1.) What percent recovery could be expected if 1.000 g of caffeine was initially

Question

1.) What percent recovery could be expected if 1.000 g of caffeine was initially dissolved in 120 mL of water and then extracted with a single 80 mL portion of ethyl acetate?

2.) What percent recovery could be expected if 1.000 g of caffeine was initially dissolved in 120 mL of water and then extracted as per the protocol given in the lab manual?

3.) Percent recovery and percent yield are occasionally used as if they are interchangeable terms. They do, however, have different meanings. Why is it more correct to use percent recovery here?

Use hot water to extract the caffeine from the tea leaves. The solubility of caffeine is 22 mg/mL (25 oC) and 670 mg/mL (100 oC). While cellulose is insoluble in the water, the tannins and chlorophyll will extract along with the caffeine into the water. Use an organic solvent to extract the caffeine and related compounds from the water. The tendency of a solute (in this case caffeine) to dissolve preferentially in one solvent over another can be represented in a partition coefficient. A partition coefficient is the ratio of solubilities of a compound between two different solvents, or P (Solvent 1/Solvent 2) [Solute in Solvent 1]/ISolute in Solvent 2]. The partition coefficient between ethyl acetate and water is approximately 2, since caffeine is (46 mg/mL)/22 mg/mL) or 2.1 times more soluble in ethyl acetate than water while the partition coefficient for dichloromethane is about 8 (the solubility of caffeine in dichloromethane is 182 mg/ml). If you wanted to calculate how many grams of caffeine were extracted into a Solvent, you Would solve for X in the following: Partition Coefficient- (X g caff./vol solvent)/((total g caff. X g caff.)/vol H20)

Explanation / Answer

1) The partiton coefficient for water - ethyl acetate system is given as 2.

Assuming 'x' g of caffeine gets extracted into ethyl acetate. So mass left in water = 1 - x grams

Using the relation given above :

Partition coefficient = 2 = (x/80) / ( (1-x)/120)

Solving we get :

x = 4/7 = 0.571 g

Thus, % recovery = x/Total mass * 100 = 0.571/1 * 100 = 57.1%

(2) As you have not provided the experimental procedure, I cannot provide an answer to this.

(3) Percent yield is more suitable in those cases where a product is actually produced by a reaction taking place under a specifed set of conditions. In this the reactants react and undergo a chemical change to yield the product.

While percent recovery is more suitable term for this extraction because here no product is formed rather the compound which has already been produced is shifting from one phase to the other, so recovery is a more proper term for this.

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