Post a New Question


posted by .

So I did a titration experiment using 0.01 M potassium permanganate. In an erlenmeyer flask I placed a mass sample of FeSO4- with enough DI water to have a total volume of 20-25ml. I recorded the volume of MnO4- it took for the solution to change color (react). I performed a total of three trials (each with different but close to-each-other masses) and recorded the volume of MnO4- for each. From these results, I have to find out the mass of Fe2+ and the mass of SO4-. How do I do that?

Here are the results I got:
Mass sample (g):
Trial 1: 0.257
Trial 2: 0.257
Trial 3: 0.255

Volume MnO4- (ml):
Trial 1: 20.83
Trial 2: 20.78
Trial 3: 20.39

  • Chemistry -

    You should have 2- for the charge on the SO4^2- above.
    MnO4^- + 5Fe^2+ ==> 5Fe^3+ + Mn^+2

    You can add the H^+ and H2O as well as K^+ and SO4^2- if you wish; however, none of that is necessary. You need what I have written above.
    mols MnO4^- = M x L = ?
    mols Fe^2+ = 5 x mols MnO4^-
    g Fe = mols Fe x atomic mass Fe
    g SO4 = mols Fe x molar mass SO4^2-

  • Chemistry -

    thank you so much! It makes more sense now!! I was clearly on the right path but scared I was making a mistake.

    For the g of SO4 is it mols Fe x mw of SO4^2-? not mols of SO4^2- x mw of SO4^2-?

  • Chemistry -

    I meant moles of MnO4^- which according to the balance reaction is 1

  • Chemistry -

    It makes no difference since mols Fe = mols SO4.
    g Fe = mols Fe x atomic mass Fe.
    g SO4 = mols SO4 x molar mass SO4 OR
    g SO4 = mols Fe x molar mass SO4. You can do it another way, also.
    g SO4 = g Fe x (molar mass SO4/atomic mass Fe) = ?

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question