determining the % of iron in an iron(II) salt

hello, having bit of trouble on how to go about this.

Mn04^- + 8H^+ + 5Fe^2+ ----> Mn^2+ + 5Fe^3+ +4H20

I know the mass of the iron salt (9.56g). i dissolved this in deionised water to make a sol'n of 250cm^3. Pipetted 25cm^3 of this in a conical flask and added H2SO4 (for a H+ source), but the vol of H2SO4 does not really matter in the calculation.

I titrated 0.02M of potassium permanganate (KMnO4) and the 25cm^3 of the Fe^2+ salt sol'n. The average titre was 27.15cm^3.

so i calculated the number of moles of permanganate:

n = 27.15/1000 x 0.02 = 5.43 x 10^-4 moles

so that must mean the same number of moles of the salt solution.

so please could you suggest what i do next??? thanks!

Reply 1

mizfissy815

That is (5.43 x 10^-4 moles) the number of moles of MnO₄^-. So to calculate the number of moles of Fe²⁺, you use the information form the balanced equation.
1 mole of MnO₄^- : 5 moles of Fe²⁺
5.43 x 10^-4 : x

x= 2.715 x 10^-3 moles of Fe²⁺ in 25 cm₃

No. of moles of Fe²⁺ in 250cm₃= 10 x 2.715 ^-3 = 0.02715 moles

Mass of Fe²⁺ in sample= 0.02715 x 56 = 1.52g

Percentage of iron in the sample= (1.52/9.56) x 100 = 15. 9%

Or alternatively,

Original number of moles in iron salt= mass/ Mr = 9.56/56 = 0.1707 moles

Percentage of iron in the sample= (0.02715/0.1707 )x 100 = 15.9%