[noiseordinance]

Hi there. I'm working on a lab write-up that requires my average experimental value, the standard deviation, the theoretical yield, and the percent error of my experimental value in comparison to the theoretical yield.

So let's say my data is as follows:

Average experimental value: 52mL
Standard deviation: +/- 2mL
Theoretical yield: 56mL
Percent error:

I know how to do percent error when we are talking about one single experimental value vs. the theoretical value, but this is my first time doing an average of my experimental values vs. the theoretical value. Would I just simple use 52mL as my experimental value and ignore the uncertainty of +/- 2mL?

In other words, would this be accurate?

(Theoretical - Experimental) / Theoretical = (56mL - 52mL) / 56mL = 7.7%

Am I doing this correctly, or is there something I need to know about using the standard deviation when calculating percent error?

Thanks for any *delete me*

[ARGOS++]

Dear noiseordinance;

I’m not sure in your case if % error is identical to % deviation, but if, then you have to include the uncertainty into the final result.

That means you should get a result in the kind of:  7.1  ±  x.y  %
It is not a difficult process to do so.

Good Luck!
                    ARGOS⁺⁺

[noiseordinance]

Hey, thanks for the help. That makes sense, although I'm not sure how I'd find the percent error of a standard deviation... is there a way? I mean, I can't imagine I'd write 7.7% +/- 2mL, would I?

Thanks!

[ARGOS++]

Dear noiseordinance;

Think about:  How much %  is  ± 2ml of the theoretical yield?

Good Luck!
                    ARGOS⁺⁺

[noiseordinance]

Hahah, sweet. Got it. That's the kinda thing I love (and hate) about chemistry: just thinking about things the right way.

Thanks a ton!

[ARGOS++]

Dear noiseordinance;

You 're welcome!   ─   Soon again.

Good Luck!
                    ARGOS⁺⁺