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Homework Help: Chemistry Mass Percent Question

  1. Sep 14, 2010 #1
    1. The problem statement, all variables and given/known data

    0.3106 g of an iron-containing compound yield 0.07017 g Fe2O3 upon oxidation. What is the mass percent of the iron in the compound??


    2. Relevant equations

    stoichiometry?


    3. The attempt at a solution

    (0.07017 g Fe203)(1 mole Fe2O3/ 159.7 g Fe2O3)(2 moles Fe/1 mole Fe2O3)(0.3106g Fe)(1 mole Fe) = 2.729468003x10^-4 grams

    Then I took 2.72x10^-4 grams / 0.07017 compound grams of Fe2O3 = .00388979 x 100 = .3889 percent .

    What did I do wrong?? Was I off on sig figs?
     
  2. jcsd
  3. Sep 14, 2010 #2
    If an oxidation happens, you have to begin in a lower oxidation state, you only have two options:
    Feº , or FeO

    In both options it are a 2:1 proportion front a Fe2O3
     
  4. Sep 15, 2010 #3

    Borek

    User Avatar

    Staff: Mentor

    OK. Number of moles of Fe2O3.

    OK. Number of moles of Fe in Fe2O3.

    No idea. It looks like ostrich, but has a head of giraffe.
     
  5. Sep 15, 2010 #4
    Ok, I find the mistake; ¿Why you multiply by the mass of Fe 0.3106?

    Compound with Fe 0.3106

    Fe in the compound "x"

    ¿How much Fe produces Fe2O3?

    0.07017 g Fe2O3 (1 mol Fe2O3 / 159.7 g Fe2O3)(2 mol Fe / 1 mol Fe2O3)---> mol of Fe

    mol Fe x (55.845 g Fe / 1 mol Fe) = 0.04907 g Fe = x

    Then

    Rock = 0.3106 g
    Fe in rock = 0.04907 g

    Percentage (0.04907/0.3106) x 100 = 15.8%
     
  6. Sep 15, 2010 #5

    Borek

    User Avatar

    Staff: Mentor

    Now it has a beak, as usual.
     
    Last edited by a moderator: Aug 13, 2013
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