1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

% of titrated sample

  1. Dec 8, 2008 #1
    1. The problem statement, all variables and given/known data

    A 0.6500 g sample of an impure FeSO4 sample is titrated with 33.25 mL of 0.0190 M KMnO4 to an endpoint. What is the % of iron in the sample?


    2. Relevant equations

    KMnO4 + FeSO4 --> KSO4 + FeMnO4

    % Fe = (grams of Fe / grams of sample) * 100%

    3. The attempt at a solution

    0.03325 L sol'n * (0.0190 M KMnO4 / 1 L sol'n) = 6.3175*10-4 M KMnO4

    g FeSO4 = 6.3175*10-4 M KMnO4 * (1 mol FeSO4 / 1 mol KMnO4) * (151.903 g FeSO4 / 1 mol FeSO4 = 0.09596 g FeSO4

    But now I'm stuck. How do I find g of Fe from 0.09596 g FeSO4?

    % Fe = (grams Fe alone/ 0.6500g FeSO4)*100% = final answer.
     
  2. jcsd
  3. Dec 8, 2008 #2

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    What kind of reaction occurs in the analytical titration? Is this an oxidation-reduction reaction? If it is, your reaction statement is wrong.
     
  4. Dec 8, 2008 #3
    Does that mean it's an acid-base reaction. If it is, then acid + base --> salt + water
     
  5. Dec 8, 2008 #4

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Sami233, how do you come to the original question that you asked? Are you enrolled in a relevant course now for which you are trying to answer the exercise question?

    KMnO4 is not necessarily an acid or base. FeSO4 is not necessarily an acid or base. What do you expect to occur in the reaction between the two? I strongly expect the reaction to be an oxidation-reduction one. Can you write a balanced reaction for this? Can you start with the two separate half-reactions? What kind of endpoint would you expect to observe?
     
  6. Dec 8, 2008 #5
    2 KMnO4 + FeSO4 --> Fe(MnO4)2 + K2SO4
    Fe + H2SO4 --> FeSO4 + H2
     
  7. Dec 8, 2008 #6
    Endpoint can be determined by:
    2MnO4 + 10FeSO4 + 8H2SO4 --> 5Fe(SO4)3 + 2MnSO4 + K2SO4 + 8H2O

    0.03325 L * (0.019 M KMnO4/1L) = 6.3175 x 10-4 mol KMnO4

    6.3175 x 10-4 mol KMnO4 * (10 mol FeSO4/2 mol MnSO4) * (151.9 g FeSO4/1 mol FeSO4) = 0.4798 g FeSO4

    0.4798 g FeSO4 * (1 mol FeSO4/151.9 g FeSO4)* (55.847 g Fe/1 mole Fe) = 0.1764 g Fe

    % Fe = (0.1764 g Fe/0.6500 g sample)*100% = 27.139%
     
  8. Dec 8, 2008 #7

    symbolipoint

    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    That reaction is not balanced.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?