I do these this way.
First, I keep H and O "normal" if I can and let C change to whatever I need.
The formula becomes C3H6O2
H = +1 each x 6 = +6
O = -2 each x 2 = -4
Which means C must be -2 for three of them or -2/3 each to make the compound zero (and all compounds are zero).
Here is a very good link.
http://www.chemteam.info/Redox/Redox-Rules.html
What are the oxidation numbers for the elements in the compound C2H5COOH?
1 answer