CM1501

Organic Chemistry for Engineers

Lab Report

Experiment 2: Separation and Identification of Two Organic Compounds

10th October 2013

Abstract

This experiment involves the separation and identification of 2 organic compounds (1 Neutral Compound and 1 Acidic Compound) in a mixture. Separation techniques used in this experiment include solvent extraction, simple distillation and recrystallization using a suitable solvent determined through a solubility test. The identification method used was through Melting Point Determination. By obtaining data of the melting points of the 2 purified compounds and cross-referencing from a list of possible organic compounds, the 2 organic compounds were

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---------- Equation 1 ---------- Equation 2

Using Equation 2 and subbing in known values, we can obtain ---------- Equation 3

Solving, we get [A]w = 1.11 X 10-2 g/mL

b) Using Equation 2,

For 1st Extraction, ---------- Equation 4 Solving, we get [A]w = 2.00 X 10-2 g/mL For 2nd Extraction, ---------- Equation 5 Solving, we get [A]w = 4.00 X 10-3 g/mL

c) Repeating the above method as shown in (b) for multiple extractions, we can obtain

For 4th Extraction using 20 mL of dichloromethane,

[A]w = 1.23 x 10-3 g/mL

For 8th Extraction using 10 mL of dichloromethane,

[A]w = 3.91 X 10-4 g/mL

d) Based on calculation above, while the total amount of dichloromethane used in (a), (b) and (c) remained constant at 80 mL, the values of [A]w obtained decreases with increasing number of extractions done with smaller amounts of dichloromethane used during each extraction.

Multi-extraction allows more of the solute to dissolve in the solvent. This leads to a higher amount of solute extracted.

e) To access if excessive extraction (8 times) is necessary, the yield difference of solute extracted between doing 4 and 8 times extraction must be considered.

Mass of

Using Equation 2 and subbing in known values, we can obtain ---------- Equation 3

Solving, we get [A]w = 1.11 X 10-2 g/mL

b) Using Equation 2,

For 1st Extraction, ---------- Equation 4 Solving, we get [A]w = 2.00 X 10-2 g/mL For 2nd Extraction, ---------- Equation 5 Solving, we get [A]w = 4.00 X 10-3 g/mL

c) Repeating the above method as shown in (b) for multiple extractions, we can obtain

For 4th Extraction using 20 mL of dichloromethane,

[A]w = 1.23 x 10-3 g/mL

For 8th Extraction using 10 mL of dichloromethane,

[A]w = 3.91 X 10-4 g/mL

d) Based on calculation above, while the total amount of dichloromethane used in (a), (b) and (c) remained constant at 80 mL, the values of [A]w obtained decreases with increasing number of extractions done with smaller amounts of dichloromethane used during each extraction.

Multi-extraction allows more of the solute to dissolve in the solvent. This leads to a higher amount of solute extracted.

e) To access if excessive extraction (8 times) is necessary, the yield difference of solute extracted between doing 4 and 8 times extraction must be considered.

Mass of