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Consider a percent error calculation where the measured value has 1 significant figure (XmL), and the actual/true value has 3 significant figures (0.XXXmL). Typically I would think that the percent error would end up being 1 significant figure, but what about the percent conversion factor, 100? This is an exact number, so I assume we don't this into account when thinking of significant figures... Is it really correct to use 1 significant figure in this case? Or is there a special rule for percent error calculations because it's a percent? I remember my lab instructor last year saying that you simply go to the one's place (no decimals) for percent error, but I'm not sure if that was actually correct/proper or just her specific instructions for us.

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You have a number of concepts conflated.

Consider a percent error calculation where the measured value has 1 significant figure (XmL), and the actual/true value has 3 significant figures (0.XXXmL).

If you only have one significant figure then you only have one significant figure. The "true" volume for any quantity of liquid would have a lot more than 3 significant figures. 1 ml of water has $3.45\times10^{22}$ molecules of water.

Typically I would think that the percent error would end up being 1 significant figure, but what about the percent conversion factor, 100? This is an exact number, so I assume we don't this into account when thinking of significant figures... Is it really correct to use 1 significant figure in this case? Or is there a special rule for percent error calculations because it's a percent?

There is no special rule for % conversion. So if you only have one significant figure then you'd have say $7\%$ or $7\times10^1\%$. You can't get $74\%$ conversion out of one significant figure. The whole point behind significant figures is too keep the experimenter from making up extra digits.

I remember my lab instructor last year saying that you simply go to the one's place (no decimals) for percent error, but I'm not sure if that was actually correct/proper or just her specific instructions for us.

That was just some instruction for that lab. If I figure that $\ce{CaCO3}$ weighs 100.1674 grams but the accepted value is 100.0869, then my % error is 0.0802%.

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