Part A
2.014 g of Panacetin was measured and put inside a 125 mL Erlenmeyer flask. 35 mL of Dichloromethane (DCM) was added to the 125 mL Erlenmeyer flask. After addition of DCM the Panacetin lumps were crushed with a stirring rod. Next a fluted filter paper was pre-weighed. The filter paper weighed around .860 g. Gravity filtration was then used to filter the mixture into a 125 mL collecting flask. The mixture was filtered in order to separate the crude sucrose from the mixture. The original container was then rinsed through the filter paper again with 5 mL of DCM and the mass of filter paper containing sucrose was measured after it was dry. The last step was to perform the isolation of Aspirin. See image below.
Part B
The mixture of DCM and Panacetin were poured from the collecting flask into a seperatory funnel and two extractions were performed, using a total of 30 mL of 5 % NaHCO3. In the first extraction 20 mL of 5 % NaHCO3 was measured and poured into the seperatory funnel. A stirring rod was then used to stir the mixture until fizzing stopped. The seperatory funnel was then stoppered and shaken and the valve was gradually turned to allow the carbon dioxide to escape. The release of carbon dioxide made a hissing sound. This step was repeated once; the seperatory funnel was then placed into the lab hood and through a metal ring holder. The seperatory funnel was used to separate the organic layer from the aqueous layer which were found to be immiscble. Finally, the stopper was removed and the organic and aqueous layers were extracted into two different beakers. See image below.
The Essay on Melting Point Flask Caffeine Layer
The purpose of this laboratory assignment was two-fold, first, we were to demonstrate the extraction of acids and bases, finally, determining what unknowns were present. Second, we were to extract caffeine from tea. These two assignment will be documented in two separate entities. Introduction: Acid / base extraction involves carrying out simple acid / base reactions in order to separate strong ...
Part C
The remaining 10 mL of 5 % NaHCO3 was added to the seperatory funnel in the second extraction. The mixture was stirred until the fizzing stopped and the seperatory funnel was stoppered and shaken, then the carbon dioxide was gradually released from the seperatory funnel by turning the valve. Two layers formed, the top layer was the unknown and it was denser than the aqueous layer that was underneath it. The organic layer which contained the unknown was poured into the same beaker as the organic layer from the first extraction and the aqueous layer which contained DCM was poured into the beaker containing the aqueous solution from the first extraction. The reason the extraction was performed was to separate the aspirin and the unknown.
Part D
3.5 mL of 6 M HCl was measured and 3.5 mL of HCl was added drop wise to the flask to make the pH of the aqueous solution less than 2, meaning the pH paper appeared as a pinkish color. During addition of the HCl, the aqueous solution formed a bubbly and cloudy white layer on top. Second, the mixture was placed into a cold ice bath for 10 minutes and the small filter paper was measured before using it for vacuum filtration. The purpose of using vacuum filtration was to collect the aspirin. The aspirator was left running until the aspirin was dried and the filter paper containing aspirin was weighed as 0.83 g. Next was the isolation of the unknown analgesic.
Part F
In part F Sodium sulfate was added to the organic mixture. The sodium sulfate was collected into a 100 mL round bottom flask by gravity filtration and the original beaker was then rinsed with 5 mL of fresh DCM. At the end the DCM mixture was placed to the rotary evaporator and the solution was concentrated to dryness, once the gauge reached 25 mm Hg which was .84 g of the unknown.
The Essay on Molecular Weight of Volatile Liquid Using Dumas Method
We have to stir the water in beaker to get the perfect temperature. * The rubber band which is used to tight the flask may release more vapour because the rubber band is not properly tight. The Dumas method is the method used to calculate the molecular weight of the substance. * This method has a condition that the substance used must be a liquid. The liquid that has boiling point above room ...
Discussion:
Obtained data:
Weight of Panacetin
2.014 g
Weight of dried sucrose
.3 g
Weight of Aspirin
.83 g
Weight of Unknown
.84 g
The initial amount of panacetin used was 2.014 g. The final product of one component, aspirin was .83 g after subtracting the weight of the filter paper. The weight of unknown component was .84 g after subtracting the weight of the 100 mL round bottom flask. After addding 35 mL of panacetin to DCM , the resulting solution was gravity filtrated because DCM is nonpolar and the component had many oyxgens on the strucure which made it polar. This difference in polarity does not allow sucrose to dissolve in DCM which is why the solution was separated by gravity filtration. 20 mL of The DCM was added to extract the product from the aqueous layer and when the DCM was mixed with Panacetin it dissolved the aspirin and the unknown component leaving a solid behind. During the extraction the organic layer and aqueous layer were isolated into two different beakers.The organic layer, containg DCM extracted the product, while the salts and impurities remained in the aqueous layer.
