Percentage Error?

Can anyone tell me the error involved in:
1) 100cm3 gas syringe
2) 5cm3 syringe
3) 2dp stopwatch
4) 10cm3 volumetric pipette

Thank you!

Edit: Oops! Wrong forum! Er... let all you clever sciency university people show off... or can the Mods just move it

To calculate the percentage error for the apparatus you need to know the error margin for each piece of equipment. The way I calculated the percentage error for the appartus is by the following formula:

Percentage = Error Uncertainty / Reading x 100

The typical uncertainties for the equipment you mentioned are ( you may need to check these, because I don't know the accuracy of the equipment you used):

1) 100cm³ gas syringe ± 0.5cm³
2) 5cm³ syringe ± 0.05cm³
3) 2dp stopwatch ± 0.01s
4) 10cm³ volumetric pipette ± 0.2cm³

The percentage error for these is therefore:

1) 100cm³ gas syringe ± 0.005%
2) 5cm³ syringe ± 0.01%
3) 2dp stopwatch ± 0.01s (T. The human reaction times can vary and but I estimate around ±0.1s. The error in judging the reading is therefore bigger that the error from the stop clock. Therefore this error can be ignored.
4) 10cm³ volumetric pipette ± 0.02%

If you have any more problems, don't be afraid to ask

To calculate the percentage error for the apparatus you need to know the error margin for each piece of equipment. The way I calculated the percentage error for the appartus is by the following formula:

Percentage = Error Uncertainty / Reading x 100

The typical uncertainties for the equipment you mentioned are ( you may need to check these, because I don't know the accuracy of the equipment you used):

1) 100cm³ gas syringe ± 0.5cm³
2) 5cm³ syringe ± 0.05cm³
3) 2dp stopwatch ± 0.01s
4) 10cm³ volumetric pipette ± 0.2cm³

The percentage error for these is therefore:

1) 100cm³ gas syringe ± 0.005%
2) 5cm³ syringe ± 0.01%
3) 2dp stopwatch ± 0.01s (T. The human reaction times can vary and but I estimate around ±0.1s. The error in judging the reading is therefore bigger that the error from the stop clock. Therefore this error can be ignored.
4) 10cm³ volumetric pipette ± 0.02%

If you have any more problems, don't be afraid to ask

How did you get your answers?

There was this question in my mock...
What is the level of accuracy of a burette in each reading? Use your answer to calc the percentage error in the titre volume of 11.30cm³

My answer:
± 0.05cm³
(0.05*2)/11.30 * 100 = 0.88% which was correct

If I try it with 2), I do (0.05*2)/5 *100= 2% (assuming the volume measured is 5cm³) Not sure what you did, can you go through it please?

hi,

when u did "0.05*2)/11.30 * 100 = 0.88%" why did u multiply 0.05 by two?

Thanks.

Hello, I'm Annika and I'm really confused about the uncertainty ±0.05 being multiplied by 2 so as to find the percentage error. May I know what that actually meant and how to be so sure of a uncertainty being multiplied by 1 or more? Because some of them from my experiments are multiplied by 1. (Sorry for posting this 6 years and 16 days later)