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Molar Mass Percent Error

  1. Oct 13, 2013 #1
    1. The problem statement, all variables and given/known data
    I did a lab on the collection of butane over water to calculate the molar mass. In one trial, the measured molar mass was 20 g/mol, however, the accepted value for butane is 58.14 g/mol. This means the percent error is -70%. Why is it so high, assuming there is no relevant human error causing it?


    2. Relevant equations
    n=PV/RT
    (measured value-accepted value)/accepted value


    3. The attempt at a solution
    I think it's so high because the accepted value is gathered at STP, and although the vapor pressure was accounted for, it still does not entirely account for the error.
     
  2. jcsd
  3. Oct 14, 2013 #2

    Borek

    User Avatar

    Staff: Mentor

    Huge error for a single trial is meaningless. Speculating over what have happened when you can't eliminate human error is a waste of time.
     
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