[111]

HI everyone,
I have the experiment on the experiment of HPLC. My experiment ,inject the same conc. of standard solution =0.1%w/v,and i have to use their peak height and pesk area to calculate the conc. of the unknown phthalate in the tygon sample.
I found that the formulas as the follwing:
 Conc of unknown component =conc. of it corresponding component of standard x Peak area(sample)/Peak area(standard)

My sample contain 2 different components of tested standard in the sample and now I need to calcuate the total 2 components' concentration in the sample tygon,is that right if i only add up the 2 component for peak area (sample), and add up 2standard componentsfor peak area (standard ),but use the same conccentration value for calculate the total concentration of the total 2 component in the sample?
For example,the conc. of standard =0.1, and peak area of a standard=a1,peak area of b standard=b1, peak area of tested sample a=a2 ,and peak area of tested b=b2,then
conc.of total 2 component in the sample =0.1x (a2+b2)/(a1+b1)
Is my calculation correct?

[marquis]

I assume that you extracted the plasticizers out of the tygon in some solvent.  The volume is important, because it affects your calculation.

Your base calculation of the unknown concentration looks good. The general procedure from here is as follows:

If the units you are using are ppm (ug/ml), you need to multiply by the solvent volume (in ml) and any dilution done (by way of a dilution factor).  Divide this by the weight of the sample you extracted (preferably in grams).  The answer will show up in ppm, which can be converted to percent, if needed.  These calculations assume linearity, which you need to prove by injecting multiple standards, in the range you are analyzing, and seeing if the curve is linear or not.

Please look over your extraction procedure.  Different extraction procedures could change the calculation. 

If you have no filler and the polymer is also soluble, there are simpler ways of calculating.  These are usually percent plasticizer methods.

[111]

we don't have any calibration curve, as we use the same amount and same concentration as the standard testing
so the diluting factor may not be applied

[marquis]

If no dilution of the sample is needed, the dilution factor is one.

You need to prepare standards of different concentrations.  Inject a standard amount of sample for a variety of concentrations that bracket your measured concentration.  Then plot the graph of concentration vs peak area.

The curve maybe a line.  This doesn't always happen, and many of the curves are very bent.  The calculation assumes a straight line.  If you don't have that, the answers will be wrong.

If the different concentrations produce a linear curve, you can accurately measure unknown concentrations to the upper level of your linear curve.