Question

back again with another analytical Q,

the question tells you to prep a 50 mL solution of a 750 ppm iron solution using ammonium iron III oxalate trihydrate (MM: 428.06 as a source of iron. Determine the mass of compound required.

I'm assuming you would start with a molarity calculation, M = mol / L, converting 750 ppm iron to moles of the ammonium compound and then to grams. Is this the way to go?

Answer 1

I would not go that way. When dealing with ppm I remember one very nice factor to keep in mind; i.e., 1 ppm = 1 mg/L.
So 750 ppm = 750 mg/L. You want 50 mL so that is 750 x 50 mL/1000 mL = 37.5 mg Fe = 0.0375 g Fe you want. Convert that to the oxalate trihydrate and you have it. I think it's way simpler that way.

Answer 2

Thanks sir! :)