Chromatography Forum

Dear Rod,

Thanks for your encouragement, I surely update the results! I have a question regarding how to estimation of LOD & LOQ by using this method?

Waiting for the positive communication.


Thanks,
Srinivas

Experimentally is the best way to determine this, rather than doing so mathematically from the regression line data.

The noise level of your GC and the purity of your water and DMAc, the amount of HS sample you inject and the ratio of your split (if you use a splitter) all affect the LOD and LOQ.

You saw from the data I sent you from my paper that my blank values were sometimes zero and sometimes a few hundred counts. One of the problems I encountered was the amount of solvent in the lab air. My lab contained several LCs and the floor of my building had several labs. Thus the building air contained low levels of methanol, acetonitrile, and some of the other solvents. In the spring and fall when the windows were open I would get traffic passing by my building and the exhaust from the shipping and delivery trucks would contaminate the air with hexane and toluene so my blanks for those solvents would not be zero.

And heaven help me if I drank a beer with my pizza at lunchtime !

My ethanol breath would affect the ethanol blanks the rest of the day !

Prepare blank samples with smaller amounts of your std addition solution (compensating the reduced volume with pure DMAc) until you are only adding 10µL of pure DMAc. When you reach the lowest level of solvent in which you can see the solvent peaks then you have reached your LOD and usually about 3x that level is your LOQ. Be sure to check that the addition of your drug to that concentration of solvent does not affect the LOD, or at least you measure how much the drug affects that value, either higher or lower.

best wishes,

Rod

Srinivas

It has been over a week since we have heard from you.

Anything to report?

best wishes,

Rod

Dear Rod,

Standard addition method worked out for calcuation of recovery study @ residual solvents with matrix effect. I followed the exact procedure as you followed in your excel sheet (controls values included for linear equation). But I have a quick question @ how to calculate the amount of residual solvent in ppm for Test sample that is having matrix effect????I have to options:

1. Got the Equation of straight line for spiked samples vs areas ( excluding controls)
If the area of residual solvent in Test sample was calculated for equation of straight line above with slope and intercept value. The Amount (ppm) was not even close to Certificate of analysis.

2. If I calculate the Amount in ppm by using external std formula:

Amount ppm = Area of test sample/ area of stdX Std wt XPurity factor/vol taken X dilution factor X (ml taken in HS= 2ml)/ wt of sample X1000000

Amount calculated was close to certificate of analysis, but can we use this formula for calculation as area of residual solvent in test sample undergoes matrix effect????

Regards,
Srinivas

I am sorry if I do not understand your post.

You performed the std addition and got a linear response for the spiked samples.

You calculated the amount in the sample from the std additions.

This value did not agree with your external std calculation.

Well, you may have samples which are not homogeneous (each portion taken from the bulk has a different amount of residual solvent contained within it).

When given a comparison of methods: ext std against a std addition, std addition wins every time (as long as the results are linear and have low RSD values.).

Assuming you did 4 or 5 points in your std addition, if you got a linear response, then your analysis of the sample performed 4 or 5 times gave the same answer.

An ext std does not have the same matrix as the sample. The partitioning into the headspace (the concentration of solvent in the headspace) will be different.

Share your results with me and I can review them.

You have my email address.

Rod

Your R square values are a bit lower than they should be.

Your higher spikes are starting to drop below the regression line, indicating that your recovery at the higher levels are deviating from linearity. (your line is curved downward)

I would lower the solvent in the vial from 2mL to 1mL or 0.5mL.

best wishes,

Rod

Dear Rod,

Methanol (150 ug/ml= 3000ppm) area @ 250
Acetone and Ethyl Acetate (250ug/ml =5000ppm) areas @ 4000 and 9000
Chloroform (3ug/ml=60 ppm) area @ 50

I cannot go for 1ml or 0.5ml bcz the specification of bcz areas of acetone and ethyl acetate for spiked samples will be more than 10K is that fine??
Please let me know how to calculate the amount of residual solvent in ppm in test sample/control from the linear equation (straight line).

Regards,
Srinivas

I made changes in the spreadsheet and performed the calculations for you. I hope a close examination of the changes will make the process clear to you.

Once you have a valid slope of the regression line, simply take the area count of the solvent in an unspiked sample and divide by the slope to calculate ppm solvent content in the unspiked sample.

best wishes,

Rod

I don't think vdnsrinivas will object if I share his final data,

but his results for the HS analysis that was giving him problems

are typical AND impressive, in my opinion:


regression line for EtoAc 0 - 1% = 0.99985

regression line for CHCl3 0-130 ppm = 0.99621

regression line for Acetone 0 - 1% = 0.99986

regression line for Methanol 0 - 0.5% = 0.99908


All the results are GREAT and the only issue (and a very minor one at that, given the EXTREME range of levels he is measuring) is his chloroform values are all consistently 3 ppm low. That is a 'problem' many would love to share.

Nice work.

Rod

Rod,

You know what this forum needs? A "Like" button. You rule. I've developed and validated a number of different headspace residual solvents methods, but I have always used internal standard. I have always luckily been able to easily hit 90-110% recovery for most solvents, so it can be done, but there are obvious deficiencies, and your way is so much more elegant. I have been increasingly thinking about going in this direction for doing residual solvents, and I just happen to have a product that we are about to develop a new method for. I'm going to show everyone your posts!

Stephen

You might like to show them the original data I gathered for my paper in Analytical Chemistry June 1997. I will send you the spreadsheet with the plots and raw data if you email me.

It was an extreme test. I was using a drug sample of 1 mg in weight and looking for residual solvents from 0.5 ppm to 1000 ppm. I also accelerated the analysis so it could be done with a water soluble sample in 6 minutes, or 15 minutes with non-water soluble samples.

Solvent R squared 0.5 to 1000 ppm recovery of 3 injections at 50 ppm

Methanol 0.99959 95.2%
Ether 0.99985 97.0%
Ethanol 0.99981 98.7%
Acetone 0.99982 96.3%
IPA 0.99982 96.6%
DCM 0.99974 96.1%
Hexane 0.99981 96.5%
n-PrOH 0.99977 95.5%
EtoAc 0.99988 96.0%
THF 0.99988 96.4%
CHCl3 0.99969 94.4%
Benzene 0.99974 96.1%
IBA 0.99982 95.8%
TCE= 0.99953 94.8%
Dioxane 0.99977 95.2%
MIBK 0.99988 96.2%
Toluene 0.99952 96.1%
Pyridine 0.99517 88.5%


Once I actually tested a 1 mg sample, split in half (2 vials) and did a three point regression with a vial with no sample, looking for ethyl acetate. I repeated the analysis with another 1 mg sample.

I found ethyl acetate at 1335 ppm and 1342 ppm in the two 1mg samples. Why only 1 mg? Each 1mg sample cost $100 in 1996. The work was done in 2 hours, including writing the report.

I used this method for septic sewer samples, blood and urine samples, as well as Tylenol and cocaine.

I usually used 100-200 microliters of solution and 5 to 10 mg samples.

Send me a note and I will send you my data. chromatographer1 at the ya hoo web site.

Rodney George

Stephen,

PS

I also am available for consulting.

Rod

Thanks, Rod. I'll pop you a note.