How to calculate the mass of iron in .275 grams of iron iii sulfate?

Molar mass of iron(III)sulfate that is Fe₂(SO₄)₃ =399.88 g mol⁻¹  

Mass given of  iron(III)sulfate that is Fe₂(SO₄)₃= 0.275 g

Number of moles of iron(III)sulfate that is Fe₂(SO₄)₃=  

                                                                                  =0.00068 mol

there are two moles of Fe in 1 mole of Fe₂(SO₄)₃

So, in 0.00068 mol of Fe₂(SO₄)₃=2 ×0.00068 mol of Fe

                                                  = 0.00136 mol of Fe

molar mass of Fe=55.8 g mol⁻¹

So the mass of iron in 0.00136 mol of Fe= 55.8 g mol⁻¹ × 0.00136 mol

                                                                  =0.075 g

Mass of iron  =

The chemical formula of iron (III) sulfate =

First, calculate the number of moles of

Number of moles of iron (III) sulfate =

Molar mass of iron (III) sulfate = 399.88 g/mol

Number of moles of iron (III) sulfate  =

=

In one mole of iron (III) sulfate there are two moles of iron according to the chemical formula.

So, in of iron (III) sulfate there are = of iron.

Number of moles of iron  =

Molar mass of iron  =

Mass of iron  =

=

=

=

Hence, mass of iron =