Hi everybody.
Recently I had GC-MS analysis for the butter volatiles done. The results they sent me several conflics and I have no idea how I can manage with them. I have no previous experience in GC-MS. But after they sent me the choromatograms I am able to hande chromatograms.
Briefly, the choromatograms were obtained in Shimatsu GC/MS with a Zorba Wax plus coloumn. N-alkanes (C4-C20) for retention index and 3 different internal standards for peak area calculations were used. Almost 100 peaks were obtained in a single choromatogram. The program library correctly gave Kovats RIs for C4 to C20 as 400 to 1000 and 3 internal standards.
Here is my problem. The person performed the analysis named the peak at KI 683-687 as n-haxene. The library however did not give it. Rather the most similar compounds are pentane, 2-methyl (92%), n-haptane (91%), haxane ,3-methyl (90%) and butane, 2-3dimethyl (89%). Definitly n-haptane is wrong since I found it at KI 700.
I really appriciate if anybody explain me to evaluate and determine the compounds from the library easly.
Thanks in advence for responses.
Sinan Colakoglu, Ph.D.
So, now when we know for sure that we talk about hydrocarbons… (Salvador - grate job – bravo for clearing this situation) :)
One more mistake – for C20 you obtain RI 2000 not 1000.
The MS identification, which is based on the MS spectrum sometimes is not good. This particularly takes place when you have to differ between isomers. The spectra for n-alkanes, iso-alkanes do not differ so much. For baseline separation the program can give you good answer, but you have very complex mixture – how can you ensure that all compounds are baseline separated? I’m afraid that in many cases you get co-elution.
The column used is not good for this type of mixtures – you use very polar PEG stationary phase – this phase is good if you need to analyze more polar compounds in hydrocarbon matrix. The hydrocarbons are eluted first often as a big not well separated complex of peaks. For example benzene is eluted after n-nonane.
To analyze properly your mixture you should use 100% PDMS stationary phase (such as DB-1 or HP-1). The column should have 60 or 100 m an have very good efficiency (i.e. the thickness of the stationary phase for 0,25 mm ID column should be 0,2-0,25 um). At such conditions you obtain almost perfect elution in order of boiling point of the hydrocarbons. For good separation you can use MS, but there is a special application (even standard test method) ASTM D5134 which allows to identify all hydrocarbons only by its retention indices (remember that Kovats RI are calculated for isothermal program and the formula is different than for Liner Retention Indices).
Best regards,
G
By definition an n-alkane has a retention index of 100 times the number of carbon atoms. Therefore any identification of a peak whose RI is not divisible by 100 must be wrong. Since you say that the alkane peaks were correctly assigned their RIs this must mean that there were two n-hexane peaks an the chromatogram and two n-heptane peaks- which is impossible Since you are analysing butter volatiles and not crude oil you can ignore the advice about using a very long non-polar column.
Dear Mr. Boczkaj and Apps,
Thanks for your kindly response. You are correct that KI for C20 was 2000. As you wrote, many peaks are nested and their identifications are difficult. There is no definite baseline and I evaluate each peak one by one. As you imagine it takes too much time.
Here is the method as they have just sent me “Extraction of volatiles was carried out using 2 cm 50/30 μm divinilybenzene-carboxen- polydimethylsiloxane (DVB/CAR/PDMS) fiber (Supelco, Bellefonte, PA, USA) and the volatile compounds were separated on a ZB-Wax plus column (60 m × 0.25 mm × 0.25 μm; J&W Scientific, Folsom, CA, USA). Chromatographic conditions were the same as described in by Cakmakci (2012).[19] The oven was held at 40 °C for 2 min and then increased at 5 °C per min to 70 °C, where it was held for 1 min. The temperature was then raised at 10 °C per min to 240 °C, to give a run time of 30 min. The mass spectrometer was set to record at 33–450 amu at a sampling rate of 1.11 scans/s.”
As given example in my previous mail, at KI 685, do I need to look compounds that contain 6 carbon atoms? But İ can see from your response that this idea does not work, since the compound with R groups (ex. Methyl) elute before or after than its related n-alkenes. I added an excel sheet that shows the gc-ms chromatogram and library search. You see that pentane, 2-methyl are the most probable compound (92%), however, its RI is given as 584 in DD2012.lib, and 554 in W9N11.lib and NIST11.lib. Two questions arise here: 1. According to KI value of the peak (684), is pentane, 2-methyl (KI 554 or 584) correct compound? 2. Which is/are library/ies reliable in terms of KI values? I need definitely your advices and recommendations.
Thanks your consideration and appreciate spending your time for my problem. Sincerely. A. Sinan Colakoglu, Ph.D.
In any GC-MS analysis based on library searches there will be some wrong identifications and some peaks with no identification that makes sense. This is just a fact of analytical life. You can spend a lot of time, money and trouble, on identifying every peak and confirming with standards, but that is a waste unless the unknown peak has some significance to the question about butter that you are trying to answer.
A 90% confidence level library hit is good but not enough to identify any molecule. it puts you in the ball park. A triple quad or a qtof would provide higher resolution data and therefore a better idea of what you are looking at. Absent one of those two instruments, try running standards based upon the library hits. Even with a triple quad or a qtof you would still want to run standards to validate your results.
Sinan,
Check for what type of stationary phase you obtain the RI from the library (is it for WAX phase?).
For your column the RI for molar polar compounds will shift to higher RI values comparing to DB-1 or HP-1.
The RI will be never the same as it is in the library.
Peter,
RI don’t have to be “divisible by 100” the calculation is based on interpolation of RI based on retention time in relation to tr of n-alkanes.
Regards,
G
Branched alkanes are always eluted before the n-alkane containing the same total number of carbon atoms, i.e. 2-methyl-pentane will be eluted before the tr of n-hexane.
Dear folks
i have unkown plant materials. I have done GCMS, few compounds are available on lib but not most of them so how i can found it on the bases of M.wt. My interest is to find out the polysaccharides. kindly help me out .
regards
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