Significant Figures - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-17T23:18:55Z https://chemistry.stackexchange.com/feeds/question/101494 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/101494 4 Significant Figures George https://chemistry.stackexchange.com/users/67714 2018-09-09T01:41:15Z 2018-09-10T22:33:50Z <p>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?</p> https://chemistry.stackexchange.com/questions/101494/-/101496#101496 4 Answer by MaxW for Significant Figures MaxW https://chemistry.stackexchange.com/users/22102 2018-09-09T03:41:51Z 2018-09-10T22:33:50Z <p>Significant figures are a sloppy method for doing error propagation. The point with significant figures is to maintain "reasonable" precision in the answer.</p> <blockquote> <p>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. </p> </blockquote> <p>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:</p> <p>$$\dfrac{21.590}{3} = 7.1966666666666666666666666...$$</p> <p>Now considering significant figures you round the answer to three significant figures after the decimal yielding 7.197. <em>The whole point with significant figures is to keep you from going wild and suddenly turn this average into say 27 significant figures.</em> </p> <p>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 <em>try</em> 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. </p> <hr> <p>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.</p> <hr> <p>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. </p> https://chemistry.stackexchange.com/questions/101494/-/101505#101505 3 Answer by porphyrin for Significant Figures porphyrin https://chemistry.stackexchange.com/users/30424 2018-09-09T08:48:31Z 2018-09-09T08:48:31Z <p>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.</p> https://chemistry.stackexchange.com/questions/101494/-/101508#101508 1 Answer by Oscar Lanzi for Significant Figures Oscar Lanzi https://chemistry.stackexchange.com/users/17175 2018-09-09T10:31:57Z 2018-09-09T10:31:57Z <p>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, $(7.195+7.198+7.197)/3=7.197$.</p>