Comparing Three 20 mL Extractions

# Comparing Three 20 mL Extractions - If we do the same...

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Comparing Three 20 mL Extractions to a Single 60 mL Extraction You were given 3 Vivarin tablets that contain 200 mg of caffeine each. Therefore you have 0.600 grams of caffeine that you dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted with CH 2 Cl 2 is known to be 8.4 so, You have the following equilibrium, Caffeine (Aq) ↔ Caffeine (MeCl) The equilibrium expression would be, Caffeine (MeCl) Caffeine (Aq) = Keq = 8.4 Initially, there is no caffeine in the MeCl, it is all in the water, so some of the caffeine must move from the water to the MeCl. Given that we have 0.600 grams of caffeine originally present, how much of it would move into the MeCl in a single 60 mL extraction? x/60 mL (0.600 x)/100 mL = 8.4 x = 0.500 grams of caffeine in one extraction
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Unformatted text preview: If we do the same extraction using three 20 mL portions of CH 2 Cl 2 we would have, x/20 mL (0.600 – x)/100 mL = 8.4 x = 0.3761 grams of caffeine in first extraction, 0.2239 grams left x/20 mL (0.2239 – x)/100 mL = 8.4 x = 0.1403 grams of caffeine in second extraction, 0.08356 grams left x/20 mL (0.08356 – x)/100 mL = 8.4 x = 0.0430 grams of caffeine in third extraction, 0.04053 grams left. So, after three 20 mL CH 2 Cl 2 extractions 0.3761 g + 0.1403 g + 0.0403 g = 0.5567 grams of caffeine would have been extracted as compared to 0.500 grams of caffeine extracted using a single 60 mL sample of CH 2 Cl 2 . Therefore, about 11% more caffeine is extracted using three 20 mL portions of CH 2 Cl 2 than a single 60 mL sample....
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