Calculating Percent Yield: C₉H₈O₄ ± 5.8%

[Saturnfirefly]

*Actual yield of C₉H₈O₄: 0.11 g ± 5,8%
*Theoretical yield of C₉H₈O₄: 0.25 g

*Percent yield: (0.11/0.25)x100 = 43% ± 5,8%

Have I done anything wrong when taking the percent error down?
Help...=/

 

[dlgoff]

Thought this might help.

When dealing with product and division uncertainty propagation I think you are using relative or percentage uncertainties, so the effect of a constant 2 with no uncertainty, should have no overall effect on the relative uncertainty of the result.

https://www.physicsforums.com/showthread.php?p=1883849#post1883849"

 

[drdizzard]

The equation to find % error is

(abs(theoretical - measured/actual)/theoretical)*100=%error

abs=absolute value