[julixtta]

Hello! I am completing an online laboratory and I am having difficulty in calculating the percent of iron by mass in my unknown iron(II) compound.

Here are the procedure instructions:
Procedure:
1. Add just over 0.2g of unknown iron(II) salt to an Erlenmeyer flask and mass it using the analytical balance. Dissolve the salt in DI water and then add 5mL of 3M sulfuric acid and 3mL of concentrated phosphoric acid to the flask. The phosphoric acid will be found in the fumehood along with a Pasteur pipet marked to show the volume of 1mL.
2. Titrate the solution very slowly initially (drop by drop) with the 0.01M KMnO4 for roughly the first 2mL. When the color from each drop disappears immediately then you can increase the rate of addition to that of a normal titration and finish the titration.
3. Complete as many titrations as possible.

The results that my teacher gave us:
Permanganate Solution where 0.3140g KMnO4 was titrated in 200mL solution
            Mass Iron Salt (g)  Initial Buret Volume     Final Buret Volume
Trial 1      0.1982g                  0.01mL                        10.03mL
Trial 2      0.2137g                10.03mL                         20.85mL
Trial 3      0.2074g                20.85mL                          31.32mL

My thought process and calculations:
I know that I need to find the number of moles of iron in the sample.
mol Fe 2+ = (5 mol Fe 2+/ 1 mol MnO4-) * M * V

And then I should multiple that by the mass of iron in the sample in grams (55.85 g mol^-1), and then finally divide that number by the grams used and then multiply by 100%. So the full equation would be:

mass % Fe = (5 * M * V * 55.85 g mol^-1)/grams * 100%

The five is from the balanced chemical reaction of 5 Fe{2+} + MnO4{-} + 8 H{+} → 5 Fe{3+} + Mn{2+} + 4 H2O.

I am just confused about the numbers that I should be plugging into the formula. Is that the correct formula to use?
Is V the final-initial? Is M the moles?

For moles of KMnO4, I did the following calculation:
molecular weight = 158.034 g/mol      
mass used = 0.3140 grams      
grams * 1/molecular weight = moles      
0.3140 grams * 1/158.034 g/mol       
0.0019869142 moles of KMnO4   

I've also found a similar lab online and the attachment is what the calculations suggested to do.

Thank you for the help in advance!

[Borek]

M is molarity.
V is volume of the titrant used to titrate the sample.

2. Titrate the solution very slowly initially (drop by drop) with the 0.01M KMnO4

Molarity is mentioned here, but it is probably not the exact concentration you should use, as you are given this information too:

Permanganate Solution where 0.3140g KMnO4 was titrated in 200mL solution

Titrated, or dissolved? If dissolved - what was the concentration of the titrant solution?