Calculating Molar Mass Percent Error in Butane Lab Results

[SkittlesGal]

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.

 

[Borek]

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.