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

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-

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-?

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

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) = ?

To find the mass of Fe2+ and the mass of SO4- in your experiments, you need to use the stoichiometric ratio between potassium permanganate (KMnO4) and the reactants (FeSO4 and sulfate ions). Here's how you can calculate it step-by-step:

1. Calculate the number of moles of KMnO4 used in each trial:
- Convert the volume of KMnO4 used to liters by dividing by 1000 (since 1 ml = 0.001 L).
- Multiply the volume in liters by the molarity of KMnO4 (0.01 M) to get the number of moles.

For example, for Trial 1:
Volume KMnO4 = 20.83 ml = 0.02083 L
Moles KMnO4 = 0.02083 L * 0.01 M = 0.0002083 mol

2. Use the balanced chemical equation to determine the stoichiometric ratio between KMnO4 and FeSO4.
The balanced equation is as follows:
5FeSO4 + MnO4- + 8H+ -> 5Fe3+ + Mn2+ + 4H2O + SO4^2-

From the equation, you can see that 1 mole of KMnO4 reacts with 5 moles of FeSO4.

3. Find the moles of FeSO4 reacted for each trial:
Multiply the number of moles of KMnO4 by the stoichiometric ratio (5 mol FeSO4 / 1 mol KMnO4).

For example, for Trial 1:
Moles FeSO4 = 0.0002083 mol * 5 = 0.0010415 mol

4. Calculate the mass of Fe2+:
Multiply the moles of FeSO4 by the molar mass of FeSO4 (Fe atomic mass + S atomic mass + 4 * O atomic mass).

For example, for Trial 1:
Molar mass of FeSO4 = (1 * atomic mass of Fe) + (1 * atomic mass of S) + (4 * atomic mass of O)
Molar mass of FeSO4 = 55.845 g/mol + 32.06 g/mol + (4 * 16.00 g/mol) = 151.906 g/mol

Mass of Fe2+ = 0.0010415 mol * 151.906 g/mol = 0.1581 g

5. Calculate the mass of SO4-:
Multiply the moles of FeSO4 by the molar mass of SO4- (S atomic mass + 4 * O atomic mass).

For example, for Trial 1:
Molar mass of SO4- = 32.06 g/mol + (4 * 16.00 g/mol) = 96.06 g/mol

Mass of SO4- = 0.0010415 mol * 96.06 g/mol = 0.100 g

Repeat these calculations for Trials 2 and 3 using the corresponding values.