Calculating Percent Yield: C₉H₈O₄ ± 5.8%

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The discussion focuses on calculating the percent yield of C₉H₈O₄, with an actual yield of 0.11 g and a theoretical yield of 0.25 g, resulting in a percent yield of 43% ± 5.8%. The method for calculating percent error is clarified, emphasizing the use of absolute values in the formula: (abs(theoretical - actual) / theoretical) * 100 = % error. The conversation also addresses the propagation of uncertainty, noting that constant values without uncertainty do not affect the overall relative uncertainty.

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*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...=/
 
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Thought this might help.
LowlyPion said:
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"
 
Last edited by a moderator:
The equation to find % error is

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

abs=absolute value
 

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