How many moles of sulfate ions are in 100 mL of a solution of 0.0020 M Fe2(SO4)3

this is all i got so far
first i find the mole by 0.0020 M Fe2(SO4)3 *0.1L=0.0002 mole Fe2(SO4)3
then percent of (SO4) in Fe2(SO4)3 which 288/399.87=72.02%
find mole of SO4 0.0002 mole Fe2(SO4)3 *.7202=1.4*E-4
(this is still a wrong answer thought, ;(

Molarity x Volume = Moles
Ratio of Ion Moles to Total Moles x Number of Total Moles = Number of Ion Moles

So...

0.0020 moles/L (0.0020M) x 100mL = 0.0002 moles

Fe2(SO4)3 is 2 parts Iron to 3 parts Sulfate, or in other words, 3 parts Sulfate to 5 parts Total

3/5 x 0.0002 moles = 0.00012 = 1.2 x 10^-4 moles