Transcript
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Due Date: Oct 23rd, 2012
CHY224 LAB #4 Report Questions
Separation of Organic Compound by
Liquid-Liquid Extraction
1. What was your actual experimental percent recovery for 3-nitroaniline and benzoic acid? Briefly explain why these values differ from 100%.
For the 3-nitroaniline, the actual experimental present recovery that obtained was 105% which is above the 100% by 5%. That’s because the mass recovery 3-nitroaniline that extracted was not dry completely when it’s transferred from the Vacuum Filtration (Büchner funnel) for the vacuum to glass watch so the mass of the recovery that was weighted and measured was combination of the mass of the compound plus water.
For benzoic acid; however, the actual experimental percent recovery that calculated was 90% which is less than 100% (difference of 10%) that indicates that some of the mass of the compounds were lost while transferring from the vacuum filtration device to watch glass in order to calculate the mass.
Calculation showing how the Actual experimental percent recovery was computed:
The actual mass recovery % = (mass of recovered compound / the mass of compound started)*100%
The actual mass recovery % for 3-nitroaniline = (2.1/2.0)*100%
= 105%
The actual mass recovery % for benzoic acid = (1.8/2.0)*100%
=90%
2. Discuss the purity of each of the recovered compounds (are they pure? how do you know?)
Physical properties of a chemical compounds are one of the best techniques that used to determine if the compound is pure or impure. Otherwise, pure compounds can be identifying by their physical properties such us boiling point, solubility, density and melting point. However, melting point is one of the important methods that used in this organic lab in order to determine the purity of a chemical compounds. Thus, in this experiments the melting point of the two organic compounds (3-nitroaniline is 114° C and benzoic acid is 122.41°C) were obtained from the CRC Handbook of Chemistry and Physics had a sharp melting point range. After, measuring experimentally the melting point of 3-nitroaniline and benzoic acid using Fisher-Johns Melting Point Apparatus, the melting range were ( 110° C -118° C) and (119°C -123°C ) respectively which indicate that the compounds are pure because pure compounds have sharp melting range which means melt with a range of 2 or less.
3. Use information available in the Merck Index, and show your calculations to answer the
following questions:
a) What is the partition coefficient for 3-nitroaniline between diethyl ether and water (at
room temperature and pH > 8)?
Solubility of 3- nitroaniline from the Merck index is:
In water 1g/880mL
In ether 1g/18mL
P= (1g/18mL)/ (1g/880mL) =48.89
b) What is the partition coefficient for benzoic acid between diethyl ether and water (at
20 °C and pH < 2.8)?
Solubility of benzoic acid from the Merck index is:
In water at 20C=2.9g/L
In ether=1g/3mL
P= (1g/3mL) / (2.9g/103mL) =114.94
4. Suppose that you have a reaction mixture which consists of 14.2 g of CsCl and 12.0 g of an interesting organic compound dissolved in 100 mL of water. You would like to have the pure organic compound, separate from water and salt, so you decide to carry out an extraction. You add 100 mL of chloroform and vigourously shake in a separatory funnel.
a) After settling, both layers are clear colourless liquids. How do you know which
layer is the organic layer?
In order to determine which layer is the organic layer or aqueous is to drip a drop of water into the mixture in the separator funnel. If the water drop dissolve as it hit the upper layer of a mixture then the bottom layer is organic. Also, if the upper layer is organic and the bottom is aqueous then the drop will travel through the upper liquid then dissolve when it reaches the bottom layer. Otherwise, if the upper layer is organic then the water drop will not dissolve and will sink to the bottom layer. However, you can determine if the solution is (organic or aqueous) is by looking at the density of the water and the organic compound that mixed in the solution. If the density of the organic compound is less so it will be in the upper layer and water is lower layer and vice versa because compound with less density will float.
b) After separating the organic from the aqueous layer, you evaporate all of the
chloroform and find that you have recovered 7.0 g of the interesting organic compound. What is this compound’s partition coefficient with chloroform?
