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# Percentage error problem watch

1. Each balance reading has an uncertainty of ±5.00 mg.
Calculate the percentage error in the initial mass of ore used.
The initial pass of the ore had a mass of 6.26g.

Markscheme answers.
0.01/6.26 * 100 =0.160% Why isnt it 0.005/6.26*100?
2. (Original post by Sam1500)
Each balance reading has an uncertainty of ±5.00 mg.
Calculate the percentage error in the initial mass of ore used.
The initial pass of the ore had a mass of 6.26g.

Markscheme answer.
0.01/6.26 * 100 =0.160% Why isnt it 0.005/6.26*100?
5.00mg = 0.005g

As the error is ±0.005g , the error can be both +0.005 and -0.005
This means that when you did 0.005/6.26*100, you had to double the 0.005, thus producing: (0.01/6.36)*100 = 0.160%
3. (Original post by FusionNetworks)
5.00mg = 0.005g

As the error is ±0.005g , the error can be both +0.005 and -0.005
This means that when you did 0.005/6.26*100, you had to double the 0.005, thus producing: (0.01/6.36)*100 = 0.160%

Thank you very much. I been trying to figure this for past 6 hours. Do we always have to double?
4. EDIT: - forget what I said, I believe that I have miss lead you

Can you post the full question please.
You only double the uncertainty when you have to measure twice.

If the person measured 6.26g (i.e. only used the weighing scale ONCE), then indeed, the answer would be 0.005/6.26*100.

However, if the person got 6.26g by taking TWO measurements (initial & final masses) and then subtracting them to get 6.26g, then if this was the case, you WOULD have to double the uncertainty (i.e. 0.01/6.26*100), because you measured an initial mass and a final mass, which introduces two uncertainties into the reading!

But we really need to see the full question to deduce this.
5. A good example for the doubling thingy is when you calculate the percentage uncertainity of a titre.
Say that you use 30 cm^3 of NaOH from a burette, that has a percentage error of +- 0.05cm^3.

So at first you might have done: 0.05/30 * 100 = 0.17%
....... but this is WRONG!
To calculate the titre, 30cm^3, you would had to have done: initial volume - final volume.
So you actually took TWO readings from the burette! So, you have to double your uncertainty (0.05*2)

therefore, percentage uncertainity = (0.05*2)/30 * 100 = 0.33%

Sorry for the long answers, but I really struggled with this at AS; just want to make sure you don't!
6. (Original post by Sam1500)
Each balance reading has an uncertainty of ±5.00 mg.
Calculate the percentage error in the initial mass of ore used.
The initial pass of the ore had a mass of 6.26g.

Markscheme answers.
0.01/6.26 * 100 =0.160% Why isnt it 0.005/6.26*100?
0.005g is the uncertainty in each recording. In the usual case where a difference is involved, you will have this uncertainty at the beginning of the reading (do we know if the scale was exactly 0g at the beginning?) and this uncertainty will also be present in the reading you get after you place your ore onto the scale.

Hence in our reading we have the uncertainty of 0.005g twice, so our total absolute uncertainty becomes 0.005 + 0.005 = 0.01g

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