Chromatography Forum

Hi Guys, please help me with this

At the beggining of my analysis i prepared standards in methanol ( 0.05,0.07 until 10 ppm ) then i have calibration curve .

I use the formula from This forum ( LOD=3.3 SD/slope) and ( LOQ=10 SD/slope)

note: ( It is the standard error of the calibration curve divided by the slope ) .I found that my LOQ = 0.07 .BUT THE STARNAGE thing is that i can determine " 0.01 " with precision R.S.D% LESS THATN 1 % , that is less than my LOQ oh come on ! .

I thing if i go down i may reach ppb level not sure but i am really confuse until what i go down and down .Do you thing if i inject 0.001 ppm will work ? and what is the acceptable R.S.D % .


Please don't leave alone i need your help please

Thanks

The ICH Guidelines define the SD as the standard deviation of the response,

I use the SD of my lowest Cal STD injections (n=6) in the equation to estimate LOD/LOQ the test at that level with an actual solution

http://www.ich.org/LOB/media/MEDIA417.pdf

Sounds like you are not using weighted least squares, which is the reason the standard deviation of the slope is making your calculated LOQ way higher than it should be.

could you post your concentrations and areas, so that we can take a look at the actual data instead of guessing?

Ace

Sorry guys i should post my result at the beginning

well my analyte concentration is at the following :

HIDDEN

i plot the calibration curve ( from 0.05 until 1 ppm ) i didn't include the 0.01 because it will reduce the LOQ i don't know why .When i ploted i go Regression in excel the i plot the residules > i take the Standard error of the calibration curve *10/slope= LOQ

10 *1680/233230 = 0.07=LOQ

Any help will appreciated plz

I will give you here the lowest concentration i measured right now .

0.01 ppm

the Standard deviation from ( n=10 ) =24.74402285
R.S.D% = 1.278364479

MEAN =1935.6

................................

How will you do it here ? as i spend now two monthes with each time someone tell different story ! so i said i must ask scientists like you

Sounds like you are not using weighted least squares, which is the reason the standard deviation of the slope is making your calculated LOQ way higher than it should be.

You should search the forum about weighted least squares...there is some good info in here. The reason is that in HPLC the standard deviation is not constant but the relative standard deviation is.

There are other approaches to finding LOQ, such as calculating your baseline noise and determing the concentration that gives you a peak at 10 times the noise, or going by RSD as mentioned above.

We normally use 5-8 points, starting from our expected LOD (or below), going to about 10-20 times higher.
We also use all our collected data.
Why did you average all your results and use this for the calibration line?
Or is it just coincidence that you have n=10 with mean 1935.6 while this is exactly the area you posted for 0.01 ppm.
Also make sure that you keep everything as much the same as possible when making your calibration line, and checking your LOD.


Ace

This is probably a detail: I believe you excluded your 0.01 point because it reduced the LOQ. I don't think you should; if you miss out low concentration points when they reduce the LOQ, it's not surprising if you end up with a LOQ that is higher than it should be! You are, in effect, selecting the worst data and making your situation look worse than it is!

There's an area of overlap between aceto_81 and Noser222. What's relevant is the s.d. at very low concentration (near LOQ), not the s.d. at much higher concentrations. I also tend to use a calibration curve in the vicinity of expected LOQ (as aceto_81) but I think Noser222 is trying to do something similar by arranging that s.d. doesn't vary with concentration. Rather than trimming out the 0.01 point, you are probably more justified to trim out the higher points, if you want.

As a matter of interest, if you have, indeed, carried out the entire calibration process 10 times, there is also another fall-back procedure to assess the s.d. at low concentration: you can plot the s.d. of each calibration point versus concentration, and then interpolate to find expected s.d. at any concentration. Now calculate the concentration at which s.d./signal is adequate. Personally I don't like this one much because the scatter on the s.d. measurements at low concentration (i.e. the s.d. of the s.d.!) is large.

Guys , you lost me because you speak at high level . Can you show me an example ? from my above peak areas what is LOQ ? can you tell me ?

now my question : i will use signal to noise ration but now someone confuse me he said take the hight of bigest peak and divided by the difference of baseline noise ! but i thing the correct is to take the high of the signal of the lowest concentration and this much come 3 to LOD

AM I CORRECT ?

As you might surmise from all of this, "LOQ" is a very large "can of worms".

Most people will cite "signal/noise = 10", but this gets *very* ambiguous:
- where do you measure the noise?
- over how long a time do you measure the noise?
- what if there are other peaks present?
- do you measure peak height from the midpoint of the baseline noise or from the top?
- how do you account for noise at the top of the peak?

FDA / USP / ICH allow hint at statistics-based estimation from the standard error of the calibration line. I personally think this is more justifiable, but it has its own issues, several of which have been touched on in this thread (e.g., ordinary least squares is really inappropriate for wide-range chromatographic data).

In fact any specific value for limit-of-quantitation ignores a relatively large elephant in the room: how precisely do you need to quantitate. I can virtually guarantee that you will not get 1% RSD from a peak that is 10x the noise.

To my mind, the most reasonable approach (unfortunately, also the most work!) is to use a "CV vs. A" plot. CV is the coefficient of variation (%RSD), and A is the "amount" of analyte. What you do is run replicates at each level (minimum of 3 replicates; 5 is better) and plot the log(CV) vs. the log (A). As long as you are well within the linear range, the CV will be fairly constant, because the largest contribution to error is likely to be sample workup and injection, and that will be the same regardless of the mass of analyte. That means that the right-hand part of the plot will be fairly flat. As the analyte level decreases, however, the error contribution from baseline noise increases, and the curve drifts upward. Define the required CV for your purposes, look at where the curve crosses that CV, drop down to find the corresponding amount of analyte, and "Bob's yer uncle".

Dear Tom , please if you have spare time can you see how is the LOQ from my previous data , so that i will use your example

thanks dear

Dear Tom , please if you have spare time can you see how is the LOQ from my previous data , so that i will use your example

thanks dear

Not without 2-4 more injections of at least the lower half of your concentrations, he can't...

(whacking the double post too, while I'm here)

DR answered for me.

In order to do a CV vs. A plot, you need to establish the CV (standard deviation divided by the mean value) at various levels, which means running at least three replicates at each level. Since the plots are usually log-log, you could cover your range with runs at 0.01, 0.03, 0.1, 0.3, 1, 3, and 10 ppm (actually, as DR implied, you probably would not need the 10 ppm run).

Ok guys

i have done it but why most of my result after the log appeared negative

Do you know someone in the litreature who did it , come on help me more


( Thanks god that i am not working in the industry otherwise i will be fired )