[mrstrh]

CH3COOH
CH3CNO
CH2CO
C2H5OH

i am trying to find the oxidation number for carbon in each of these. in the first one i think the oxidation is
-3 and the secondproblem it is -3 and the third problem is -2. i dont know if i am doing this right but if some one can help i would appreciated it i also would like someone to explain how to find the oxidation numbers on  larger problems like these.thanks

[Tetrahedrite]

Carbon almost exclusively has an oxidation state of +4 in simply organic compounds

[mrstrh]

yes it has+4 in normal states up it number can change because it goes from +4 to -4

[Tetrahedrite]

Yes, it can have a -4 state eg CO2, but in all examples u have given the oxidation state is +4. The third compound doesn't seem like a compound that would be very stable, is this a typo?

[AWK]

Oxidation state is a formal number because we should use a few rules, sometimes contradictory with our understanding of periodic table and electronegativity.
Only two rules are absolutely valid:
1. Sum of oxidation states is equal to 0 for molecule (or to charge of ion for ion).
2. Oxidation state for elements is always 0.
The other rules based on periodic table and electronegativity are not absolutely valid, so I may found my private rules just for one or more reactions, assuming that rules 1 and 2 are fullfilled.
Usually, For O oxN=-2 and for H OxN=+1 (on the basis of periodic table and electronegativity)
CH3COOH for C OxN=0
CH3CNO for C OxN=+1 (for N oxN=-3 assumed)
CH2CO for C OxN=0
C2H5OH for C Ox=-2

But someone who solves equation, eg:
Cr₂P₂O₇ + I₂ = Cr₂O₃ + CrP₃O₉ + PI₃
using periodic table and electronegativities can be called Master of Oxidation Numbers
With private rules it is almost as easy as C+O₂=CO₂

[movies]

I thought CO₂ was C⁴⁺?  Isn't oxygen always 2- except in ozone?

[Mitch]

CARBON IS NOT ALWAYS +4 IN ORGANIC CHEMISTRY!!!

Take something like C₂H₄, a double bond. The Carbon has an oxidation state of +2. Then look at ethane C₂H₆. Now carbon has an oxidation state of +3. Carbon lossed an electron, so it was oxidized. That is why we say we oxidized a molecule when we break-up the double bond. And that is why we say, "we reduced" when we form double bonds.

[Tetrahedrite]

You are correct, my apologies!