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I need to measure out 70 ml of liquid.

I have a 10 ml measuring cylinder with a maximum uncertainty of 0.1 ml and a 100 ml measuring cylinder with a maximum uncertainty of 1 ml.

I've calculated that using the 10 ml cylinder 7 times would give a 0.7 maximum uncertainty, or a 0.7/70=1% percentage uncertainty. Meanwhile, using the 100 ml cylinder, I get a 0.1 ml uncertainty, or a 1/70=1.4% percentage uncertainty. Clearly, the smaller cylinder approach is more accurate (1% < 1.4%)

Meanwhile, the mark scheme says that one must use the 100 ml cylinder, for the 10 ml cylinder would give a total 7*0.1/10=7% percentage uncertainty, compared to the 1.4% from the 100 ml cylinder. (1.4% < 7%)

Isn't adding percentages in this way fallacious? There's no way to accumulate a 4.9 ml error using a 10 ml cylinder 7 times, right? I'd appreciate if someone could clarify which approach is correct and why.

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  • $\begingroup$ "Maximum uncertainty" is a weird concept. See propagation of error and the treatment of combining variances from multiple sources. You're case is with errors strongly correlated with one another. $\endgroup$
    – Zhe
    Jan 4 at 18:30
  • $\begingroup$ @Zhe maximum uncertainty is calculated as half a division on a cylinder. The small one has divisions of 0.2 ml, while the large one has divisions of 2 ml. Hence, their maximum uncertainties are as above. I'm fairly certain that we are meant to assume that each error has no effect on the next one, so I don't think it has to get this complicated. $\endgroup$ Jan 4 at 18:42
  • $\begingroup$ Unlike problems on homework assignments, life is complicated. The standard deviation of your measurement is going to increase as you make the same measurement a bunch of times (assuming that your errors are strongly correlated). $\endgroup$
    – Zhe
    Jan 4 at 18:49
  • $\begingroup$ @Zhe I appreciate the extra insight you have provided, however my only question is how to calculate maximum uncertainty in a more or less logical way, assuming that the errors are not correlated. That is, which one of the approaches described above better suits our, perhaps crude, model. $\endgroup$ Jan 4 at 18:55
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    $\begingroup$ One thing is cylinder resolution, other thing is cylinder precission ( repeatability ) and yet another cylinder accuracy ( correctness of mean value ) $\endgroup$
    – Poutnik
    Jan 4 at 19:47
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You have three types of errors:

  1. The errors in the accuracy of your measurement. Assuming you are reading the water levels consistently (at eye level at the bottom of the meniscus), these should be random.

    To add random, uncorrelated errors, the standard method is to sum their squares and take the square root. So, for the 0.1 mL average error for the 10 mL cylinder:

    $$\text{average error for seven combined measurements} = \sqrt{7 \times (0.1 \text{ mL})^2} = 0.26 \text{ mL}$$

    As you correctly intuited, it wouldn't make sense to simply multiply the 0.1 mL by 7, since this would be saying you have either a +0.1 mL error every time, or a –0.1 mL error every time, but that's not how random errors work—they are just as likely to be positive as negative. In addition, they don't all have a magnitude of 0.1 mL; rather, their magnitudes follow a distribution. Thus essentially what you have is a random walk, with a random distribution of step sizes, where the average step size is 0.1. The square-root-of-the-sum-of-the-squares method gives you the average distance you'd end up from your starting point with 7 of these steps

  2. Error in the calibration of the cylinder. This would be a systematic error, and it would add to the error of each measurement. Higher qualities of cylinders (e.g., Class A glassware) have lower errors.

  3. Error in the consistency with which the cylinder delivers the measured amount. These cylinders are probably calibrated as "TD", which means "to deliver". That is, it is assumed not all the water will drain out, so the markings on the cylinder are calibrated to account for that. However, there will be some random error in how closely the amount that drains out corresponds to the assumed amount that drains out. The only way to determine this error would be to contact the manufacturer (or to test it yourself using a scale). In addition, if your cylinder is not clean, more water may adhere to the walls, which could cause a systematic error in the amount of water delivered.

Note also the comment by Andrew Morton that there's an additional important practical consideration, which is that if you have to use seven measurements, you could miscount and thus end up being off by $\pm$10 mL! And this error is much easier to make when measuring with a graduated cylinder than, say, a tablespoon, because each measurement with the cylinder takes focus to get the total volume to 10.0 mL, and that focus can cause you to lose count. There are of course ways to address this, e.g., making a mark on a piece of paper for each 10.0 mL added.

