Percentage errors (chemistry) Watch

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LS.
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#1
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I'm writing up my chemistry titration evaluation.

For the percentage error of a burette is it:

(0.05cm x 100 )/ the average titre = percentage error(?)

and...

How do you add up percentage errors to find the +/- % error in the concentration found?

Thanks in advance
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john !!
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yes, but you might also want to take into account the mean error:

mean error = (max value - min value)/2

so if you have titres of:

20.05, 20.15, 20.25, 20.15

then the mean error is (20.25-20.05)/2 = 0.1
add this to the instrument error of 0.01

total error: +/- 0.11cm^3

% error = 0.11 / 20.15 * 100 = 0.55%
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john !!
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How do you add up percentage errors to find the +/- % error in the concentration found?

concentration = moles/volume

add the error values of moles and volume together.

say you had the equation:

x = y²z³a√b

then call X, Y, Z, A and B the uncertainties of x, y, z, a and b respectively.

X = 2Y+3X+A+0.5B

So you multiply the error by the power of the corresponding letter in the actual equation.
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LS.
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Thanks mik1a

The first post was very helpful, but unfortunately I didn't understand the second one ( I don't know if it's because it's higher than AS level, or beecause we havent really done percentage errors), but thanks for trying
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john !!
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I learnt that in physics, so that might mean it's not on the chemistry syllabus, but I'll try to make it clearer:

Say you are finding the volume of a cylinder. The equation is:

V = pi*r^3*h

If you measure the radius and height with a ruler, whose uncertainty is +/- 0.5mm, because it could be upto half a milimetre above or below the measures value, and get, for example:

r = 5cm +/- 0.05cm
h = 10cm +/- 0.05cm

Then you use the equation to find the volume of the cylinder, it is obviously:

1250Pi

But to find the uncertainty of that, you need the uncertainties of the raw data (r and h), and you add them together. But there's a catch - if one of the values is to the power of 2, you have to multiply that specific uncertainty by 2. If one is to the power of 3, like the radius is on this example, then you have to multiply the uncertainty by 3. Likewise, if the answer is squarerooted, you halve it (because root x is the same as x to the power 0.5). Constants are not included, because 4 or 5 do not have uncertainties - Pi does, but we don't include it becuase the accuracy of Pi on anyone's calculator is far smaller than anything else (usually).

So the uncertainty of V here is:

unc. V = unc. h + 3(unc. r)
= 0.05 + 3*0.05
= +/-0.2cm^3

So you can finally say that the volume is 1250Pi cm^3 +/- 0.02 cm^3
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LS.
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That makes sense now :cool:

[I do physics too... Just had a look in my text book and found some more information on errors in the appendix]

Thanks very much for taking the time to help me.
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