Chromatography Forum

Can anyone tell me what is wrong with this equation? thanks.


%Dissolved = (AreaofsampleSolution)(slope)+ y-intercept

Pretty much everything. It's meaningless.

To expand further here is how the dissolution procedure was made.

Initially, the vessel contains 500ml of Disso Medium A. After 1hour, 6ml of sample solution was withdrawn and replaced with exactly 6ml of disso medium A. Adding disso Medium B was made.

After 4, 8 and 24 hours, 6ml of sample was withdrawn and for every withdrawal, 6ml of disso medium B was being added.

1. For standard preparation, various concentrations were prepared at a 5%, 20%, 50%, 80%, 100% and 120% solution. Linearity curve was established by plotting each percentage (on the y-axis) and average percent area (for the x-axis).

2. As per normal way of establishing linearity curve, concentration of standard (is on the x) while peak area (is on the y).

The reference of the procedure was using Method 1 instead and generated a formula something like the following:

%dissolved = (peakareaofsample)(slope) + y-intercept + (6ml * X/volume)

where:

peakareaofsample: peak area of sample from the withdrawn solution
X: Percent Amount dissolved from the previous interval, corresponds to 0.0 for 1 hour
volume: corresponds to 500ml for the 1 hour and 1000ml for the remaining of the test.


I followed the above formula and compared the result generated by using Method 2 way of calculating as follows:

SampleConcentration = (samplearea - intercept) / slope
percent dissolved = (sampleconcentration*vesselvolume*100)/LabelClaim


After evaluating the results, I find both of them was calculating same values but my colleague doesn't want to accept Method 1 way of calculation because it somehow violates the way of creating linearity plot and that is concentration (as the x) and peak area (as the y). Just wondering, if Method 1 is not correct, How come that every time I'm comparing the results generated from both Methods, they are just the same? Do I have to accept the fact that Method 1 is not right? Thanks a lot for responses.

Instead of blindly applying cookbook formulae, pause a moment to think about what you are actually doing.

"percent dissolved" is (mass in solution)/(total mass)
we'll assume that "total mass" in this case is what you're referring to as "LabelClaim"
(mass in solution) = (concentration)*(volume)
once you have your calibration curve, (concentration) = [(area)-(intercept)]/(slope)
So: your "Method 2" calculation makes sense -- although it doesn't account for the effect of withdrawal of previous samples.

Your "Method 1" calculation is meaningless. Even the units don't match up:
- "peakareaofsample" probably has units of "area" (more accurately, millivolt-seconds)
- slope has the units of area/(mg/mL) = area*mL/mg
- y-intercept has the units of "area"
- X is a percentage, so is dimensionless
- volume has the units of mL
In that first term, when you are multiplying area * slope, the dimensions end up being area^2*mg/mL
The second term is area
The third term is mL/mL and so is dimensionless
So, what you're doing is adding up three terms, each of which has different dimensions. That's like calculating my height by adding my weight to my birthdate.

Note that if you want to correct your Method 2 calculation for the withdrawn sample, you would have to add back in the mass in solution from the withdrawn aliquots: (concentration)*(6 mL) for each aliquot.

Instead of blindly applying cookbook formulae, pause a moment to think about what you are actually doing.

"percent dissolved" is (mass in solution)/(total mass)
we'll assume that "total mass" in this case is what you're referring to as "LabelClaim"
(mass in solution) = (concentration)*(volume)
once you have your calibration curve, (concentration) = [(area)-(intercept)]/(slope)
So: your "Method 2" calculation makes sense -- although it doesn't account for the effect of withdrawal of previous samples.

Your "Method 1" calculation is meaningless. Even the units don't match up:
- "peakareaofsample" probably has units of "area" (more accurately, millivolt-seconds)
- slope has the units of area/(mg/mL) = area*mL/mg
- y-intercept has the units of "area"
- X is a percentage, so is dimensionless
- volume has the units of mL
In that first term, when you are multiplying area * slope, the dimensions end up being area^2*mg/mL
The second term is area
The third term is mL/mL and so is dimensionless
So, what you're doing is adding up three terms, each of which has different dimensions. That's like calculating my height by adding my weight to my birthdate.

Note that if you want to correct your Method 2 calculation for the withdrawn sample, you would have to add back in the mass in solution from the withdrawn aliquots: (concentration)*(6 mL) for each aliquot.

Now I fully understand why you think method 2 is meaningless, because you still apply the fact that you plot the curve by making the concentration (as x-axis) and peak area (as y-axis). What I did on method number 2 actually was to plot peak area of the std (on x-axis) and percent concentration of the standard (on the y-axis) making its formula as follows:

%dissolved = (peakareaofsample)(slope) + y-intercept + (6ml * X/volume)


peakareaofsample = mv.sec
slope = percentstdconc/mv.sec
y-intercept = percentstdconc
X = percentdissolved from previous aliquot

by means of doing this, I was able to compute the same values generated from method number 1 itself.

I know that what is more appropriate is to do the curve by plotting concentration as your independent variable and peak area as the dependent variable (cause and effect relationship in a way). Method 2 on the other hand make it different, what causes your percent dissolved calculation is your peak area generated from your sample.

Question: Would method number 2 still considered as erroneous even by doing what I did on that method was generating the same results as method number 1? Thanks.

The only thing would be to add an additional 6X/vol term at each time point, where X was the concentration at the previous point.

As a statistical nit-pick, plotting area on the X-axis and concentration on the Y-axis is improper because all of the measurement uncertainty is imputed to the dependent variable. In practice, it doesn't matter, but most people *do* expect to see it the conventional way.

The only thing would be to add an additional 6X/vol term at each time point, where X was the concentration at the previous point.

As a statistical nit-pick, plotting area on the X-axis and concentration on the Y-axis is improper because all of the measurement uncertainty is imputed to the dependent variable. In practice, it doesn't matter, but most people *do* expect to see it the conventional way.

Thank you very much Mr. Tom Jupille. You are such a great help on my part