2-EHA calculation

[mohammad]

Dear All,

I am doing analysis for 2-Ethyl Hexanoic Acid by GC as BP 2009/EP 7.0. Now the problem for me is the calculation part. In EP/BP the calculation part is given as

[Area of sample x Internal std area in standard x standard weight in g x 2] [/area of std x internal std area in sample x sample weight in g ].

Now my question is what is the role of "2" in the calculation part

Plz explain.........


Regards
Mohammad.

[Mr.Buffer]

I do not have that monograph at the moment but it is probably dilution of your sample.Probably, it is last extraction volume of your sample.
if you can share your method details , I can answer your question precisely.

Best Regards.

[mohammad]

Please note the analysis details:

Inernal Std Preparation: Dissolve 100mg of 3-cyclohexylpropionic acid in 100ml cyclohexane. (1.00mg/ml).

Test solution: 0.3g of substance+4.0ml of conc.Hcl+1ml Internal std. shake vigorously for 1 min. allow the phase to settle. Inject upper layer 1.0µL.

Standard Preparation:Dissolve 75.0mg of 2-EHA in internal std & dilute to 50ml with internal standard.
To 1ml of this solution add 4.0ml con.Hcl. shake vigrously for 1 min, allow the phase to settle. Inject upper layer. (1.0µL).

Capillary Column: Stable wax or DA-FFAP.

Calculation:
(Area ratio of sample/ Area ratio of standard) X (standard wt / sample wt) X 2.

[kalidassa]

Hi this is Kalidass in the calculation part the 2 is given for dilution of standard and the multiplication of the purity of the 2-EHA. In the method dilution of standard is 0.075g/50 mL. = 100/50=2

Where 100 is the potency of the 2-EHA Standard and 50 is the dilution of the standard 2-EHA solution.

Regards
A.KALIDASS

[mohammad]

Thnx Mr.Kalidass for the info....

I would like to ask, instead of taking 1 ml of internal std in std and smaple as describe in the monograph, if i take 3 ml, will there be any change in the results, bcz the concentration of std will be different?

Regards
mohammad.

[kalidassa]

Hi,

Adding 3 mL of Internal standard in both standard and sample will not differ in the result. Both are resulting is same concentration of 2-EHA. If you want to deviate these things from the monograph then you should validate the same.
I will giving below how the result remains same.

Sample & Standard with 1 mL of internal standard

2-EHA (%) = [(0.075/50)*1/1*1/0.3*100] (Potency of 2-EHA)
= 0.5 %

Sample & Standard with 3 mL of internal standard

2-EHA (%) = 0.075/50*1/3*3/0.3*100 (Potency of 2-ERA)
= 0.5 %

If you adding 3 mL of Internal standard to the sample and 1 mL to standard then the resultant concentration of 2-EHA is now 1.5 %

2-EHA (%) = 0.075/50*1/1*3/0.3*100 (Potency of 2-ERA)
= 1.5 %

Treating the Sample and Standard same will not change in the resultant concentration.
Only the µg/mL will change the µg/g remains as such.

The relation between µg/mL and µg/g is

µg/g= (µg/mL/sample wt in g)* dilution of the sample.

With Best Regards
A.KALIDASS

[mohammad]

Once again thnx for clarifying the doubt.....

With respect to test concentration the standard concentration is 0.5% only, where as the limit is 0.8%m/m.
Can you please explain this........... :


Best regards
Mohammad.

[kalidassa]

Good morning,

I don't know about the specification limit what ever you having. But for the limit 0.8 % we can perform the standard injection at 0.5 %, because if you keeping the limit 0.8% then while doing validation you are going to show the linearity from LOQ Level to 125% or 150% depend on your guideline. Hence we proving our method is linear from LOQ Level to 125% or 150% of specified concentration, we can compare the test solution with any of the standard concentration between LOQ Level to 125% or 150%. It is not compulsory to inject the standard at only specified concentration.

[mohammad]

Greetings..

Once again thnx for the info....


Regards
Mohammad.