Calculate the molar mass of Iron(III) sulfate pentahydrate,Fe2(SO4)* 5H2O giving the correct units.?

Question

Show work please! Thank You

Answer 1

It is 56[Fe]*2+(32[S]+4x16[O])x3+5x(2x1[H]+16[...

You forgot. It must be Fe2(SO4)3.5H2O

Answer 2

The correct formula for Iron (III) sulfate penta hydrate is

Fe2(SO4)3.5H2O

(Iron (III) has a valency of 3+)

molar mass is obtained by summing the atomic mass of all atoms in the molecule. The atomic mass of an element is the weight in g of 1 mole of that element so the units are g/mol

So molar mass = (2 x 55.85 g/mol ) + (3 x 32.07g/mol)  + (17 x 16.00g/mol ) + (10 x 1.008g/mol) = 489.99 g/mol

Answer 3

Anhydrous Iron (III) sulfate is 399.87 g/mole.

The pentahydrate is 489.94 g/mole.