CH5NO = methoxyamine
4CH5NO + 9O2 --> 4CO2 + 10H2O + 4NO