PERCENTAGE ERRORS..for buretts and other stuff!!!!!..PLEASE HELP ME!!!!!!!!!!!!?

Question

hi guys...well my dilemma is that i NEED TO WORK OUT THE PERCENTAGE ERRORS.....well , firstly the minimum error for a burette is 0.05 so, eg. (0.05/50mL x 100 = 0.1)...ok thats the begining...but i have also found somewhere that...

"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...(0.2/watever measurement x100)...BUT..in my past courseworks..i divided (0.05/ measurement x 100)..and i REALLY DO NOT KNOW WHICH ONE IS RIGHT???

i dont know whether i shud do % errors for every repeated experiment i have dne for that 1 test??or just for one...i just dont know which error to USE 0.05 or 0.2 AARGH!!! PLZ HELP ME !!?

HEY BEAUTIFULNESS FAIRY.....that is EXACTLY WAT I MEAN..BUT what it is ..i'm doing titration with a pH meter...so i actually have something like 15 readings for ONE concentration...so...like u said about finding the average...dose that mean i have to do it for every single 15 results i have taken...PLUS the repeated ones too??



eg. for conc 1.00mol i have



rough:11.0, 11.2, 11.6 ,......

trial 1: 11.2, 11.5,........(13/15 MORE results for all of them...)

trial 2:11.2, 11.5...........

trial 3: 11.1 , 11.4............

HEY BEAUTIFULNESS FAIRY.....that is EXACTLY WAT I MEAN..BUT what it is ..i'm doing titration with a pH meter...so i actually have something like 15 readings for ONE concentration...so...like u said about finding the average...dose that mean i have to do it for every single 15 results i have taken...PLUS the repeated ones too??

eg. for conc 1.00mol i have

rough:11.0, 11.2, 11.6 ,......

trial 1: 11.2, 11.5,........(13/15 MORE results for all of them...)

trial 2:11.2, 11.5...........

trial 3: 11.1 , 11.4............

SO...can u help me out a little...plzzz...i have soo mny results dose it mean i have to do it 4 EVERY SINGLE ONE OF THEM???..helllppp!!! pretty please??..

Answer 1

OK, Well if you've taken 3 readings from the burette (as they need to be concordant , for example:
First reading: 27.0 cm3
Second reading: 26.6 cm3
Third reading: 26.7 cm3

You can take the second and third reading to get an average (because second and third are concordant), so the average is 26.65

As during titration, the burette must be read at the start and again at the end, so the % error of the burette is
2*0.05/26.65*100 = 0.37%

I don't know what you've meant by repeating experiment but I'm guessing you mean you've repeated the experiment to take different readings for the burette.

If that didn't help, tell me and I might solve it to you ;)

More information:

Hey ang3lbluff, it was good to provide me with some more information :). Anyway As you said you had 15 readings for 1 concentration thats mean you need to find out the volume so that you can work out the number of moles by using the equation: n = conc. * vol.

I think you need to get the the concordant readings for each trial, for example:

rough: rough:11.0, 11.2, 11.6 - the concordant readings are 11.0 and 11.2 so you need to get an average for them two. I think you need to do that for each trial until you get 1 final reading for each of them, and then take th average from them by adding them all up and dividing them by how times you've repeated it.

OR

You can just take 3 results that are concordant enough for each trial that would be good for your experiment instead of calculating the whole results.

I'm just wondering how many readings did you take for each trial? so that We can know how to calculate the % error for the burette.

*let me know you didnt understand it :))

Answer 2

First, make sure the bananas are not oranges. The 0.05 mL burette error sounds OK as long as you are using the "fairly standard" burette. You have that twice,as you observed. The "judgement error" is also a volume error, since the "drop" has a certain volume, and some students are sloppier than others.

This being said, you should compute the sum of the errors in mL first, divide by the measurement, and multiply by 100. Also realize that this may be a somewhat pessimistic number, since it assumes that any error being made is compounded by another, when in some cases, they cancel each other out.

The moral of the story is not to get your self so shook-up about errors that you interfere with your work.

Answer 3

Hmmmm scientific errors:

Random errors - those that you can overcome by repeated measurement.

Systematic errors - errors which you cannot overcome unless you can actually work out what they are - most cannot be overcome without using quite different analysis methods and comparing across a range of techniques - possible? depends.

To start with, you have to consider reading error - this what I suspect you are calling burette error.

You have two readings, start and finish. You should be able to confidently read a burette to 2 decimal places, probably ±0.01 dm3. With the correct reading technique, surely you can detect between 0.03 and 0.05, so 0.04 ± 0.01 is realistically pushing the reading error down.

This gives you an uncertainty in your titre of ±0.02 cm3. Skilled burette reading involves having the meniscus at eye level, using a shade, etc.

the titration end-point is itself a systematic error, because you often may not know how far from the true equivalence point of the titration it is. So you are only able to determine the endpoint as per the indicator (or, if practical, pH meter or conductivity meter can get you much much closer to equivalence where such resources are available and can be used for the particular analysis).

One thing, though - unless your titre is close to the capacity of the burette, you should conduct your different titrations from different parts of the burette to overcome any systematic error in repeatedly using the same section of burette.

You then repeat the titration to achieve three or more concordant results. Often, the experimental procedure will give you the average difference between results that is required for concordancy.

Beyond this point, I am not prepared to say how to calculate your percentage error, as it depends a little on which way you are taught to calculate it.

Were it me teaching (and I used to teach titration calculations this way - others may disagree), I would tackle it as follows:

Three concordant titrations: 20.26; 20.35, 20.38 (all ± 0.02 reading error)
Mean = 20.33
mean deviation or uncertainty or error:
add the differences of each reading from the average.... 0.07; 0.02; 0.05 has mean 0.14/3 = 0.05 (rounded off)

So the titre is 20.33 ± 0.05 dm3

other people may say that you should use 20.33 ± 0.07 or even 20.33 ± 0.09 (i.e. the sum of the maximum error in the titre calculated PLUS the total reading error per titre)

Go with what you are taught

The percentage error is then 100 x (E/20.33)
E = error/uncertainty to use according to how you have been taught - for me, 100 x (0.09/20.33) = 0.5% about 1 in 200

E is not an official symbol; should use δv (delta v)

My advice is to save the calculation of the percentage error until the end of the process. In that way, you would obtain concordancy, then calculate the uncertainty in that and convert to a percentage error.

You need percentage arror at this point in order to compound the percentage errors that are included in calculations BEYOND this first level. When you start of consider mass and volume (and concentration and number of moles.....), to determine overall error or uncertainty, percentage errors then are the only meaningful ones to work with.

Hope this helps without being too long winded

Answer 4

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