Question

Quantitative analysis by selected ion monitoring. Caffeine in beverages and urine can be measured by adding caffeine- ${\mathit{D}}_{\mathbf{3}}$as an internal standard and using selected ion monitoring to measure each compound by gas chromatography. The figure shows mass chromatograms of caffeine (m/z,194)and caffeine-D${\mathbf{D}}_{\mathbf{3}}\left(m-z.197\right)$, which have nearly the same retention time.

Suppose that the following data were obtained for standard mixtures:

(a) Compute the mean response factor in the equation

role="math" localid="1664874599903" $\frac{\mathbf{Area}\mathbf{}\mathbf{of}\mathbf{}\mathbf{analyte}\mathbf{}\mathbf{signal}}{\mathbf{Area}\mathbf{}\mathbf{of}\mathbf{}\mathbf{standard}\mathbf{}\mathbf{single}\mathbf{}}\mathbf{=}\mathbf{F}\mathbf{\left(}\frac{\mathbf{concentration}\mathbf{}\mathbf{of}\mathbf{}\mathbf{analyte}\mathbf{}}{\mathbf{concentration}\mathbf{}\mathbf{of}\mathbf{}\mathbf{standard}}\mathbf{\right)}$

(b) For analysis of a cola beverage, 1.000mL of beverage was treated with $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{\mu L}$of standard solution containing 1.11 g/Lcaffeine-${\mathbf{D}}_{\mathbf{3}}$ in methanol. The combined solution was passed through a solid-phase extraction cartridge that retains caffeine. Polar solutes were washed off with water. Then the caffeine was washed off the cartridge with an organic solvent and the solvent was evaporated to dryness. The residue was dissolved in$\mathbf{50}\mathbf{\mu L}$ of methanol for gas chromatography. Peak areas were 1144 for m/z197 and 1733 for m/z194. Find the concentration of caffeine (mg/L ) in the beverage.

Short answer

(a) As a result, F=1.041 is the mean response factor for the given equation.

(b) The caffeine content of a particular beverage is 80.8mg/L.

Step-by-step solution

Definition of caffeine

Caffeine is a methylxanthine alkaloid that is chemically connected to the adenine and guanine bases of deoxyribonucleic acid (DNA) and ribonucleic acid (RNA), respectively (RNA).

Determine the mean response factor in the equationArea of analyte signalArea of standard signal=F(concentration of analyteconcentration of standard)

(a)

The mean response factor must be calculated from the following equation.

The following equation is provided.

$\frac{\mathrm{Area}\mathrm{of}\mathrm{analyte}\mathrm{signal}}{\mathrm{Area}\mathrm{of}\mathrm{standard}\mathrm{single}}=\mathrm{F}\left(\frac{\mathrm{concentration}\mathrm{of}\mathrm{analyte}}{\mathrm{concentration}\mathrm{of}\mathrm{standard}}\right)$

Now, let's find the response factor for the first set by inserting the given numbers into the above formula and doing some math, and we'll obtain the following result.

$\frac{11438}{2992}=F\left(\frac{13.60\times {10}^2}{3.70\times {10}^2}\right)=F=1.040$

The response factor for the next two sets of data may be found in the same way, and the results are as follows.

F=1.020 and

F=1.064

As a result, the response factor is F=1.041 and the mean value for these three is F=1.041.

Determine the concentration of caffeine (mg/L  ) in the beverage.

(b)

It is necessary to determine the caffeine content of a certain beverage.

To begin, we must determine the concentration of internal standard in the caffeine mixture D$D_3$as well as cola, as follows

$\left(1.11\mathrm{g}/\mathrm{L}\right)\times \frac{0.0500\mathrm{mL}}{1.050\mathrm{mL}}=52.86\mathrm{mg}/\mathrm{L}$

Let's look at how to calculate the caffeine concentration in the chromatography solution.

role="math" localid="1664878629758" $$\begin{gathered} \frac{\mathrm{Area}\mathrm{of}\mathrm{analyte}\mathrm{signal}}{\mathrm{Area}\mathrm{of}\mathrm{standard}\mathrm{single}}=\mathrm{F}\left(\frac{\mathrm{concentration}\mathrm{of}\mathrm{analyte}}{\mathrm{concentration}\mathrm{of}\mathrm{standard}}\right) \\ \frac{1733}{1144}=1.041\left(\frac{\left[\mathrm{caffeine}\right]}{52.86\mathrm{mg}/\mathrm{L}}\right) \\ \mathrm{Ϸ}\left[\mathrm{caffeine}\right]76.9\mathrm{mg}/\mathrm{L} \end{gathered}$$

The unknown beverage had been diluted from 1.000 to 1.000 when the standard was applied 1.50 mL. As a result, the caffeine concentration in the original beverage can be computed as follows

$\left(\frac{1.0500\mathrm{mL}}{1.00\mathrm{mL}}\right)\left(76.9\mathrm{mg}/\mathrm{L}\right)=80.8\mathrm{mg}/\mathrm{L}$

As a result, the caffeine content of a given beverage is$80.8\mathrm{mg}/\mathrm{L}$