If you want to know the amount cleared at any one time, then rearrange the clearance equation so that the left-hand side of the equation now equals the amount.
(CA - CV) × Q = Clearance × CA
As an example, determine the amount cleared over one minute when you have system with a flow of 2 L/min and the concentration going into the clearing organ is 5 mg/L and the concentration coming out is 3 mg/L. The extraction ratio would be:
((5 mg/L - 3 mg/L) / 5 mg/L) = 0.4.
The clearance is then 0.4 × 2 L/min = 0.8 L/min
So the amount would be:
(5 mg/L - 3 mg/L) x 2 L/min or 0.8 L/min x 5 mg/L = 4 mg/min
If one adds all the cleared amounts over the full time the drug is present in the body, this must be exactly equal to the available dose.
If you want to know the concentration at any one time, begin with the same equation (CA - CV) × Q = Clearance × CA
From this it follows that the amount of drug that is removed from the body (note that this is the absolute amount) continuously diminishes. If we divide the amount by the volume of distribution we can calculate the concentration at any one time point.
((CA - CV) × Q) / Vd
If one adds up all the concentrations that occur over the full time the drug is present in the body, this is exactly equal to the Area Under the Curve (AUC).
A patient (75kg) with gastric cancer is treated with an intravenous infusion of 5-fluorouracil (5-FU). The infusion rate is set to 30 mg/h, and the steady-state concentration is aimed at 0.8 mg/L. What is (in mL/min) the expected clearance of 5-FU in this patient?
Extra info: CL = infusion rate / Css = 30 mg/h / 0.8 mg/L = 37.5 L/h = 625 mL/min