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Empirical Formula of a Compound with only percentages given

  1. Mar 26, 2009 #1
    1. The problem statement, all variables and given/known data
    Determine the Empirical Formula of a compound which, when analyzed, is shown to consist of 62.07% carbon, 27.59% oxygen and 10.34% hydrogen.
    The answer is supposedly C3OH6

    2. Relevant equations



    3. The attempt at a solution
    First,I started by writing the ratio: C : O : H
    Then, under each, I wrote the respective percentages.
    I divided everything by the lowest percentage (10.34), giving me a ratio of 6:2.668:1, (C6O3H).
    I have no real idea of what to do or where I went wrong. Thanks so much for your help!
     
  2. jcsd
  3. Mar 26, 2009 #2
    I'm really sorry- worked it out from another thread:)

    Percent to mass,
    Mass to mole,
    Divide by small,
    Multiply 'til whole.

    Assume 100g of compound. Change % to g. Divide these masses by their molar masses to get moles. Divide each mole amount by the smallest mole quantity of the group. You'll get numbers that are either close to a whole number or close to half (such as 1.5, 2.5, etc.) multiply all of the numbers by a common factor until they are all whole numbers, and this will be the empirical formula.
    Thanks to Gannon on thread: https://www.physicsforums.com/showthread.php?t=224348 :)

    PS: how do I delete a thread once I found the solution (before any posts)?
     
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