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  • $\begingroup$ Thank you for your answer. If I understand correctly, 0.026 is the most likely uncertainty for seven measurements. However, wouldn't a maximum uncertainty specifically involve looking at the extremities, i.e. -0.1 or +0.1 every time? Although 70±0.026 ml is the most likely volume, aren't 69.3 ml and 70.7 the, though unlikely, physical boundaries to how inaccurate the volume ultimately is? For the sake of simplicity, the question disregards errors 2 and 3. But thank you for pointing those out. $\endgroup$ Jan 5 at 2:45
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    $\begingroup$ Note that, in the above analysis 0.1 mL is the average error, not the maximum error. If 0.1 mL were the maximum error, and you made the maximum error each time, and you made it in the same direction each time then, yes, of course you'd get a maximum 0.7 mL error. But how likely is that? You'd need to take the probability of getting the max error once, and raise it to the 7th power to get the probability of getting the max error 7 straight times. $\endgroup$
    – theorist
    Jan 5 at 2:53
  • $\begingroup$ I agree, it's quite unlikely. The question does say that 0.1 is, in fact, the maximum error. I guess I have ended up with the maximum, however unlikely, yet possible error, while you have provided the maximum most likely error. $\endgroup$ Jan 5 at 3:10
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    $\begingroup$ Also, there is the possibility of an off-by-one error if the counting to seven is not accurate. It might seem to be a simple task, but a distraction could lead to an error. $\endgroup$ Jan 5 at 9:42
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    $\begingroup$ With the small cylinder, it's really easy to be off by 10 ml as you've miscounted the number of transfers. $\endgroup$
    – D Duck
    Jan 5 at 11:43
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Keep in mind that a graduated cylinder is never meant for accurate volume transfer. You would probably use a Class-A 100 mL buret to transfer 70 mL to a vessel if accuracy were that critical say, 70.0 mL was desired. All these general chemistry rules of thumbs are not useful for real analytical work. Experimental numbers matter the most.

Let us assume you don't have a 100 mL burette. This is how a practical scientist would do it. You would like know whether 7 transfers from a 10 mL cylinder are better or 1 transfer from a 100 mL cylinder is better? You calibrate your cylinders. You know the density of the liquid. You would transfer 10 mL, 7 times and weigh it on an analytical balance. Note the mass and from the mass-density-volume relationship, determine the volume.

Then repeat the same with a 100 mL cylinder, and transfer 70 mL in one go. Measure the mass and hence volume. Whatever delivers the right volume, is the way to go.

Practically, 7 transfers will generate more errors because you do not know the drainage time of the cylinder, i.e., how long should you wait till all the film of liquid is out of the cylinder as the last drop of liquid. Also see the rules for the propagation of error.

By error propagation rules, uncertainties cannot be added like you have shown. It is an incorrect approach. The cylinder with 0.1 mL error will not add up to 0.7 mL by any means. if you make 7 additions.

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    $\begingroup$ A practical scientist would know that graduated cylinders are to contain, not to deliver. So if you have seven of the 10 mL cylinders, you could fill them pretty accurately with 70 mL total. It would just be difficult to remove the entire volume, and you would have to do a lot of dishes. $\endgroup$ Jan 5 at 2:30
  • $\begingroup$ And if you have a balance, you would use the 100 mL graduated cylinder to figure out the density, and then use the balance to measure out the 70 mL into the container you need it in (unless it is too heavy, in which case you would empty the 100 mL graduated cylinder, add 70 mL via the balance, and then empty it again to deliver roughly 70 mL). $\endgroup$ Jan 5 at 2:32
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    $\begingroup$ Karsten, There are TD (to deliver) and TC (to contain) graduated cylinders. It would be foolish to use a TC cylinder for transferring volumes. $\endgroup$
    – M. Farooq
    Jan 5 at 2:34
  • $\begingroup$ Thank you for the detailed answer. Perhaps I should have clarified that my question makes several unrealistic assumptions for the sake of simplicity. It is assumed that liquid is transferred entirely (disregard drainage time and the such) and that each error/uncertainty has no effect on the next one. (Frankly, I fail to see how these measurements even interfere with each other, but ok) $\endgroup$ Jan 5 at 2:38
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    $\begingroup$ @Karsten, Yes, but high viscosity issue applies to all TD volumetric glassware. This is why I recommended weighing to actually see how much is transferred. See reverse pipetting technique en.wikipedia.org/wiki/Reverse_pipetting $\endgroup$
    – M. Farooq
    Jan 5 at 14:34
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Meanwhile, the mark scheme says that one must use the 100 ml cylinder, for the 10 ml cylinder would give a total 7*0.1/10=7% percentage uncertainty, compared to the 1.4% from the 100 ml cylinder. (1.4% < 7%)

Let's simplify the problem and say the 10 mL cylinder has a +1% systematic error (and there is no random error), i.e. delivers 10.1 mL when filled to the 10 mL line. So filling it up to 10.1 mL seven times will deliver 70.7 mL instead of 70.0 mL. That is a 1% systematic error.

So the relative systematic error does not change comparing delivering a single volume to delivering the sum of multiple measurements. The absolute error, of course, increases (i.e. the difference between the intended and the actual volume).

In the calculation of the marking scheme, the denominator is 10 (mL), but it should be 70 (mL). If you look at the other answers, which are excellent, that is not the only thing the marking scheme got wrong. The bigger problem is that the question does not make sense in face of the reality of measuring volumes in the lab.

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