You don't need to use molar mass of cation or anion. You know the molarity of salt. You also know (for the formula (NH4)2SO4) that for one mole of you salt you receive (after dissolution) two moles of cation and one of anion. On the basis of that easy ratio calculate molarity of both ions.
Ah I see. So then I can write:
21.9675g (NH4)2SO4 * (1 mole / 132.140g/mol) * (2 moles NH4 / 1 mole (NH4)2SO4) = (0.33248827 / 500 mL) * (1000 mL / 1 L) = 0.66497654 M NH4+
And then for S:
21.9675g (NH4)2SO4 * (1 mole / 132.140g/mol) * (1 moles SO4 / 1 mole (NH4)2SO4) = (0.33248827 / 500 mL) * (1000 mL / 1 L) = 0.166226524 mole
(0.166226524 mole / 500 mL) * (1000 mL / 1 L) = 0.332488 M SO4
Which doesn't make sense to me because how can SO4 anion have the same Molarity as the entire compound?
P.S. Sorry for the double post.