Chromatography Forum

Hi Guys. I need help with my calcualtion. Let me first explain my procedure;
1-Three grams of meat mixed with 9 mL of water and homoogenized and centrifuged.
2- One mL dried and redissolved in 125 uL and injected into the GC. Also, 125 uL as standareds is injected directly into the GC for comparison and aslo to generate a standard curve.
how can I calculate my analyte?. Thanks
Also, can I use this equation:
Internal STD nanogram added*analyte peak area)/(GC reponse factor*3 gram of meat* internal STD peak area)
and hoe I can calculate my internal STD recovery.

Rather than try to blindly apply a "cookbook" formula, you need to understand how calibration works. Here's a link to a tutorial buried on our web site (it's written for HPLC, but applies to any form of chromatography):
http://www.lcresources.com/resources/ge ... ction4.htm
Note that it refers to "sensitivity factor", which today would be called "response factor".

Once you have worked through that (which deals with "external" standardization), internal standardization is easy. You add the internal standard *at the same concentration* to all of your calibrators and samples. When you do the chromatography, measure the area of both your analyte peak and the internal standard peak and then compute the ratio (area of analyte)/(area of internal standard). From there, the math is identical to what you did with external standardization; just use the area ratios in place of the areas.

Hi Guys. I need help with my calcualtion. Let me first explain my procedure;
1-Three grams of meat mixed with 9 mL of water and homoogenized and centrifuged.
2- One mL dried and redissolved in 125 uL and injected into the GC. Also, 125 uL as standareds is injected directly into the GC for comparison and aslo to generate a standard curve.
how can I calculate my analyte?. Thanks
Also, can I use this equation:
Internal STD nanogram added*analyte peak area)/(GC reponse factor*3 gram of meat* internal STD peak area)
and hoe I can calculate my internal STD recovery.

At no point in this description do you say you actually add an "internal STD" into the mix. What you are describing is an "external standard" method.

For an Internal STD method you should be adding a known quantity of the IS to the samples and standards immediately prior to analysis.

Internal STDs added prior to extraction are now often referred to as "surrogates). Their recovery would be determined against unextrcted spikes. For example in the above extraction, if you spiked a sample with 100 ng of the surrogate, you should compare the spiked extract with a 10ng/125µL solution to estimate the recovery.

This is the formula for the analyte concentration recovery:
Ci/Cis=Ai/(Ais*slope) => Ci=Ai*Cis/(Ais*slope), where Ci is analyte concentration, Ai area of the analyte, Cis- concentration of the internal standard, Ais - area of internal standard, slope of linear relation of the external standards. After findidng of Ci in the solution you've made you can find the beginning concentration of the analyte solution which you diluted by multiplying Ci with dilution factor.

[quote="xeim"]This is the formula for the analyte concentration recovery:
Ci/Cis=Ai/(Ais*slope) => Ci=Ai*Cis/(Ais*slope), where Ci is analyte concentration, Ai area of the analyte, Cis- concentration of the internal standard, Ais - area of internal standard, slope of linear relation of the external standards. After findidng of Ci in the solution you've made you can find the beginning concentration of the analyte solution which you diluted by multiplying Ci with dilution factor.[Thanks Xeim, so I have to multiply the calculated Ci after applying your equation with the dilution factor, In my case I am mixing 3 grams of meat with 9 mL of water and centrifuge this mixture and I do take 1 mL of the supernatant. My dilution factor will 3 or 4 times]