On NIST Avogadro's Number, $N_A = 6.022\;141\;29 \times 10^{23} \text{ mol}^{-1}$ has 9 significant figures and a standard uncertainty of $0.000\;000\;27 \times 10^{23} \text{ mol}^{-1}$.

First, can it be written as the following?

$$ N_A = 6.022\;141\;29 \times 10^{23} \pm 0.000\;000\;27 \times 10^{23} \text{ mol}^{-1} $$ If yes, consider the following quote from Rice's lab guide on error analysis (emphasis added):

Whenever you make a measurement, the number of meaningful digits that you write down implies the error in the measurement. For example if you say that the length of an object is 0.428 m, you imply an uncertainty of about 0.001 m. To record this measurement as either 0.4 or 0.42819667 would imply that you only know it to 0.1 m in the first case or to 0.00000001 m in the second. You should only report as many significant figures as are consistent with the estimated error. The quantity 0.428 m is said to have three significant figures, that is, three digits that make sense in terms of the measurement. Notice that this has nothing to do with the "number of decimal places". The same measurement in centimeters would be 42.8 cm and still be a three significant figure number. The accepted convention is that only one uncertain digit is to be reported for a measurement. In the example if the estimated error is 0.02 m you would report a result of 0.43 ± 0.02 m, not 0.428 ± 0.02 m.

and the following quote from A Project of Joliet Junior College (IL), Lee College (TX) and the National Science Foundation (emphasis added):

The absolute error places bounds telling you that the actual value is expected to fall between x - D x and x + D x. The absolute error is an easy way to report the error and makes it easy to check that you are not reporting too many significant figures. For example, suppose you had a measurement that was reported to be 10.02 ± 0.3 m. Because the reported value is uncertain in the tenths place, it is wrong to report it to the hundredths place. It should be reported to the same decimal place as the absolute error: 10.0 ± 0.3 m. Another point is that the absolute error and the reported value must have the same units.

If there is such a convention (cf. Rice's lab guide) and it is wrong to report more significant figures than is warranted by the absolute error (cf. the joint project's guidelines), and NIST should have adhered to such convention and striven to be correct, is it true that NIST adheres and is correct because the standard uncertainty has two least significant digits (i.e., $27$)?

Had the standard uncertainty been $0.000\;000\;2\text{ mol}^{-1}$, would NIST have reported Avogadro's Number as $N_A = 6.022\;141\;3\times 10^{23}$ with 8 significant figures?


2 Answers 2


NIST has its own more sophisticated guidelines for reporting uncertainty of measurments. http://physics.nist.gov/Pubs/guidelines/TN1297/tn1297s.pdf

There is nothing wrong with reporting two digits in an uncertainity, and many peer reviewed journal articles do. This is especially true when the first digit of the uncertainity is $1$. If you report $\pm 1$, $1$ could mean anything between $0.5$ and $1.5$, which is not reasonably specific for the purpose of a published physics measurement. On the other hand $\pm 9$ is much more specific. Another way to think of this is, what if you express the uncertainty in base $2$ instead of base $10$? Then, $\pm 1$ would be $1$ digit and $\pm 9$ would be $4$ digits.

  • 1
    $\begingroup$ This difference in the precision of significant figures is why that method is a weak indicator of certainty, and also responsible for the old slide-rule rule about leading ones not counting as significant. $\endgroup$ Mar 4, 2014 at 17:42
  • 1
    $\begingroup$ Really enjoyed the bit about re-expressing in binary. $\endgroup$
    – Kyle Oman
    Mar 4, 2014 at 18:27
  • $\begingroup$ Thank you for the link to the NIST TN. I've learned some important things from it. And, thank you for telling me from your experience that many peer-reviewed journals report two digits in an uncertainty. But, why does +/- 9 is much more specific when it can mean anything between 0.5 and 9.5? $\endgroup$ Mar 5, 2014 at 1:18
  • $\begingroup$ Sorry, I made a calculation error in the question. It should have been: "But, why does +/- 9 is much more specific when it can mean anything between 0.5 and 18.5?" Do you mean +/- 0.9 instead? $\endgroup$ Mar 5, 2014 at 1:28
  • 2
    $\begingroup$ when you say +/- 9, "9" could mean anything between 8.5 and 9.5 $\endgroup$
    – DavePhD
    Mar 5, 2014 at 1:33

First, can it be written as the following? $$N_A=6.022\,141\,29\times10^{23}\pm0.000\,000\,27\times10^{23} {\rm mol}^{-1}$$

Yes. A cleaner way to write it is $$N_A=6.022\,141\,29(27)\times10^{23}{\rm mol}^{-1}$$ where the $(27)$ indicated the uncertainty in the last $N$ digits (where $N$ is the number of digits inside the parenthesis).

If I have a meter stick that is demarcated down to the 1mm mark, I can estimate the length of some object to within, say, 0.2mm. That means I could measure something to be 0.3456 m (345.6 mm) with an uncertainty of 0.0002 m (0.2mm). I cannot say that I measured a length to be 0.34558m (345.67 mm) because I do not have that accuracy, I can only estimate the 0.6mm part of the measurement. As far as I know, NIST (and every lab) adheres to that practice

Had the standard uncertainty been $0.000\,000\,2 {\rm mol}^{−1}$, would NIST have reported Avogadro's Number as $N_A=6.022\,141\,3\times10^{23}$ with 8 significant figures?


  • $\begingroup$ In that light, would the author of the Rice's lab guide be inconsistent if the emphasized part of Rice's lab guide were put as the following? "The accepted convention is that only one uncertain digit is to be reported for a measurement. In the example if the estimated error is 0.02 m you would report a result of 0.43 ± 0.02 m, not 0.428 ± 0.02 m. However, reporting 0.428 ± 0.020 m is allowed given that the evaluation of uncertainty (e.g., using standard deviation) obtains you 0.020." $\endgroup$ Mar 5, 2014 at 1:39
  • $\begingroup$ I don't see what the point of doing that would be, but it would be consistent with what we know about uncertainties. $\endgroup$
    – Kyle Kanos
    Mar 5, 2014 at 1:40
  • $\begingroup$ The point is to check my understanding on reporting a measurement. I learned that when reporting a measurement, 9.0 is different from 9.00 because the latter is less uncertain. Then, I was introduced to the concept of significant figures that tells me that only the last figure is estimated (this is in agreement with Rice's lab guide on the accepted convention and illustration). But then, I found the NIST page that gives significant figures with 2 estimated digits, which contradicts the lab guide. So, I tried to solve the contradiction with my last comment using 0.020. What do you think? $\endgroup$ Mar 5, 2014 at 1:50
  • $\begingroup$ Through statistical measures and extra calculations, NIST is able to make the 2 digit significant figure. Unless you are doing similar statistical measures, then I think it is wrong for you to add an extra decimal place out of nowhere. $\endgroup$
    – Kyle Kanos
    Mar 5, 2014 at 1:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.