Question

Calculate weight of ferrous ammonium sulfate needed[ Fe(NH4)2(SO4)2 x 6H2O ] to prepare 250 mL of 100 ppm ferrous ion?

Answer 1

So i know 100ppm = 100 mg/L , so 100 mg/L of ferrous

100mg/L x (0.250L) = 25mg of Fe

then with the 25mg = 0.025g of Fe I have to find how much Fe(NH4)2(SO4)2 x 6H2O is needed but I'm having a brain fart as to where to head from here

Answer 2

Okay, if we take your calculations so far as correct, you need to know how much FAS you need to have 0.025 g of Fe content, yes? Can you find the molar mass of FAS, and of Fe? The molar mass of Fe/molar mass of FAS will give you the fraction of any quantity of FAS that is Fe.

Answer 3

So basically 55.845/284.05=0.197 then use this ratio to find the value of FAS that contains 0.025g of Fe right

So equation will be like
Mass FAS x Ratio Fe/FAS = Mass Fe
and solve for Mass FAS ?

@whpalmer4

Answer 4

Well, you didn't count the molar mass of that 5H2O, it appears, so your answer will not be correct, but the procedure is right.

Answer 5

sorry, 6 H2O, not 5 H2O

Answer 6

Ah true so 393.1388 g/mol thanks for helping me through this I would have gone mad because I knew my steps made sense. :)