I'm working out the percentage errors of some titrations i did for my chemistry investigation. My problem is the burette readings, I know that:

One drop from a burette has a volume of approx 0.05ml. All burette readings should include 2 decimal places in which the second is either 0 or 5. An error of one drop in a volume of 25ml gives a percentage error of 0.2% for each reading

Does this mean that in a 50ml burette the % error is 0.1% for each reading?

My teacher told me to divide 0.05 by my reading and multiply this by 100, which is what you do for other % errors. But the volumes on the burette increase as you go down, to show how much solution was used up. Also if the burette reading is one 1.00 then the % error would be 5!! doesn't make sense..I'm thinking it's just a % error of 0.1 for ever reading...please help...my teacher must have told me wrong...
That is all correct what you say.

In your last example, the error of 5 you got is a %% error. But you need to remember that in that titration you only let through 1.00ml, a very small amount of which an extra drop of 0.05ml is 5% more, a large error.

Isn't it true that your final reading on the burette thingy is the amount of liquid you've let out the end? (Assuming you start the level at 0ml at the top).

So again, if you have a final reading of 50ml, then you should have let out 5oml of the liquid. Therefore a one drop error would be (0.05/50)x100 = 1% error
So are you saying that i should only work out a % error of the total amount of liquid used in the titration? most of the volumes were around 17ml, so the percentage error of that would be (0.05/17)x100= 0.2941176471%

My teacher told me to take a percentage error of my initial and final readings and then add the two together to get the final % error. For example if the initial reading was 0.5, the percentage error would be 10! and if the final was 17.5, the percentage error would be 0.28571... this wuold give a total % error of 10.28571... whereas if i just took a % error of the total volume it would only be 0.29411...

Thats is big difference..and I know % errors aer oly very small...But I'm sure the errors should be taken from readings...ahh I'm so angry that I may have been told wrong! This project needs to be in on monday and it's half term now, so its not as if i can even ask him again!
I'd do things just from a total of the liquid (ie the final reading), but please don't take my word alone for it as it is now over 4 and a half years since I did a titration....maybe someone else here could help.....I really don't want to tell you wrong if what I've said is so different from what your teacher said...
Ok I have found that:

Burettes are also more accurate than measuring cylinders. They have graduations every 0.1 cm3, so when you take a reading it should not be more than 0.05 cm3 too high or too low.

However, when you use a burette you take a reading at the start and the end , so you have two errors of 0.05 cm3 i.e. total error = 0.10 cm3. If you are using your burette to do a titration there may be another error of one or two drops which is due to your judgement of when the indicator changes colour. This means that in a titration (as opposed to just using a burette to measure a volume) you may have an error of 0.2 cm3.

So from this, the 2ml error in a 50ml burette, would give a % error of 0.4% [(0.2/50)x100]

Am I right? would I just put down 0.4% for the burette % error for every titration?
dingaling
Ok I have found that:

Burettes are also more accurate than measuring cylinders. They have graduations every 0.1 cm3, so when you take a reading it should not be more than 0.05 cm3 too high or too low.

However, when you use a burette you take a reading at the start and the end , so you have two errors of 0.05 cm3 i.e. total error = 0.10 cm3. If you are using your burette to do a titration there may be another error of one or two drops which is due to your judgement of when the indicator changes colour. This means that in a titration (as opposed to just using a burette to measure a volume) you may have an error of 0.2 cm3.

That all sounds good and you are doing what your teacher said in one way (probably the way she meant you to do it).

So from this, the 2ml error in a 50ml burette, would give a % error of 0.4% [(0.2/50)x100]

Am I right? would I just put down 0.4% for the burette % error for every titration?
The first part here is correct if you let out 50ml of the liquid during the titration. However the percentage error in the result will nopt be the same for every experiemet as you may get different amounts of liquidlet out in differnet experiements.

You will however have the same volume error each titre of 0.2ml.

You'll need to work out the percentage error each time separately from this in the way you've been doing above:

(0.2/measuremnt)x100 = percentage error
hey Dingaling??lol..(i'm not sure what ur name is..sorry.) i was wondering whether you cleared that dilemma of % errors...because i am sort of stuck my self. well..in one of the post you sent, it talked boutt % error of the burette being
(0.2/watever measurement x100)...BUT..in my past courseworks..like u mentioned in ur first post...i divided (0.05/ measurement x 100)..and i REALLY DO NOT KNOW WHICH ONE IS RIGHT???

....but the part where u mentioned

"However, when you use a burette you take a reading at the start and the end , so you have two errors of 0.05 cm3 i.e. total error = 0.10 cm3."

where it equalled to 0.10..i wanted to know how u gt to that answer....?..what did u divide 0.05 by??.lol.....

other than that i had also repeated my experiment 3 times...dose that also mean i have to do % errors for them as well???.........

HEYY guys, well i was searching teh net and i found this information oN % ERRORS....but the problem with it is..i just can't get aenswers AS THEM???...can u try and see if u get the same as them..and please tell me how????..i'm goin mental here!!!???...well here is it...and the site:

Some measurement uncertainties are given below:

Equipment Measurement to the nearest:
Balance (1 decimal place) 0.08 g
Balance (2 decimal place) 0.008 g
Balance (3 decimal place) 0.0008 g
Measuring Cylinder (25 cm3) 0.5 cm3
Burette (50 cm3, Grade B) 0.08 cm3
Stopwatch (digital) 0.01 s

Calculating the percentage uncertainty (often called percentage error) ...

percentage uncertainty= uncertainty/actual measurement made x100

Now try calculating the following percentage uncertainties...
1) 1.00 g on a 2 decimal place balance
2) 10.00 g on a 2 decimal place balance
3)1.00 g on a 3 decimal place balance
4)10 cm3 in a 25 cm3 measuring cylinder
5)25 cm3 in a 25 cm3 measuring cylinder
7)25 cm3 in a 50 cm3 burette (Grade B)
9)50 s on a digital stopwatch

1) 8%
2) 0.8%
3) 0.08%
4) 5%
5)2%
6)0.16%
7)0.32%
8)0.08%
9)0.02%