Section Name CHEM 108 Determination of Caffeine in Commercial Tablets Prelaborat

Question

Section Name CHEM 108 Determination of Caffeine in Commercial Tablets Prelaboratory Assignment Please read the lab handout. Please refer to the graphing with excel module for instructions on graphing. Graph a calibration curve for caffeine using the following data in Excel. Fit the data with a straight line. Display the equation of the line and R2 value on the graph. Print out the 1. graph and submit it as a part of your prelab assignment. Absorbance at 273 nm Caffeine Concentration (ppm) 10.0 15.0 20.0 25.0 30.0 0.8513 1.2837 1.7162 2.1486 25811 Based on the data below and the graph that you prepared in Question 2, determine the number of milligrams of caffeine in a tablet labeled as having 200 mg of caffeine. 2. A caffeine tablet and approximately 150 mL of distilled water were added to a 250 mL beaker. Next a stirring bar was carefully added and the beaker was sealed to prevent evaporation. The beaker and contents were then stirred vigorously for 30 minutes to extract the caffeine out of the tablet and into solution. This cloudy "extract" was allowed to settle until the top layer was clear. The solution was then filtered using gravity filtration into a 200.00-mL volumetric flask, and the resulting solution was diluted to the mark by the addition of distilled water as described in the experimental procedure. This is referred to as the "extract". A 2.00-mL sample of the above extract was added to a 100.00-mL volumetric flask and enough water was added dilute the solution to the mark. The absorbance of this diluted extract at 273 nm in a 1.000 cm cell was found to be 1.5343. Determine the number of milligrams of caffeine in the tablet.

Explanation / Answer

1) when a graph is ploted with the above data it is obtained

Y = 0.08629 *C - 0.01362

2) the calibration data is used for the calculation of sample. absorbance of sample at 273 nm is 1.5343

A = 0.08629 *C - 0.01362

1.5343 = 0.08629 *C - 0.01362

1.5343 + 0.01362 = 0.08629 * C

1.54792 = 0.08629 * C

C = 17.93 ppm

17 ppm means 17 mg in 1000 ml

(x / 250 * 1000) * 2 /100 = 17.9

x = 223.75

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