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Offline mrivera3

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GC quantification
« on: December 03, 2010, 10:05:32 AM »
I have to quantify chloroxylenol with some GC Chromatograms, I did not do the experiment, I only have the data obtained by another chemist. I know the concentration that was injected and the area of the peak, I just want to figure out how can I obtain a precise amount of concentration of the sample from the data that I have and if there is a way to calculate the difference between the Known and the unknown. There was no calibration curve performed. I am lost, the only equation I found in the notebook was: std. area/std concentration = sample area/sample concentration, is this accurate for trying to obtain the sample concentration?? and what is %LS???

thanks!!

Offline cil-baha

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Re: GC quantification
« Reply #1 on: December 03, 2010, 11:17:19 AM »
From what you said I gather you have all the data required for a one-point calibration curve, for which you cannot do a %LS (Percent Least Squares, or, more appropriately/commonly accepted as the statistical standard, Correlation Coefficient) "std. area/std concentration = sample area/sample concentration". Your printout of the chromatogram data should have the standard area = std. area, and all the data you need to enter into the calculation.  Yes, it is acceptable for a one point calculation, your correlation coefficient will be 1.000000000 for a single point calibration.

Offline cil-baha

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Re: GC quantification
« Reply #2 on: December 03, 2010, 11:27:38 AM »
Note for precise measurements you need to bracket your samples with a standard.  If you got the std calibration point from another instrument, then it is unlikely to be effective, it is unclear from your paragraph what exactly you have and from exactly where.

Offline mrivera3

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Re: GC quantification
« Reply #3 on: December 03, 2010, 12:17:22 PM »
Sorry I know it is very confusing, I am as confused as you can see, that is maybe why it is unclear what i wrote. Thanks for your help, the data I have is the Following:

I have a sample that supposedly contains 1% Chloroxylenol

Apparently they prepared three standards of Chloroxylenol, 0.04, 0.08 and 0.1 mg/mL and ran them on a GC and I have the data with the three chromatograms all with the area and height of each peak. (what I did with this data was to make a calibration curve, or what i think a calibration curve is plotting the concentration as 'x' and area as 'y', is this correct???)

The other thing I have is some data of samples of chloroxylenol at the same concentrations as before, filtered and unfiltered samples with their chromatograms and what the chemist did was calculate the sample concentration with the equation of before, is that correct??? They are asking me to accurately give them the concentrations of chloroxylenol but I have no clue as to how I can obtain it as of right now... Maybe I am overlloking something that is right there in my face but I just cant seem to reach a conclusion

Offline cil-baha

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Re: GC quantification
« Reply #4 on: December 03, 2010, 02:29:18 PM »
You have more than one calibration data point - Very Good!
You now have a calibration curve using the Stds - Very Good!
Say the 0.04 has area=40000, 0.08 Area=80000 and 0.1 Area=100000, and the Area of your sample analyte has Area of 50000, then the concentration will be 0.05, don't forget the units, in this example, the same as your stds, mg/mL!

Now to confuse you:
It is straightforward using the linear relationship:
y= mx + b, yes, usually the y-axis is for the Detector Response (Height or Area of the response) which is assumed to give a directly proportional relation to the concentration of analyte, when the Correlation Coefficient of the plot is close to unity (0.99-1.0000).  The unknown can be matched/fitted on the curve, &/or even better, by just substituting the values of y1=mx1+b, y2=mx=b2, and so on.  Concentration in mg/mL in your example is used as the other axis.
Remember that you were asked to use the Area, which I use as well, you might be confused by having two graphs - one is a chromatogram, the other is your calibration curve for Chloroxylenol std.
Don't be confused, this is elementary, you have to get it right!

A reference from
http://en.wikipedia.org/wiki/Calibration_curve
seems to have a nice graphing calculator you can use!
http://terpconnect.umd.edu/~toh/models/CalibrationCurve.html
http://terpconnect.umd.edu/~toh/models/CalibrationLinear.GIF
Do you see R^2=0.9997?  That R^2 is your Correlation Coefficient, and is very good in this image.
Following this, the
Instrument reading is your Area (y-axis) and the Concentration of analyte in Dettol (oops, Chloroxylenol std) is your x-axis (it shouldn't matter too much which is x or y in this example, as long as you are consistent).
I expect you do know how to plot a graph, and use it!  That is your Calibration Curve to which you are to fit-your-data of Area of Analyte in sample, and get the unknown concentration.

www.google.com/search?q=calibration+curve

Additional help you may need,
http://en.wikipedia.org/wiki/Parts-per_notation
1% solution = 0.01part solute or analyte in the '1-entire solution' => [1/100]*100% = 1%,
therefore, 1mg analyte/{100*(100%)}=0.0001ppm = 0.0001mg/L,
www.onlineconversion.com/forum/forum_1130512020.htm
We can now forget about the above - you may need it to know it for your Multiple Choice Tests, and continue with your Chloroxylenol std solutions of 0.04mg/mL=>40mg/L, 0.08=80mg/L and 0.1 mg/mL=100mg/L,
or 40, 80 & 100ppm Std Solutions, respectively.
Your 1% Chloroxylenol solution is 10,000ppm and much more than your calibration bracket, you need to dilute your Chloroxylenol, say 166.6667 (=500/3) to get your Xchloroxylenol to fit within the calibration bracket of 40, 80, 100 mg/L.
You could read further - Skoog, West, Holler, et al. in your science library/online.

Now go back up to the third line and do it!  What is your answer?  ____________mg/mL Chloroxylenol in sample.

Offline cil-baha

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Re: GC quantification
« Reply #5 on: December 03, 2010, 03:14:26 PM »
yes, you need to dilute your sample so it fits within the calibration curve, and then multiply your result from the fit to get the true concentration in the original sample, yes, that is correct.

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