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- Posts: 95
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%Dissolved = (AreaofsampleSolution)(slope)+ y-intercept
Basic questions from students; resources for projects and reports.
tom jupille wrote:
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Note that if you want to correct your Method 2 calculation for the withdrawn sample, you would have to add back in the mass in solution from the withdrawn aliquots: (concentration)*(6 mL) for each aliquot.
tom jupille wrote:
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.
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