[WhiteSkyRising]

I'm having difficulty studying for my final exam. I'm reviewing old finals, and this question stumps me:

The percentage of water in an unknown hydrate was determined by heating the sample and driving the water off the sample. Two independent measurements gave values of 19.564 and 21.731%. Its percentage should be reported as:
A. 20.648%
B. 20.65%
C. 20.6475%
D. 21%
E. 20.6%

The answer key says D is the correct answer, but I don't know why. Aren't there 5 significant figures in the question? The average of the two is C, but the correct amount of significant figures would be A. The only way I can see getting to C from A is to continue rounding up. Am I supposed to do that?
Thank you.

[Borek]

No idea what the real answer is - but don't worry. Sig figs have no meaning in real chemical life. I understand your worry when it comes to exam, but IMHO there is no reasonable way to explain why 21% is the correct answer - I mean, show me the explanation and I will show it is wrong.

Check this site:

http://www.av8n.com/physics/uncertainty.htm

While the author is notorious for giving ultimate answers that not everybody accepts, he is IMHO  right about the sig figs issue.

[WhiteSkyRising]

Hahaha... thanks a lot.

"Significant figures are a convenient
way of getting the wrong answer."

Love that phrase... anyway thank you for your time, I won't worry about it then.

[Cuttlefish]

An old professor once told me to only ever give percentages as a whole number. Given that the average of the two to the nearest % is 21%, I would have gone for D (and got it correct, as it happens). Other than that, I have no idea either.

[Borek]

An old professor once told me to only ever give percentages as a whole number.

Sure. What about 60.66% of chlorine in NaCl?

[Yggdrasil]

Here's how I would approach the problem.

Every reported measurement should have two components, a value and an uncertainty.  In the case when only a specific value is reported, the concept of sig figs allows you to infer the uncertainty.  For example, if someone reports 20.0%, you can infer than the uncertainty is +/- .1.  If someone reports 20.000%, then you can infer that the uncertainty is +/- 0.001.

So, in the case of these two measurements, the average is 20.6475 and the uncertainty is 1.0835 (since this is the average distance of your measurements from the mean).  So, your measurement is valid to the ones digit and you should report 21% as your percentage (although 21 +/- 1% would be a better way of putting things).

[Borek]

and the uncertainty is 1.0835 (since this is the average distance of your measurements from the mean).

You may as well throw dice or even ask palm reader (www.handanalysis.com) to tell the uncertainty. All numbers will be of the same statistical value

[Yggdrasil]

I may not be an accurate estimate of the uncertainty, but it is the best estimate that you can get from the data available.

[enahs]

If someone reports 20.000%, then you can infer that the uncertainty is +/- 0.001.

So, in the case of these two measurements, the average is 20.6475 and the uncertainty is 1.0835 (since this is the average distance of your measurements from the mean).  So, your measurement is valid to the ones digit and you should report 21% as your percentage (although 21 +/- 1% would be a better way of putting things).

I do not see how you go from 0.001 to the "ones digits"?

[Yggdrasil]

The 0.001 does not refer to the part about the "ones digit."

For the data set {19.564, 21.731}, the mean μ= 20.6475 and the standard deviation σ = 1.0835.  Since σ is of order 10⁰, the ones digit it the significant digit.

[Borek]

I may not be an accurate estimate of the uncertainty, but it is the best estimate that you can get from the data available.

Technically you are correct ("best estimate") but the result of such calculations is reliable just like saying that in the case of 1 measurement average equals this measurement result.

This is one of these questions that should be never asked, as they do more harm than good