# Error propagation Titration

I hope I am right in this section.

I am unsure with error propagation. When calculation the error in a titration, many errors has to be taken into account:

Error in Glassware/ Error in Balance/ Error in Burette etc.

I learned that the absolute and relative error have only 1 significant figure and that the total amount is rounded to the decimal place of the error.

Therefore 5.34532g ± 0.001428g would be 5.345g ± 0.001g

The relative error 0.001g/5.345g = 0.00018709 = 0.0002

If there is an experiment with a lot of steps and error propagation wouldn't the rounding of all the errors in every single step change the result a lot? Wouldn't be rounding the error just in the end make more sense?

Many thanks in advance

• – Loong Oct 28 '15 at 22:49
• If you (or any future reader) want to go for a deeper dive, not necessarily for the original problem, but for something in the future, check this out: EURACHEM/CITAC Guide, “Quantifying Uncertainty in Analytical Measurement”, 3rd Ed., 2012. Just search on the web. This is just for future reference, since @MaxW gave a nice answer years ago. – Ed V Jul 3 at 18:22

## 2 Answers

I learned that the absolute and relative error have only 1 significant figure

That really isn't true. In your example you showed 5.34532g ± 0.001428g which could result from some kind of experiment. The point being that the digits 1428 could all be significant.

You then simplified to 5.345g ± 0.001g which is fine. My personal preference would be more like 5.345g ± 0.0014g since ± 0.00067g would round to ± 0.001g as well. So I'd prefer one more digit in the ± "precision" than the measurement itself.

If there is an experiment with a lot of steps and error propagation wouldn't the rounding of all the errors in every single step change the result a lot?

Yes absolutely. Carry "extra" significant figures until the final step.

Wouldn't be rounding the error just in the end make more sense?

That is the preferred method.

We typically use a calculator now to do the calculations. You can set the calculator to show 3 significant figures but it carries more internally.

There are various statistical techniques which allow all the measurements used in the calculations to have random errors injected and then the calculation is simulated many times to see how the errors propagate through the calculations.

Your instinct about not rounding early is good. Before I get to that, I wanted to address one thing in another answer:

"> I learned that the absolute and relative error have only 1 significant figure

That really isn't true. In your example you showed 5.34532g ± 0.001428g which could result from some kind of experiment. The point being that the digits 1428 could all be significant."

Theoretically possible, but pretty unlikely. To the best of my knowledge, even the National Institute of Standard and Technology never give more than 2 digits of uncertainty because the number of measurements required to give a 3rd digit any meaning is unreasonably large.

The 1 significant figure in error "rule" is pretty good. There are two exceptions. 1) If the uncertainty is very well known (because it was calculated based on a large number of measurements), two sig figs can often be justified. 2) If the first digit in the uncertainty is a 1, two sig figs are usually justified. Think of it this way: Rounding ±0.16 to ±0.2 changes the uncertainty by 25%, while rounding ±0.86 to ±0.9 changes the uncertainty by only 5%. In many cases, the uncertainty is known to better than 25%. It's rarer to know it to 5%.

So, one or two sig figs are justified in the answer, but what does that mean for rounding? A good guideline is to keep at least one more digit than is significant until the last step. There's nothing really wrong with keeping even more, they just are unlikely to make a difference in the final (rounded) result.