After allowing the aqueous mixture to cool the aspirin was collected by means of vacuum filtration and the weight was recorded, which was previousy mentioned. The aqueous layer after that was separated now contained aspirin and NaHCO3. To extract the aspirin from the basic solution an acid had to be added. The addition of HCl, a strong acid, to the aqueous solution made the solution more acidic. This allowed all of the aspirin to be removed during gravity filtration. After, the drying agent (Sodium Sulfate) was added because it has high capacity and it was a smaller amount which is reactive and in granular form, and is also very extract from liquids. The mixture was put in a round bottom flask and placed into the rotary evaporator. The rotary evaporator was used to evaporate the solvent dichloromethane and any impurities from the unknown analesgic. To find the weight of the unknown the weight of the empty round bottom flask was subtracted from the weight of the round bottom flask containg the unknown. The uknown inside the flask appeared small, white, round, and solid due to the evaporation of the liquids, more details shown in the followng:
The Essay on Water 5
Thesis Statement: Water is the most crucial part of life its self, and must never go unnoticed. I. People frequently overlook the importance of water in the body. In order to keep the body healthy, people must consume water. A healthy body is a well-hydrated body. Without the constant consumption of water the body becomes dehydrated. Perhaps people overlook waters importance, simply because its ...
•Percent Composition of Panacetin based on weight of Sucrose: (.3/2.014)*100 = 14.90 %
Percent Composition of Panacetin based on weight of Aspirin: (.83/2.014)*100 = 41.21 %
Percent Composition of Panacetin based on weight of Unknown: (.84/2.014)*100 = 41.71 %
B) % Recovery = [wt. of aspirin obtained/Starting wt. of Panacetin] x 100. % Recovery of Panacetin = [1.97 g/2.014 g] x 100 = 97.82 %. One possible reason why the % recovery is lower than 100 % is that some of the Panacetin could have been lost during the processes of filtration and purification. C) Once the two layers were in the seperatory funnel and it is stoppered and the valve is tightly sealed the seperatory funnel can be shaken and the air pressure should gradually be released by turning the valve. Improper operation of the seperatory funnel can lead to loss of product. If the valve of the seperatory funnel is not properly sealed when initially pouring in the mixture there is a chance that some of the mixture will be lost. In addition if the valve is turned too fast when performing the extraction the products can remix and this will make it difficult to separate the organic and the aqueous mixtures into two different containers. Conclusion:
In this experiment the method of liquid-liquid extraction was utilized. In order to obtain a pure compound from Panacetin, the desired compound had to be separated from the other components to by taking advantage of differences in physical and chemical properties. In this instance extraction was used due to the very different solubility’s of the aqueous and organic layer. Post-Lab Questions
(a) Isopropanol is completely miscible in water; as a result it would be impossible to perform a liquid-liquid extraction with isopropanol and water since there would not be separate layers to extract. (b) Although kerosene and water are immiscible, together these solvents would be a poor choice to use for liquid-liquid extraction. These two would be a poor choice due to the very polar nature of water and the very nonpolar nature of kerosene. If kerosene were added to extract a particular constituent in water liquid-liquid extraction could not be achieved. Water and Kerosene are opposite extremes of one another. In order for two liquids to be suitable for carrying out a liquid-liquid extraction, both liquids must not be too polar or too nonpolar, but must be in the middle.
The Term Paper on Extraction: Extraction With Acid And Alkali
Objective 1. To recover the benzoic acid and p-dichlorobenzene from its mixture from its mixture by using acid-alkali extraction. 2. To determine the percentage recovery and melting point of the recovered benzoic acid and p-dichlorobenzene. Introduction Acid-base extraction is a process which purifying the acids and bases from mixtures based on their chemical properties. Acid-base extraction is ...
•(a)
(aq) + NaHCO3(aq) (aq) + CO2(g)+H2O(l)
Weak acid weak base weak acid (b)
(aq) + HCl(aq) (s) + NaCl(s)
Weak acid Strong acid weak acid
•K = Solubility in organic solvent (g/mL)/ Solubility in water (g/mL) Y = amount of compound X in DCM
20 – Y is the amount of compound X in water
First Extraction-
6.0 = (Y/100 mL)/[(20 – Y)/200 mL]; (Y/100 mL) * (2/2) = (2Y/200 mL) 6.0 = (2Y/200 mL)/ [(20 – Y)/200 mL] invert the problem by multiplying the numerator by the denominator to cancel out the volumes. 6.0 = 2Y/(20 – Y)
2Y = 6.0 *(20 – Y) = 120 – 6Y
Y = 60 – 3Y
4Y = 60
Y = 15 g in DCM
20 – Y = 5 g in Water
Second Extraction-
6.0 = 2Y/(5-Y)
2Y = 30 – 6Y
Y = 15 – 3Y
4Y = 15
Y = 3.75 g in DCM
20 – (15 + 3.75) = 1.25 in Water
Weight of compound X removed after the two extractions is 18.75 g.
Sources:
•http://www.convertunits.com/molarmass/Dichloromethane
•http://avogadro.chem.iastate.edu/MSDS/NaHCO3.htm
•http://chemicalland21.com/industrialchem/inorganic/HYDROCHLORIC%20ACID.htm
•http://msds.chem.ox.ac.uk/DI/dichloromethane.html
