Aqueous Layer Water Add 10 Ml example essay topic

Acid Base Extraction 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 organic acids, weak organic acids neutral organic compounds and basic organic substances.

The procedure for this laboratory assignment are on the following pages. 3) Separation of Carboxylic Acid, a Phenol and a Neutral Substance The purpose of this acid / base extraction is to separate a mixture of equal parts of benzoic acid (strong acid) and 2- (weak base) and 1, 4- (neutral) by extracting from term-butyl methyl ether (very volatile). The goal of this experiment was to identify the three components in the mixture and to determine the percent recovery of each from the mixture. 4) Separation of a Neutral and Basic Substance A mixture of equal parts of a neutral substance containing either naphthalene or benzoin and a basic substance containing either 4- ethyl 4-amino benzoate were to be separated by extraction from an ether solution. Once the separation took place, and crystallization was carried out, it became possible to determine what components were in the unknown mixture, by means of a melting point determination. ResultsProcedure ObservationsInferenceDissolve 3.05 g Phenol Mixture was a golden-Neutral acid in 30 ml brown / yellow color t-butyl methyl ether in Erlenmeyer flask and transfer mixture to 125 ml separator y funnel using little ether to complete the transfer Add 10 ml of water Organic layer = mixture aqueous layer = water (clear) Add 10 ml saturated aqueous Sodium bicarbonate NaHCO 3 dissolves in solution sodium bicarbonate water. to funnel and mix cautiously with stopper on Vent liberated carbon Carbon dioxide gas dioxide and shake the mixture was released three times thoroughly with frequent venting of thefunnelAllow layers to separate Layer = H 2 O +NaHCO 3 completely and draw off lower layer into 50 ml Erlenmeyer flask 1 Add 10 ml of 1.5 aqueous NaOH Flask 2 = H 2 O + NaHCO 3 (5 ml of 3 M and 5 ml H 2 O) to separator y funnel, shake mixture, allow layers to separate and draw off lower layer into a 25 ml Erlenmeyer flask 2.

Add additional 5 ml of water to funnel, shake as before Add 15 ml Na Cl to funnel. Shake Bottom layer is white and Na Cl was added to the mixture and allow layers to separate gooey. wash the ether and draw off lower layer, which is layer and to remove NaOH and NaHCO 3 Pour ether layer into 50 ml Erlenmeyer flask from the top of funnel (not allowing any water droplets to be transferred) Flask 3 Add anhydrous NaSO 4 to ether extract until it no longer clumps together and set it aside Acidify contents of flask 2 Litmus went from Acidification was now by drop wise addition of blue to pink. Flaskcomplete concentrated HCl while 2 = creamy color testing with litmus paper and cool in ice Acidify contents of flask 1 Litmus went from Acidification was now by adding HCl drop wise blue to pink. Flaskcomplete while testing with litmus 2 = white solution paper and cool in ice Decant ether from flask 3 into a tared flask Boil ether with boiling chips Do a vacuum filtration and Solution turns to a Crystallization is now recrystallize ether by dissolving it solid. complete in 5 ml, taking out boiling chips, adding drops of Ligroin until the solution was cloudy and cool it in ice Isolate crystals from flask 2 by Crystals = creamy-white Dried crystals are now vacuum filtration and wash with powder ready for melting point a small amount of ice water determination and recrystallize it from boiling water Repeat the above for flask 1 Crystals = white powder Flasks number 4 and 5 were done by the following diagram. Results: As a result of this acid / base experiment, the following results were obtained: Flask 1: 31.113 g -30.223 g. 890 gFlask 2: 36.812 g -36.002 g.

810 gFlask 3: 90.789 g -90.114 g. 065 g% yield = experimental weight x 100% theoretical weight Flask 1: . 890 g x 100% = 89% 1.00 gFlask 2: . 810 g x 100% = 81% 1.00 gFlask 3: . 675 g x 100% = 67.5% 1.00 g When taking the melting points of the unknowns, flasks 4 and 5, I came to the conclusion that the samples contained, benzoin, melting point of 136-137 Degrees (C) and 4-, melting point of 67-80 degrees (C), respectively.

