Chromatography Forum

Hy Im a student,and I need help in understanding the below procedure.

This is the experimental setup:

(1) Qualitative analysis of organic waste samples

Internal Standard is Ethyl Benzene
Test solvents include: methanol,Acetonitrile,Toluene and Acetone

1) Prepare each of the test solvents and waste samples by the addition of IS.This is achieved by the addition of 1 mls of IS to 1 ml of solvent.

2) Dilute each of the samples and test solvents with dichloromethane in a ratio of 50:50 (v/v)

3) Inject each test solvents and record their retention time's.Also inject the waste samples

4) Qualitatively determine which solvents are present in the waste samples.Use the chromatograms to determine the % composition.

My question here is how to determine the % composition?

(2) Quantitative analysis

Here they asked me to find the Response Factor (F) to find the concentration,of the unknown mixture.

concentration of std/conentration of IS=Fx area of std //area of IS.

My question is


how to find the concentration of std and concentration of IS if there concentrations are not given?
Should i calculate it,if so how?

Thanks

Use area normalization
(individual area / total area) x 100 = % composition

For ISTD calculate RRF (relative respond factor)
(Ci / Ai) x (As / Cs)
Where: Ci = Concentration of individual component
Ai = Area of individual component
As = Area of ISTD
Cs = Concentration of ISTD

Hopefully this information can help you.

M Azmi

Well the concentration is actually given or rather can be calculated in 2 ways.

You know that you add 1ml. So if doing it accurately you can always look up the density for each solvent and calculate the final concentration. Should do for a student project.

Alternatively but a bit tricky if due to evapouration: add the IS first to a vessel (less volatile), put on a scale and zero it, then add 1ml of the solvent of intrest and you get the weight.

How do i do that?
For example take Methanol.It has a density of .79130 g/ml.I used 1 ml.Therefore the Mass i used will be .79130 x 1 =.79310g.

After finding the mass I used,whats the nxt step?
How do I express this in g/ml or in any other concentration terms?

Sanju,

If I understand your dilution scheme correctly, you have 0.79g/4ml total volume of sample or ~0.2g/ml. That gives you a response for that concentration of that solvent.

By the way, it should come out pretty close to the area % of that component relative to the other components if you are using an FID. Not exactly because you are not dealing with straight hydrocarbons, but close.

Best regards.