Molar Mass Percent Error

  • #1

Homework Statement


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?


Homework Equations


n=PV/RT
(measured value-accepted value)/accepted value


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.
 

Answers and Replies

  • #2
Borek
Mentor
28,676
3,167
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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