- #1
SkittlesGal
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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.