Formal charges on CH2O and H2SO3

Question

I was taught thus way valence electrons subtracted from addition of dots plus bonds
For CH2O

Carbon 4-(0+3) = + 1
. Oxygen 6-(4+1) = - 1

For H2SO3 I am having issues figuring out the dots and bonds

Answer 1

I think you're having trouble with the CH2O, too. I think by your formula and the way I do it that the formal charge on C is 0 and the formal charge on O is 0. (and the formal charge on each H is 0).
Here is the dot structure.
http://www.thegeoexchange.org/chemistry/bonding/Lewis-Structures/CH2O-lewis-structure.html

The way you've been taught is
For C. valence e = 4
- additional dots = 0
- bonds and there are 4 (1 CH, another CH and 2 for C=O making 4)
so 4-4 = 0

The way I do it.
valence e = 4
1e from the CH bond belongs to C; ( that's 1)
1e from the other CH bond belongs to C(that's 2)
There are 4 electrons shared with the C=O and half of them belong to C.(That's 2 more to make a total of 4)
So C should have 4, it has 4, charge zero.