[Primus]

I don't know what molar mass to use since there is apparently an anhydrous form and a hexahydrate form. The instructions for the experiment did not specify. All it said was that 1L of the solution contains 0.07g Iron(II) ammonium sulfate. I know how to do the calculations. I just don't know the molar mass. Is one form more common than the other?

[Arkcon]

The solution is described with a molecule at a certain molarity.  So that's what you have to make:  the solution at that molarity.  How you get there doesn't matter.  If you have the hexahydrate, you'll need a greater mass of solid, but once its dissolved in a minimum amount of water, and worked up to one liter, it will be the correct molarity of solute. 

I went around the block a few times to get the same point in that explanation didn't I?  Sorry.  But I hope you get the point.  Often, the anhydrous salt isn't soluble in water, even if the hydrate is.  That's not a joke.  The anhydrous salt just sits there, gradually hydrating and gradually dissolving.  So often, we use the hydrate.

[Primus]

The solution was prepared by my lab instructor. I just need the amount of iron to do the calculations in my lab report. I'm just going to guess that he used the hexahydrate to prepare the solution which would make it 1.785 x 10^-4 M Fe²⁺.

[Arkcon]

No.  Sorry if I wasn't clear.  If it says it has 0.07g Iron(II) ammonium sulfate per liter, then it does.  Again, how it got there may have been with the hydrate, but the reported g/L is as if it was anhydrous.  Try to realize, that when a hydrate dissolves in water, the attached water molecules leave and become part of the solvent water, so their mass no longer "counts" once dissolved.

[Primus]

OK Thank you very much! That means it is 2.464 x 10^-4 M Fe²⁺.