I need to prepare a 100 ppm iron from ferrous ammonium sulfate and don't know how much to weigh out.

I've been trying to do the calculation but keep getting it wrong.
Once again, I have ferrous ammonium sulfate and need to get a 100 ppm Iron from it, how much will I have to weigh out and how do you do the calculation?

I will assume that you are using the anhydrous form. A similar calculation follows for the hydrate.

(NH4)2Fe(SO4)2 molar mass = 284.05 g/mol

1 mole of (NH4)2Fe(SO4)2 contains 1 mole of Fe

Atomic mass of Fe = 55.845 g/mol

Equivalent mass ratio of (NH4)2Fe(SO4)2 to Fe = (284.05 g/mol)/(55.845 g/mol) = 5.0864 g (NH4)2Fe(SO4)2/g Fe

Wanted: 100 ppm Fe
1 ppm = 1 mg/L
Therefore, 100 ppm Fe = 100 mg Fe/L

(100 mg Fe/L)*(5.0864 g (NH4)2Fe(SO4)2/g Fe) = 508.64 mg (NH4)2Fe(SO4)2/L