Flask 4: 90.912 g -89.174 g 1.738 g% yield = 1.738 g x 100% = 90.4%1.922 gFlask 5: 87.833 g -86.064 g 1.769 g% yield = 1.769 g x 100% = 87.3%2.027 gConclusion: After each procedure was complete, it became apparent that flask number 4 and flask number 5 contained benzoin and 4-, respectively. The melting point range that was experimentally determined for each was 136-137 for benzoin and 67-70 for 4-. As you can see, this experiment was not error-free, as my percentage yield was not 100%. This is expected for any experiment; for there is no way that, under the conditions, this experiment can be free of error. This error could have occurred for many reasons. The most prevalent reason, I feel that maybe not all of the substance was transferred from the flask to the vacuum, giving a slight error.

Also, some residue could have also been left in the vacuum funnel when transferring the crystal substances. Questions 2) It is necessary because nothing would come out of the stopcock- the reason for this is because of pressure. Leaving the stopper on, would decrease the pressure pushing down on the liquid and the pressure pushing upward would prevail, allowing nothing to escape. 3) I would not expect p-nitro phenol (pk a = 7.15) to dissolve NaHCO 3 (pk a = 6.4) because having a weak acid and a weak base, the reaction would favor the products, not the reactants, hence, the reaction would not proceed forward.

Would expect 2, 5- (pk a = 5.15) to dissolve in NaHCO 3 the reaction would proceed forward. 5) a) 1 g benzoic acid x 1 mol = . 00699 mol benzoic acid 143 g benzoic acid) 1 ml 10% solution NaHCO 3 x 1 g x 1 mol = . 00116 mol NaHCO 34 ml 96 g NaHCO 3.00699 moles of benzoic acid Introduction: The purpose of the second part of this laboratory assignment was to extract caffeine from tea using and then to confirm the identity of it by preparing a derivative of the extracted caffeine which has a sharp melting point, unlike caffeine itself. Once the extraction was complete, we were to test for melting point and get a HPLC reading for our derivative. Discussion: Tea leaves contain acidic, colored compounds as well as a small amount of un decomposed chlorophyll, which is soluble in.

Caffeine can be easily extracted from tea. This procedure can be done using conventional methods. Simply pouring hot water on the tea bags and steeping the bags for about 5-7 minutes would extract most of the caffeine that the tea contains. Pure caffeine itself is a white, bitter, odorless crystalline solid, therefore, it is obvious that more than just caffeine is in the liquid tea solution since tea is a brown color. Because of this, is used to dissolve the caffeine that is in the tea, which leaves the other constituents in the aqueous layer.

Using a separator y funnel, it becomes possible to extract the dissolved caffeine from the aqueous layer and the extraction is now ready for further procedure. ResultsProcedure ObservationInferenceTo a 250 ml beaker containing 7 tea bags, add 100 ml of boiling water. Allow the mixture to stand Brown aqueous solution for 5-7 minutes while steeping containing caffeine and the tea from the bags other impurities. Decant the mixture into another flask Cool solution to near Dichloromethane = room temperature and water soluble, clear, extract twice with 15 ml heavier that water. portions of using a gentle rocking motion and venting.

Drain off Dichloromethane Evaporation of the layer on first extraction; organic layer found solvent leaves crude include emulsion layer on on the bottom of the caffeine, which on the second extraction. funnel where the sublimation, yields caffeine is dissolved. a relatively pure Chlorine = top, aqueous product. solution. Drain extraction 1 and 2 back into the funnel Dry combined The solvent layer is solutions and any emulsion yellow. layer with sodium sulfate Wash the drying agent Residue of greenish with further portions of white crystalline weighs solvent and steam bath 50 mg (solid) the solvent To 5 mg of the Salicylic acid is water sublimed caffeine in water soluble. beaker, add 7.5 mg of salicylic acid and. 5 ml. Heat mixture to a boil Petroleum ether is a poor and add a few drops solvent for the product. petroleum ether until the mixture turns cloudy. Insulate beaker and allow it to cool slowly to room temperature and then cool in an ice bath Remove the solvent with Needle-like crystals are Caffeine salicylate is a Pasteur pipette while the isolated (white color) formed. beaker is in the ice bath then vacuum filter. Caffeine beaker: 51.61 g -51.56 g.

05 g = 50 mg% yield = . 05 g x 100% = 20%. 25 caffeine salicylate: 17.198 g -17.036 g. 062 g% yield = .

062 g x 100% = 25%. 25 gConclusion According to the HPLC graph that follows, my product was very pure. The actual melting point of caffeine salicylate is 137 degree (C), my product was found to have a melting point of 138 degrees (C). As before, of course this experiment was not done completely error-free, the error is due almost entirely on human error.