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Dilution Factors, Determining Molarity

  1. Oct 30, 2008 #1
    1. The problem statement, all variables and given/known data

    "A scientist prepares 100 mL of an iron (III) nitrate solution by weighing out 4.8372 grams of the solid iron (III) nitrate and dissolving it in deionized water using a 100 mL volumetric flask. She then pipettes 5 mL of that solution into a 20 mL volumetric flask and fills it to the mark, then dilutes the resulting solution by a factor of 100. What are the molarities of Fe3+ (aq) and NO3- (aq) in the final solution? Show your work."

    2. Relevant equations

    Molarity = moles / liter
    M1V1 = M2V2

    3. The attempt at a solution

    I believe there are two ways to solve this (correct me if I'm wrong)

    1. Raw calculation:
    Well it starts off with 100 mL of deionized water added to 4.8372g of Fe(NO3)3. That solution will be 0.048372g Fe(NO3)3 / mL of H2O. She takes 5 mL of that solution (0.048372g/mL x 5 mL = 0.012093 g of Fe(NO3)3), and then dilutes it by a factor of 100. A dilution factor of 100 is the same as 99 mL of H2O and 1 mL of solution correct?

    Not sure how I would apply the dilution factor .. would it just be 0.012093g of Fe(NO3)3 / 99 mL H2O?

    2. I imagine M1V1 = M2V2 could provide a shortcut..
     
  2. jcsd
  3. Oct 30, 2008 #2

    Borek

    User Avatar

    Staff: Mentor

    Yes.

    Why 99 mL? What is volume after mixing 99+1?

    And don't forget to convert it to molarity now.

    Not a shortcut - exactly the same amount of work. One part is molarity calculation (just done at the beginning), the other - calculation of all dilution steps.

    Don't forget final concentrations of Fe3+ and NO3- will be different.
     
  4. Oct 30, 2008 #3
    Oh so to account for total volume it must be 0.012093g of Fe(NO3)3 / 100 mL H2O?

    So therefore, 0.00012093g of Fe(NO3)3/mL of H2O, which is the same as 0.12093g of Fe(NO3)3/L of H2O

    Molar mass of Fe(NO3)3 is.. 241.8602 g/mol, so:

    0.12093g Fe(NO3)3 / L x 1 mol / 241.8602g = 0.0004999 mol / L

    So the Fe(NO3)3 will be 0.0004999M. Molarity of Fe will be 0.0004999M, and the molarity of NO3 will be 0.0004999M x 3 I believe since there are three NO3's for one Fe.

    Informing me of any mistakes along this pathway would be greatly appreciated :)
     
  5. Oct 30, 2008 #4

    Borek

    User Avatar

    Staff: Mentor

    I got 241.8597 for molar mass, which leads to 5.000 instead of 4.999. Otherwise looks OK.
     
  6. Oct 30, 2008 #5
    Thanks!
     
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