Software calculations differ from human calculations

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LMP
LMP
in Software Products
Hello!

I am new to chromatography and empower software, so be gentle. 
I have a problem with value (%) calculations:
1. Is there a way to see what actions dose the program do to calculate (like multiplication, division etc.)
2. The only things i enter are wight of standard and sample, and the dilution of standard. How dose the program calculate anything with just these?
3. The final concentrations of sample and standard are the same 60mg/L, and the area of sample is just a little bit smaller than that of standard, but (%) value of standard is 0.153, and sample it is 0,047. But should`nt these be close, as the concentrations are really close?

Weight of sample - 3,0640
Weight of stan. - 0,1550
Dilution of stan. - 10
Purity of stan - 99.5

Reaaaaally sorry for these dumb questions ;( 
Thank you for your time.  

Best Answer

  • HeatherLongden
    HeatherLongden
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    I'm wondering if you are using the sample field Standard weight.... where you really need to put the amount of standard into the Component Amount. Very often in chromatography there are more than one component in a standard, and you need to add the amount of each component for Empower to draw calibration curves specific to each component. This is why you cant use a sample level field. That field called sampleweight is a correction field, it is commonly used for samples but rarely used for standards

Answers

  • LMP
    LMP
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    Oh, and the calculations differ. 
    The on hand formula is - ((Area of sample)x(mass of stan.) x 100%)/((Area of stan.) x (mass of sample)x dilution)). 
  • MJS
    MJS
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    I'm not sure what exactly you are referring to by value (%).  Are you looking at a particular field in Empower or what when you refer to that?  What are you expecting that to represent...the % active in the weighed sample or just a concentration in the sample or a % agreement to the standard somehow?  If you could clarify what the analysis is really trying to report, that might help explain it better, but I've taken a shot at explaining some below.  You may also want to consult some docs from Waters.  I've attached a Waters guide on acquisition and processing...check out section 4.3.1 and 4.4.1 as they seem to be specifically what you are applying here (probably read through 4.3.7 for good measure).

    There are quite a few fields that contribute to amount values within Empower to ensure it is robust for even the most complicated analysis.  It would seem that you aren't utilizing all of those fields, which is fine so long as that is correct for your method as most will simply default to a value of 1 within Empower.  There is certainly some information on this process in the online help within Empower

    When calculating amounts in a sample, I like to start with the calibration.  Empower is going to plot the amount (x-axis) vs the area (y-axis).  Amount is calculated by the entry in the component editor (0.1550), multiplied by purity (99.5, automatically divided by 100 subsequently), multiplied by sample wt. (1 unless you changed it for some reason for the std injection), divide by the std dilution (10) which achieves a concentration (units are dependent then upon the std wt and dilution).  From there, it calculates the slope (you only have one std concentration, so even simpler as it doesn't require full linear regression).  So, slope is essentially Stdarea/StdConcentration, or more fully: (area/(stdwt*purity/dilutions)).

    For the sample then, Empower calculates an amount.  It takes the sample's area, divides by the slope, multiplies by the sample's dilution factor, divides by the sample wt. (3.0640).  Sample is samplearea/slope*sampledilution/samplewt.

    When combined, the std goes into the denominator and the the sample calculation becomes: (Samplearea * Stdwt * Purity * SampleDilution) / (Stdarea * Samplewt * StdDilution).  This seems to me to match the hand calculation you provided (with some additional detail that apparently doesn't really matter in your particular use.

    When using your actual numbers I don't understand how you are getting a 60 mg/L std solution for the standard.  0.155 * 99.5/100 / 10 = 1.54.  Am I missing something there? The 10 doesn't seem right at all for that mass regardless of units.  How are you not entering a dilution of some sort for the sample?  The 3.064 has to be dissolved in something, right?

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