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# question regarding percentage errors watch

1. Board OCR, if it matters

I have a doubt regarding calculation percentage errors in wieghts

To prepare a standard solutions we use the scale 3 times
1.mass of cake case - 0.3 g
2.mass of cake case + cuso4 (example) - 2.4 g
3.mass of case case after pouring cuso4 (to check for residue) - 0.1

Thus mass used is 2.0g

if the error is 0.01 , is the calculation of percentage error

1 x (0.1/2.0) x 100

Suppose you repeat the same procedure 3 times, will be percentage error be?

3 x (0.1/mean of 3 masses of cuso4) x 100

Now, say in titrations,

You read the burette 2 times (initial and final vaule) to compute the titre

is percentage error

1 x (error of burrette / titre ) x 100

Suppose you repeat thrice

3 x (error of burrette / means of 3 titres) x 100

Sorry for lack of latex, but i guess you guys can make out what i mean

Thanks
2. Bump,
sorry in a hurry
3. Hey

I am doing As Chemistry, so I am not as well knowledged as the guys here...but I'll try my best to help you out....if it turns out someone proves me wrong. I am sorry lol

Firstly, when you measure percentage error in a scale, the scale you are talking about is accurate to 1 decimal place, hence the value can either be 0.35 or 0.45, it is therefore that the error is .

Hence to work out the percentage error for the first reading , i.e mass of cake case, you would do
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
(\pm 0.05 \div 0.3)\times100%

Now in you experiment, you took three measurements from the balance, I would assume that the total of the percentage error will give you the possible percentage that you could obtain, seeing as that you took three measurements, there are 3 times likely to be errors of

Now for the burette....
When you use a burette, you are acutally taking two values, so if the error is , you will have two erros as you are measuring twice(the intital volume and the final volume)...hence the acutally error will me be 0.10(this is the total of 0.05+0.05).

Extra notes:
However if you are using a burette do to titration, it is possible for another error to occure, this is the human error due to the judgment in when the solution changes colour, this means that in a titration process, it is actually possible to have erros

sorry about my spelling, i had to rush to do this fast.

I hope i was somewhat helpful
4. (Original post by New Username)
Hey

I am doing As Chemistry, so I am not as well knowledged as the guys here...but I'll try my best to help you out....if it turns out someone proves me wrong. I am sorry lol

Firstly, when you measure percentage error in a scale, the scale you are talking about is accurate to 1 decimal place, hence the value can either be 0.35 or 0.45, it is therefore that the error is .

Hence to work out the percentage error for the first reading , i.e mass of cake case, you would do
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.
\frac{\pm 0.05}{0.3}\times100%

Now in you experiment, you took three measurements from the balance, I would assume that the total of the percentage error will give you the possible percentage that you could obtain, seeing as that you took three measurements, there are 3 times likely to be errors of (this is the total of 0.05+0.05).

Extra notes:
However if you are using a burette do to titration, it is possible for another error to occure, this is the human error due to the judgment in when the solution changes colour, this means that in a titration process, it is actually possible to have erros

sorry about my spelling, i had to rush to do this fast.

I hope i was somewhat helpful

Thanks a lot!
Even though it wasnt the way my teacher explained it to me...
I will fine tune this later on.
We werent told to account for a specific reading of human error..

Rep+
5. % Error = 100 x half the smallest reading / actual reading
Double this whenever you need to take a reading at the start (1x error) and and at the end (2x error).

When there's more than 1 source of error, just add them.
6. (Original post by Zygroth)
% Error = 100 x half the smallest reading / actual reading
Double this whenever you need to take a reading at the start (1x error) and and at the end (2x error).

When there's more than 1 source of error, just add them.
I understood that,
but the problem is that when you are finding the mass of anything (may it be cuso4 crystals) you need to use a cake case and hence have 3 readings. I was wondering whetheri am correct in using the answer only and times that by one.
Same for titration
7. (Original post by divinelord)
Thanks a lot!
Even though it wasnt the way my teacher explained it to me...
I will fine tune this later on.
We werent told to account for a specific reading of human error..

Rep+
No worries

I am doing OCR too btw....I just did my quantitative task today in chemistry...now I got qualitative task coming up tomorrow
8. (Original post by New Username)
No worries

I am doing OCR too btw....I just did my quantitative task today in chemistry...now I got qualitative task coming up tomorrow
Hey!
I am doing mine tomorrow!
Quantitative task, hence i want to get errors out of the way

I will go and ask my teacher tomorrow
9. (Original post by divinelord)
Hey!
I am doing mine tomorrow!
Quantitative task, hence i want to get errors out of the way

I will go and ask my teacher tomorrow
lol Yeah...i think that'll be a better idea, a teacher is probly the best person to ask

Good luck tomorrow

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