4.1: Calculations

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A medical technologist is performing the following assay for a drug whose molecular weight is 318.5 daltons. A 2 mL aliquot of serum is extracted into 5 mL of diethyl ether. It has been established that the extraction efficiency of the analyte into ether is 85%. Four mL of the ether layer is transferred to a test tube and evaporated to dryness at 37°C under a stream of nitrogen gas. The residue is taken up in 2 mL of a reagent that shifts the absorbance maxima of the analyte to 310 nm. After 10 minutes incubation at 37°C, the absorbance of the solution is measured at 310 nm using a spectrophotometer with a 1 cm pathlength. The recorded absorbance is 2.475. Since this absorbance exceeds the linearity of the instrument, the absorbance is measured on a 1:10 dilution of the sample. The absorbance of the diluted sample is recorded as 0.475. The molar extinction coefficient of the analyte at 310 nm is 18.7 x 10³ L•mol^-1•cm^-1.

QUESTION

What is the concentration of the drug in the original serum sample in mg/L.

Answer

The serum concentration of this drug is:________ mg/L.

Answer

First calculate the concentration of the analyte in the final solution using Beer’s law as follows (p. 88): A = $\epsilon$cb; where $\epsilon$ = 18.7 x 10³ L • mol^-1• cm^-1, c = unknown, b = 1 cm, and A = 0.475. $$\begin{split} 0.475 &= 18.7 \times 10^{3}\; L \cdot mol^{-1} \cdot cm^{-1}\; (1\; cm) \times c \\ c &= 0.0254 \times 10^{-3}\; mol/L \\ & \quad or \\ c &= 0.0254\; mmol/L \end{split}$$Since this absorbance was obtained on a 1:10 dilution, the actual concentration in the 2 mL sample is: $$c = 0.0254\; mmol/L \times 10 = 0.254\; mmol/L$$
Now calculate the amount of material extracted (p. 34-37). The concentration is 0.254 mmol/L in the 2 mL sample, and the total amount of analyte extracted is: $$\text{concentration} \times \text{sample volume} = \text{amount}$$
The micromoles of analyte extracted in part b were originally in a 4 mL aliquot of ether which represents 4/5 of the total volume into which the analyte was extracted. Thus, the actual total amount extracted from the 2 mL serum aliquot is: $$0.508\; \mu mol/ \; 4\; mL \times 5\; mL = 0.635\; \mu mol\; extracted\; from\; 2\; mL\; of\; serum$$
Since the extraction efficiency is only 85%, the total analyte extracted must be corrected for the 15% of the analyte that was not extracted. This is calculated as follows: $$\frac{\text{amount extracted}}{\text{efficiency of extraction}} = \text{actual amount present in sample} = \frac{0.635}{0.85} = 0.747\; \mu moles$$Thus, the analyte concentration in the 2 mL of serum is: $$\frac{0.747\; \mu moles\; extracted}{2\; mL\; of\; sample} = 0.37\; \mu mol/mL$$
To calculate (pp. 35-37) the analyte concentration in the serum sample in
mg/L use the following conversion formula: $$\frac{\mu mol}{mL} \times \frac{1000\; mL}{L} \times mol.\; wt.\; (\mu g/mol) \times \frac{1\; mg}{1000\; \mu g} = mg/L$$Thus $$\frac{0.37\; \mu mol}{mL} \times \frac{1000\; mL}{L} \times 318.5\; \mu g/ \mu mol \times \frac{1\; mg}{1000\; \mu g} = 118\; mg/L$$The serum concentration of the drug is 118 mg/L (0.37 mol/L).

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