Question

An experiment produced 0.10 g CO2, with a volume of 0.056 L at STP. If the accepted density of CO2 at STP is 1.96 g/L, what is the approximate percent error?

Answer 1

The experiment result shows that the density of CO2 at STP is 0.1/0.056=1.79 g/L. The percent error equals: |1.79-1.96|/1.96*100%=8.67%. So the answer is 8.67%.

Answer 2

The experimental density is calculated as: $\frac{0.10}{0.056} = 1.79$ g/L. To calculate the percent error, we use the equation:
$\frac{|Experimental-Theoretical|}{Theoretical}=\frac{|1.79-1.96|}{1.96}=0.0889$
Therefore, the % error is 8.89%