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Gas Chromatography of Ethanol in Alcoholic Beverages Questions (1) What is the p

ID: 702272 • Letter: G

Question

Gas Chromatography of Ethanol in Alcoholic Beverages Questions

(1) What is the purpose of n-propanol in this experiment and why is it so cruical to accuracy and precision to this experiment?

(2) Why were there only two chromatographic peaks (Ethanol + N-propyl) seen in the data collected for all of the analytical standards?

Please answer both in detail, thank you!

EDIT: Procedure below:

1) Using an adjustable pipet, transfer 1.00mL of the liquor (66% proof Fireball) and 0.500mL of n-propanol into a 10.00mL volumetric flask. Dilute to the mark with nanopure water.

2)5.00% v/v Ethanol standard (one per group): Pipet 0.500 mL of absolute ethanol and 0.500mL of n-propanol into a 10.00mL volumetric flask using an adjustable pipet. Dilute to the mark with nanopure water.

3) Set up the gas chromatograph for an isothermal anaylsis at 90 C. Inject 1 uL of the 5% ethanol standard until consistant results are obtained.

4) Sequentially analyze 1uL of both your unknown sample and the 5.00% ethanol standard.

5) Estimate the concentration of the ethanol by comparing area ratios to the 5% ethanol standard. Be sure to correct for the dilution factor used to prepare the sample. Prepare 3 addtional ethanol standards suitable for calibration of your particular alchol sample.

6) Injections of the 1uL should be used for all FOUR ethanol standards and unknown (fireball 66% proof). Each standard and sample should be run three times and use the data to answer questions and create graphs.

The graphs had 2 peaks. One smaller peak of Ethanol and one larger peak of N-Propanol.

Explanation / Answer

(1) What is the purpose of n-propanol in this experiment and why is it so cruical to accuracy and precision to this experiment?

Here in this Experiment our purpose is to estimate conc. of unknown sample of ethanol . for this estimation we took help of Chromatography.

Chromatography is a laboratory technique to separate mixtures this technique is based on differential partition.Subtle difference in partition coefficient of substance causes diference.

here we have taken one known & other one is unknown sapmple on results & behaviour of results .we could estimate the concentration of unknown ethanol.

here another compound which we took to study comapare is N-Propanol.

so its crushial for Accuracy & precession as result completely depend on it.

(2)Why were there only two chromatographic peaks (Ethanol + N-propyl) seen in the data collected for all of the analytical standards?

as there are two components of interest in chromatography so there will be two peaks in chromatography .

here we use analytic chromatography to confirm presence & to estimate concentration of unknown sample.