Final Exam: EXPT 1 - Gas Chromatography of ethanol in gasoline

General Principle of Experiment

Using Gas Chromatography, we employ a column specifically designed to separate, identify, and quantify the ethanol (the main oxygen-containing additive) in a gasoline sample.

GC offers excellent sensitivity, a wide range of applicability, and potential freedom from interferences

Layout of the Gas Chromatography Components

Mobile Phase reservoir (pressurized He cylinder)---> \[ Sample Injector (micro-syringe)---> Stationary Phase (column)---> Detector ( Flame ionization detector) \] ---> Read-out (computer) \n \n \[ \]= OVEN \n \n Mobile phase reservoir- provides mobile phase which is a chemically inert gas that serves to carry the molecules of sample through the column \n Sample injector-allows introduction of a precise sample volume into the column. \n Stationary phase- acts as a separator, substances more attracted to it will slow down and yield longer retention times \n Detector- detects components of mixture being eluted off column \n Read-out- converts detector data into a chromatogram, giving peak area and retention time

Retention Time

The length of time between when a compound is placed at the top of the column and when it comes off the bottom, and is different for each compound. TIme spent in the stationary and mobile phases of the gas column.

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Larger retention times: longer interaction with the stationary phase & temperature is not near compounds boiling point

Smaller retention time: longer interactions with the mobile phase and weak interactions with stationary phase & temperature is near or at compounds boiling point

Partition Coefficient

K = \[A\] stationary phase / \[A\] mobile phase

Calibration Curve

x - concentration

y - peak areas

Internal Standard

Why used? By adding a known amount of internal standard that is similar to the compound in question structurally and that produces a similar retention time it can be used to create a new peak in the chromatogram that by comparison can be used to quantitate the unknown peaks.

Response Factor Calculation

The factor of the internal standard to the unknown compound. The value of RF can be obtained from a single experiment in which known amounts of unknown and internal standard are injected and the respective peak areas are measured.

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\[Amount\] ethanol / \[Amount\] butanol = Area ethanol / Area butanol \* RF

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Experimental Method

* Use of FID (Flame Ionization Detector)

Why use separatory funnel?

Gasoline is a mixture of more than 300 different hydrocarbon compounds. Ethanol is one of the few that are water soluble. Therefore, it is easily isolated from the hydrocarbons by extracting it into water, which can then be analyzed by the GC.

There are fumes coming from the gasoline that can be released without messing up the separation.

In a chromatogram, what quantities are plotted on the x and y axes?

x-time \n y-detector response

What are the structures, molecular weights, and boiling points of ethanol and 1-butanol? What functional group must be present for a compound to be considered an alcohol?

ethanol- CH3CH2OH, 46.07 g/mol, 78.37 degrees C \n 1-butanol- CH3CH2CH2CH2OH, 74.12 g/mol, 117.7 degrees C \n \n -OH (hydroxyl)

Suppose that you are using GC to determine the concentration of ethanol in a sample, and 1-butanol is used as IS.

A mixture containing 8.0x10^-5 M ethanol and 6.4x10^-5 M 1-butanol is injected into the GC and gives peak areas of 122,000 and 172,000, respectively. Calculate the RF for ethanol.

Amount unknown/Amount IS=Peak Area unknown/Peak area IS x RF \n \n 1.76

Suppose that the unknown peak and the IS have an RF of 0.81. If injection of a sample containing 50 ppm IS gives peak areas of 8230 for the unknown and 12,410 for the IS.

What is the unknown concentration in ppm?

x/50=8230/12410 x RF \n \n 26.9 ppm

In choosing a compound to serve as IS for a GC analysis, what are the desirable properties to look for?

1\. cannot be a compound in the sample \n 2. want IS to be similar in structure and properties to compound looking for, this will give it a similar retention time as well

Similar size, slightly higher boiling point, same functional groups

What are the two principal advantages of the IS approach over use of a calibration curve?

1\. Any hard-to-reproduce operation in the analysis procedure that is the same for both unknown and IS should not influence the determination \n 2.possible to use a single IS compound for several unknown GC peaks, as long as a specific RF value between IS and each sample component to be determined known

Not affected by instrumental variance between injections. And concentration of multiple compounds can be determined in a single injection.

A GC peak has a retention time of 2.6 minutes and a half-width of 5 seconds. What is N for this GC system? Would a GC system with a higher N value be considered more or less efficient?

N=5.55 (Tr)^2/(w1/2)^2 \n \n MAKE SURE TO CONVERT Tr and w1/2 to same units \n \n 173.16 \n \n more

How does peak broadening affect the efficiency of a chromatographic separation? Why are narrow peaks desirable?

It lowers efficiency. Narrow peaks are desirable because they have less of a chance of interfering with another peak if they are narrow.

Broadening leads to peak overlap and won’t be able to quantitate peak areas to a single peak.

Qualitative & Quantitative Components

Qualitatively compare the neat gasoline sample, aqueous extract, and 1% aqueous ethanol solution, noting the retention times

Quantitatively, the peak areas of the 3 IS sample to determine RF and determine the amount of ethanol in samples

1. (10 POINTS) arrange the following in the proper order for a functional GC: Detector, Column, PC/printer, Helium Cylinder, Detector

Helium Cylinder → Injector → Column → Detector → PC/Printer

2. (10 POINTS) What do the x and y-axis correspond to in a typical GC chromatogram

x- Time

y- Detector Response

3. (10 POINTS) Of the following GC observable values (detector response, time, area), which

Qualitative: Retention Time

Quantitative: Area

4. (10 Points) A mixture containing 1.0 % ethanol and 1.0 % 1-butanol is injected into the GC and gives peak areas of 122,000, and 172,000, respectively. Calculate the RF.

1\.0 = 0.709 \* RF

RF = 1.41

5. (10 pts) In choosing a compound to serve as IS for a GC analysis relative to the analyte of interest, name three desirable (hint: similar) chemical and physical properties to look for? \[ie, why choose n-butanol (CH3CH2CH2CH2OH) for ethanol (CH3CH2OH) \]

Similar functional groups, similar size (molecular weight), higher but similar boiling points