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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

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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.

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