How do I find percentage iron in an ore, given KMnO₄ titration data?

How do I find the percentage of iron in an iron ore sample, given the following data?

• 0.206g of KMnO₄ is dissolved in enough water to form a 100.0 mL KMnO₄ solution.
• Then 0.238g of iron ore is dissolved in 50.0 mL of distilled water.
• To this 1.7mL of H₂SO₄ is added.
• This solution is titrated with the KMnO₄ solution until it reaches a red endpoint. It takes 10.3mL of KMnO₄ solution to reach the red endpoint.
• Calculate the M (molarity) of KMnO₄ solution.

Mike

To attack a complex problem, try a 'divide and conquer' strategy as follows:

Step 1: Choose your target. You want to get a percentage of iron in the ore. If you had the mass of iron in the ore and the mass of the ore, you'd have your answer. The mass of ore is given, but the mass of iron isn't: your target is g Fe.

Step 2: Examine the given information. The problem gives you several pieces of information; look at each one closely and try to decide how it is related to your target.

0.206 g KMnO4 in 100.0 mL KMnO4 solution	This can be used to relate moles of KMnO4 to L KMnO4 solution.
0.238 g iron ore in 50.0 mL distilled water	The amount of ore will be necessary to compute the percentage of the iron in the ore. (By the way, water won't work as a solvent here. I believe this part of the problem should specify that all of the iron in the ore is converted to Fe2+(aq)).
1.7 mL of H2SO4	This tells you the redox reaction between the KMnO4 and the Fe2+ is occuring in an acidic solution. The amount of H2SO4 is not crucial here, since the limiting reagent in the titration will be the Fe2+.
10.3 mL of KMnO4 solution at endpoint	This tells you how much KMnO4 is required to completely react with all of the Fe2+ from the ore sample. (KMnO4 is deep purple and all the other species in the titration are nearly colorless. Just one drop of excess permanganate imparts a definite pink tinge to the solution.)

Step 3: Relate the target to information given in the problem. The key piece in the remainder of the problem is relating the amount of KMnO₄ added to the amount of iron in the ore:

mL of KMnO₄ solution    g Fe

Whenever you're confronted with a problem that asks for the amount of one substance given an amount of another substance, you must look for a mole-to-mole relationship between the two substances. Look for mole-to-mole relationships in chemical formulas or chemical equations supplied in the problem. In this case, the mole-to-mole relationship must be obtained from a balanced chemical equation between the Fe²⁺ and the KMnO₄. I'll give you the half reactions involved here in aqueous solution and let you combine them into a balanced equation for the titration yourself:

MnO₄^-(aq) + 8H⁺(aq) + 5e^- = Mn²⁺(aq) + 4H₂O(l)
Fe²⁺(aq) = Fe³⁺(aq) + e^-

Once you have the balanced equation, you can look at the coefficients and convert mol KMnO₄ to mol Fe. This gives you a 'bridge' from your starting point to the target:

mL of KMnO₄ solution    mol KMnO₄    mol Fe²⁺    g Fe

The mole-to-mole relationship cuts the problem up into 3 easier pieces:

• Milliliters of KMnO₄ are converted to mol KMnO₄ via the KMnO₄ molarity that you've already calculated.
• mol KMnO₄ is converted into mol Fe²⁺ via the mole^-to-mole relationship you obtain from the balanced chemical equation.
• mol Fe²⁺ is converted into g Fe via the atomic weight of iron.

Step 4: Perform and check the calculation.Verify that all the units in the calculation we set up in the previous step 'cancel' consistently. You should get an iron mass that is some fraction of 0.238 g (the iron can't weigh more than the ore it came from, right?) Use the grams of iron and the grams of ore to obtain the percentage of iron in the ore.

Author: Fred Senese senese@antoine.frostburg.edu

Copyright © 1997-2001 by Fred Senese
Comments & questions to senese@antoine.frostburg.edu
Last Revised 02/23/18.URL: http://antoine.frostburg.edu/chem/senese/101-hidden/moles/faq/titration-fe-kmno4.shtml