P= [compound] organic / [compound] aqueous
P= [X] organic / [X] aqueous
= (7.0g/100mL)/ [(12.0-7.0g)/100mL]
=1.4
c) If you then add the remaining aqueous solution back into the separatory funnel, along
with 100 mL of fresh chloroform, and perform another extraction, separation and evaporation, how much more of the interesting compound would you expect to recover?
Assume m = the mass of X recovered from chloroform.
So the (5 g – m) is equal to the mass of X extracted into the aqueous layer.
P= [X] organic / [X] aqueous
1.4 = (m/100mL)/ [(5g-m)/100mL]
1.4(5g-m)/100=m/100 100m=100(1.4) (5g-m)
2.4m=7 m=2.9167g
m recovery total= 2.9167g+7.0g =9.9167g
d) Suppose that instead of two successive extractions with 100 mL of chlorofrom each
time, you had used all 200 mL of chloroform in a single extraction. How much of the
interesting compound would you expect to have recovered?
P= [X] organic / [X] aqueous
1.4= (m/200mL)/ [(12.0g-m)/100mL]
1.4[(12.0g-m)/100mL] = (m/200mL)
(16.8-1.4m)/100mL= (m/200mL)
3360-280m=100m m=8.8421g
e) Suppose that instead of one extraction with 200 mL, or two extractions with 100 mL,
you had performed three successive extractions with 67 mL of fresh chloroform each time.
How much of the interesting compound would you expect to have recovered?
P= [compound] organic / [compound] aqueous
Calculation for the first extraction:
P= [X] organic / [X] aqueous
1.4= (m/67mL) / [(12.0g-m)/100mL]
1.4[(12.0g-m)/100mL] = (m/67mL)
[(16.8-1.4m)/100mL]= (m/67mL)
100m=1125.6-93.8m 193.8m=1125.6
m= 5.8080g the mass of the first extraction
Calculation for the second extraction:
m=mass of the organic compounds- recovered mass (first extraction)
= (12.0g)-(5.8080g)
=6.1920g
P= [X] organic / [X] aqueous
1.4= (m/67mL) / [(6.1920g-m)/100mL]
1.4 [(6.1920g-m)/100mL]=(m/67mL)
(8.6688-1.4m)/100mL= m/67mL
100m=580.81-93.8m 193.8m=580.81
m=2.9970g
Calculation for the Third extraction:
m=mass of the organic compounds- recovered mass (second extraction)
= (6.192g)-(2.9970g)
=3.1950g
P= [X] organic / [X] aqueous
1.4= (m/67mL)/ [(3.1950-m)/100mL]
1.4[(3.1950-m)/100mL] = (m/67mL)
(4.473-1.4m)/100mL=m/67mL
100m=299.6910-93.8m
193.8m=299.6910 m=1.5464g
m total recovery=1.5464g+2.9970g+5.8080g
= 10.3514g
f) Summarize the results of your calculations above, and make a clear, concise
generalization about them.
The result of the calculation above concludes that the more accurate and proficient approach is that the repeating of the extraction process multiple times using smaller volume of chloroform. In this lab, for example, when the volume of the chloroform was changed from 100mL (used in two extraction ) to 67mL (preformed in three consecutive extraction process), the mass was slightly increased from 9.9167g to 10.3514g which indicate that the more extraction, the more efficient result and the more pure the aqueous solution can get. In the other hand, when the extraction process for an organic compound was computed using 200mL of chloroform the recovery mass was 8.8421g less than the extraction of 100mL and 67mL which in return proofed that the more efficient value of the extracted mass recovery can be obtained when the extraction process recurrent multiple times.
References:
CHY 224 Laboratory Manual Appendices, Appendix B1.
The online 14th edition of the Merck index, 2006-2012
http://themerckindex.chemfinder.com/themerckindex/Forms/Home/ContentArea/Home.aspx
The Online 92st edition of the CRC Handbook of Chemistry and Physics, 2011-2012 http://www.hbcpnetbase.com.ezproxy.lib.ryerson.ca/
