I recently did my first chemistry lab where we had to find the average mass of our aluminum cylinder. We were required to weigh and record its mass 3 times using an electric balance which gave us a value up to the thousandths place. After recording I got these values: $\pu{7.195 g,}$ $\pu{ ~ 7.198 g,~}$and $\pu {7.197 g}$. My professor asked to find the average of these. Adding them up I got $\pu{21.59 g}$. To get the average I had to divide by $3$ since I have $3$ quantities. My question here is, how many significant figures should I have in my answer?
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$\begingroup$ So would I just simply add a 0 to match 3 decimal places after my decimal since my original results had 3 places after the decimal place? $\endgroup$– GeorgeCommented Sep 9, 2018 at 2:10
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$\begingroup$ The 3 in the integer is infinite precision. It has an infinite number of 0's after the decimal place, unless you're not sure you counted properly. Otherwise, it's exactly 3. $\endgroup$– ZheCommented Sep 9, 2018 at 2:13
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3 Answers
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Significant figures are a sloppy method for doing error propagation. The point with significant figures is to maintain "reasonable" precision in the answer.
Ideally you'd assume that the measurements followed the normal distribution and you'd take many more measurements to get a better average and standard deviation. From the standard deviation of the individual values and the number of measurements you could calculate the standard deviation of the mean.
For addition or subtraction the least significant digit is the limiting factor. Since each of your three measurements had 3 significant figures after the decimal, your sum and average should too. So your sum was not 21.59 g, but rather 21.590 g. (In math the trailing zeros are not significant so that 21.59 = 21.590, but when using significant figures such zeros in the sum do matter.) Now consider the estimator for the mean:
$$\dfrac{21.590}{3} = 7.1966666666666666666666666...$$
Now considering significant figures you round the answer to three significant figures after the decimal yielding 7.197. The whole point with significant figures is to keep you from going wild and suddenly turn this average into say 27 significant figures.
In general it is wise to carry at least two extra digits in intermediate calculations and only round the final result to the proper number of significant figures to try to avoid rounding errors. The extra digits are no problem with calculators these days. However when I started chemistry we used log tables or a slide rule. With a slide rule three significant figures were the best that you could do for intermediate calculations. With a good set of log tables you could do 4 significant figures for intermediate calculations.
So for addition if you had values of 17.19 (not 17.190!), 7.198g, and 7.197 g than the average would be rounded to the hundredths place since 17.19 only has two significant figures after the decimal.
For multiplying it is the number of significant figures that matters. So multiplying $17.19\times3.198 = 54.97$ since each of of factors has 4 significant figures. Multiplying $17.19\times3.20 = 55.0$ since $3.20$ only has three significant figures.
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As an error is not explicitly given and as your first number is $7.195$ g this means that the absolute error bound implied is 0.0005, which is to say that the error (uncertainty) is in the decimal place after the last decimal place and in your example and is the same for each number. Your first number should be written as is $7.195\pm 0.0005$ and similarly for the others. The average total uncertainty is $(\sqrt{ \sigma_1^2+\sigma_2^2+\cdots })/3$ where $\sigma = 0.0005$ and the error works out as $\pm0.0003$ when rounded up so that 3 decimal places are needed in your average since the uncertainty is the fourth place.
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4$\begingroup$ Your calculation of the estimated theoretical error as ±0.0003 g is correct. However, the experimental data show a standard deviation of ±0.0015 g and thus an experimental standard error of ±0.0009 g, which is larger than your estimated theoretical error of ±0.0003 g. Therefore, there is another source of measurement uncertainty and the reported result should not be more precise than the total uncertainty. $\endgroup$– user7951Commented Sep 9, 2018 at 9:22
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$\begingroup$ @Loong where did the 0.0015 come from ? We know that the data has three decimal places as these are given so are known, thus the uncertainty has to be beyond this hence $\pm 0.0005$ is the standard assumption. $\endgroup$ Commented Sep 9, 2018 at 11:35
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4$\begingroup$ 0.0015 is the experimental stadard deviation estimated from the individual values 7.195, 7.198, and 7.197 as $\sigma=\sqrt{\frac{\sum_{i=1}^N\left(x_i-\overline{x}\right)^2}{N-1}}=0.0015$. Since the resulting standard error of 0.0009 is larger than your correct estimate of the random error 0.0003 coming from the used electrical balance, there might be an additional source of error (e.g. from handling or touching the sample, temperature drift, air pressure, humidity etc.) that is not included in your estimate. $\endgroup$– user7951Commented Sep 9, 2018 at 13:40
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1$\begingroup$ Furthermore, there might be a systematic error of the used electrical balance that cannot be eliminated by repeated measurements with the same electrical balance. We both have not taken such systematic error into account yet. $\endgroup$– user7951Commented Sep 9, 2018 at 13:42
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$\begingroup$ @Loong, I see; it is of course ok to do this but we now have uncertainty in the 3rd decimal place (and also rounding the error to 0.002) so the result has only 2 decimal places that are significant. This is also ok as it is always better to err on the cautious side with errors. $\endgroup$ Commented Sep 10, 2018 at 7:38
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The significant figures rule for averaging is to identify the lowest significant place value in each addend and then round the average to the highest of these minimal place values. Here $7.195, 7.198, 7.197$ are significant to the place values $0.001, 0.001, 0.001$ respectively. The highest of these place values is $0.001$ so you round the average to the nearest $0.001$. Thus, $(7.195+7.198+7.197)/3=7.197$.
answered Sep 9, 2018 at 10:31