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Calculating percentage by mass

  1. Sep 15, 2013 #1
    1. The problem statement, all variables and given/known data
    Many fertilizers contain ammonium phosphate as a source of phosphorus and nitrogen. A 7.225 g sample of a fertilizer is dissolved in water and mixed with excess barium chloride solution, and the following reaction occurs:

    2(NH4)3PO4(aq) + 3BaCl2(aq) -> Ba3(PO4)2(s) + 6NH4Cl(aq)

    The product mixture is found to contain 3.741 g of barium phosphate. Using this info, calculate the percentage by mass of ammonium phosphate in the fertilizer.

    2. Relevant equations

    % by mass = grams of fertilizer/molar mass of fertilizer x 100
    (unsure if that's correct, but that's what I think it should be)

    3. The attempt at a solution

    I started by taking the given amount of barium phosphate to find moles of ammonium phosphate. I think I need to use that to find grams of ammonium phosphate. However, I am also having difficulty trying to understand what the problem is specifying as the fertilizer. Is it the ammonium phosphate AND the barium chloride, or just the ammonium phosphate?
     
  2. jcsd
  3. Sep 16, 2013 #2

    DrClaude

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    Staff: Mentor

    There is no such thing as the molar mass of a fertilizer, as it is a mixture. You want to know what percentage of the total mass is due to ammonium phosphate:

    % by mass = mass of ammonium phosphate / total mass of fertilizer

    You can do that. You can also calculate everything using just masses.

    The fertilizer is a mixture of ammonium phosphate and other stuff. The barium chloride is added in the lab.
     
  4. Sep 16, 2013 #3
    Thank you DrClaude! Makes a whole lot of sense now (Answer I got: 25.65%)

    Steps:
    Grams of barium phosphate -> moles of barium phosphate -> moles of ammonium phosphate -> grams of ammonium phosphate.

    Then divided grams of ammonium phosphate by grams of fertilizer (multiplied by 100).
     
    Last edited: Sep 16, 2013
  5. Sep 16, 2013 #4

    Borek

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    Staff: Mentor

    Looks good.
     
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