Lab Report: Separation of a Mixture by Extraction. Crystallization. (Exp. 1) 40 pts/max
NAME: Shawn Vuong SECTION: 145
Unknown Numbers: 1 & 4
Mass of Unknown: 3.425 g
Mass of Neutral Compound Recovered: 2.43 g
Percent Recovery of Neutral Compound: 71%
List and Assign Characteristic IR-bands of Neutral Compound: 2900 – Alkane bond (Sp3) / 2800 –
O=C—H / 1450 –CH
2
- bending (5 pts)
Identity of Neutral Compound: Cyclohexanone (5 pts)
Mass of Acidic Compound Recovered: .048 g
Percent Recovery of Acidic Compound: 1.4%
Melting Point of Acidic Compound: 165

C (5 pts)
Identity of Acidic Compound: p-tert-butyl benzoic acid (5 pts)
1. List five desirable properties for solvents to be used in extraction procedures. (5 pts)
1.
it must be immiscible with the solvent from which the desired compound is to be extracted
2.
it should readily dissolve the compound to be extracted
3.
it should dissolve little or none of the other compounds (impurities) present in the mixture
4.
it should be easily separated from the desired compound at the end of the extraction
5.
it should not undergo unwanted reactions with the desired compound

2. Using the data in appendix 3, chose which of the following solvents should be the best choice to extract a polar organic compound from an aqueous solution? Briefly explain why each of other solvents is less desirable than the solvent you have chosen.
Acetone Cyclohexane Dichloromethane Methanol Toluene (5 pts)
Dichloromethane. Acetone is miscible in water so it will not work for an extraction. Cyclohexane has no polarity, and will not dissolve the solvent that it is extracting. Methanol is miscible in water so it will not work for an extraction from water. Toluene has too small of a dipole movement and will therefore not dissolve the solvent that it is extracting.
3. The distribution constant for suberic acid between ether and water is 4 (K = C ether
/C water
= 4). A solution contains 0.09 g of suberic acid in 50 mL of water. (a) Calculate the mass of suberic acid that can be recovered by extraction with one 50 mL portion of ether. (b) Calculate the mass of suberic acid that can be recovered by two successive extractions with 25 mL portions of ether. (5 pts) a) 4 = (x g/50 mL)/((.09-x) g/50 mL)

x / (.09 – x) = 4

x = 4(.09 – x)

x = .36 – 4x

5x = .36  x = .36/5  x = .072 g
.072/.09 = .8

80% is extracted. b) 4 = (x g/25 mL)/((.09-x) g/50 mL)

2x/(.09 – x) = 4

2x = 4(.09 – x)

2x = .36 – 4x

6x = .36

x = .36/6

x = .06 g
.06/.09 = .6666667  66.7% is extracted at first this leaves .03 g left
4 = (x g/25 mL)/((.03-x) g/50 mL)

2x/(.03 – x) = 4

2x = 4(.03 – x)

2x = .12 – 4x

6x = .12

x = .12/6

x = .02 g
.02/.09 = .2

22.2% is extracted the second time
Overall .06 + .02 = .08 g was extracted during the two extractions of 25 mL and therefore
.08/.09 = .88889

88.9% was extracted during part (b)

4. You are given a sample containing two organic compounds; a neutral compound (RH), and an acidic compound (RCOOH). All of these compounds are very soluble in ether and insoluble in water.
Using a flow chart , describe in detail how you could separate the components of this mixture by a series of extractions. (5 pts)
First the mixture of the two compounds is dissolved in ether.
The ether solution is then extracted with a dilute aqueous solution of sodium hydroxide (5-10%
NaOH)
The acidic compound will react with the NaOH and be converted into a sodium salt.
The sodium salt is an ionic compound that can be soluble in water and quite insoluble in ether
So the salt will be extracted into the aqueous phase
To recover the acid at the end of the extraction add a strong acid to the aqueous phase
The neutral compound will not react with base and remains in the ether phase.
The remaining NaOH will be neutralized, and then the added acid will react with the sodium salt and form the original acid
The original acid will precipitate because it is insoluble in water