Two 0.9mL aliquots of the same specimen are placed in test tubes. To one is added 0.1 mL of water (spec A). To the other is added 0.1 mL of 200 mg/dl urea (spec B). They are analyzed for their results: Spec A: 25 mg/dl, Spec B: 43 mg/dl. What is the percent recovery of this method?

Question

Alonzo Boyle

Chemistry

24 October, 14:39

Two 0.9mL aliquots of the same specimen are placed in test tubes. To one is added 0.1 mL of water (spec A). To the other is added 0.1 mL of 200 mg/dl urea (spec B). They are analyzed for their results: Spec A: 25 mg/dl, Spec B: 43 mg/dl. What is the percent recovery of this method?

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Ahmed Malone · Answer 1

We assume that the method that made use of urea was able to recover all of the recoverable substance. The method in question is the method that makes use of water.

The total amount of substance is 43 mg/dl. The recovered amount is 25 mg/dl. The percent recovery is

(25 mg/dl / 43 mg/dl) * 100 = 58.14%