Question

Caffeine (C8H10O2N4 • H2O, FW = 212.1) has an absorbance of A = 0.255 for a concentration of 0.500 mg per 100 ml at 272 nm. A sample of 6.50 g coffee was mixed with 400 ml of water, boiled and then diluted to a total volume of 500 ml of water. A 25 ml aliquot was transferred to a flask containing 0.1 N H2SO4, the solids removed and the final solution made up to 500 ml. This solution showed an absorbance of 0.330 at 272 nm. (All path lengths are 1 cm.)

Calculate the extinction coefficient for caffeine at 272 nm. Calculate the number of grams of caffeine per pound of coffee.

Answer #1

As per Beer's Law

A = ECl where E = extinction coefficient

C = concentration and l = pathlength which is 1 cm

0.5 mg/100 mL is 5 mg/L 5 x 10^{-3} g/212 = 2.36 x
10^{-5} mole/L

A = 0.255 = E * 2.36 x 10^{-5} * 1

E = 10812 L mol^{-1} cm^{-1} extinction
coefficient

A = ECl

0.33 = 10812 * C * 1

C = 3.05 x 10^{-5} mol/L = 0.00647 g/L in the final 500
mL

So in the 25 mL it was

N1V1=N2V2

0.00647 * 500 = N2 x 25

N2 = 0.129 g/L

So in 500 mL it was 0.129 * 500/1000 = 0.0647 g

SO 0.0647 g of caffeine in 6.5 g of coffee

So in 1 pound or 453.6g we will have 0.0647 * 453.6/6.5 =
**4.51 g of caffeine**

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