Organic chemistry lab: Extraction and IR 1. Oops you forgot to label your layers

Question

Organic chemistry lab: Extraction and IR

1. Oops you forgot to label your layers after extraction. Describe a fast method for determining which layer is your organic phase.

2. You weigh out 100mg of benzoic acid and dissolve it in 3mL of diethyl ether. This is then extracted with 3mL of water. The ether layer is removed, dried with sodium sulfate and the ether evaporated yielding 30mg of benzoic acid. What is the K(p) value (partition coefficient) for the ether/water system?

3. Tell whether the following solvents would be the top or bottom layer in an aqueous extraction

a) Methanol

b) hexane

c) benzene

d) dichloromethane

e) diethyl ether

4. When performing extractions it is often better to do multiple extractions rather than one large volume extraction. Suppose you have 200mg of a sample dissolved in 3mL of water. The partition coefficient for this compound in water and diethyl ether is 4.0

a) Calculate how much total compound you would obtain with 3 x 1.0mL extractions with ether. Show your work.

b) Calculate how much you would obtain with 1 x 3mL extraction. Show your work.

c) Which method is more efficent?

Explanation / Answer

1. A fast method would be dissolving a sample of the layer in water. If it dissolves, the OTHER layer is the organic phase.

2. log Kp ether/water = log (30/3 / 70/3) = -0.3680; Kp = 0.4286

3 a. Neither. Methanol mixes with water.

b. Hexane would be top layer.

c. Benzene would be top layer.

d. Dichloromethane would be bottom layer.

e. Diethylether would be top layer.

4.a For the first 1.0ml extraction (we use grams and mL)

K = 4 = [(0.2 - x)/1]/[(x)/(1)], ---> x = 0.04 g of the sample remains in the water, 0.16 g in the ether

For the second 1.0ml extraction,

K = 4 = [(0.04 - x)/1]/[x/1] ----> x = 0.008 g of the sample remains in the water, 0.04 - 0.008 = 0.032 g remains in the ether.

For the third 1.0ml extraction,

K = 4 = [(0.008 - x)/1]/[x/1] ----->x = 0.0016 g in water and 0.008 - 0.0016 = 0.0064 g in ether. At the end of the three extractions we will have 0.16 + 0.032 + 0.0064 = 0.1984 g of the sample in the ether and 0.2000 - 0.1984 = 0.0016 g in the water.

b. For the 1 x 3 ml extraction:

K = 4 = [(0.2 - x)/3]/[x/3] -----> x = 0.04 g remains in water, and 0.2 - 0.04 = 0.16 g extracted in the ether.

c. Clearly the 3 x 1ml extraction method is more efficient.

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