Agilent 5800 Dilution, Dilution factor

Hi hayk,

I'm not skilled enough to give you an accurate answer, but according to my short experience with an agilent ICPOES : the "Dilution" column is a ratio=final volume of solution (ml)/mass of sample (g). For the example you gave, it would be 254.97g/ml. You can check it out in filling the two left columns (weight (g) and volume (ml).

The ratio of extraction (volume of acid/mass of sample) and the dilution factor (final volume of solution/initial volume of acid) are not used by the software.

The weight would be 0.1961grams and the final volume is 50ml, so you put 0.1961 in the "Weight" column, and 50 in the "Volume" column. When you analyze the samples, you get a result in ppm typically, which is ug/ml or ppm in solution. This value gets multiplied by the volume of 50ml, and divided by the weight in grams, for a final result of ppm in the original solid or ug/g. 

Hi dear Tina

Thanks for your reply. In another forum i got answer like that:

Thanks for reply. But looks like the ratio of extraction and the dilution factor are used. I did some experience and checked the calculations with different equations. And looks like software is counting concentration as follow: Concentration in the original sample = Concentration of the measurement solution * Dilution * (Volume/Weight). However I will wait for another replies. Thank you so much. And which model of Agilent ICPOES  do you use?

this equation is good, however, you have no additional dilution, so the "dilution" factor would equal 1. 

Thank you very much