2. What volume of an organic solvent must be used to effect 90% separation in one extraction when only 2.7g of a certain compound dissolves in 100ml of water? (K=15) 3. What percentage of the organic compound could be recovered if two extractions were made, each time using half of the volume calculated 2?
Ans 2:
Expression for partition coefficient can be written as follows:
K = (Wo/Vo) / (Waq/Vaq)
Where,
K = partition coefficient = 15
Wo = Weight of solute in organic phase = (2.7 x 0.90)= 2.43g
Vo = Volume of organic solvent
Waq = Weight of solute in aqueous phase =(2.7 - Wo) = 2.7 – 2.43 = 0.27g
Vaq = volume of aqueous phase = 100mL
Thus, partition coefficient can be written as follows:
K = 15=(2.43/Vo) / (0.27/100)
15 = 2.43 / (Vo x 0.0027)
Vo = 2.43 / (15 x 0.0027)
Vo = 60mL
Thus, volume of organic solvent required to effect 90% separation is 60 mL
--------------------------------------------------------------------------------------------------------------------------------------------------------
Ans 3:
The percentage of organic compound that can be recovered after two extractions of 30mL each can be calculated as follows:
K = (Wo/Vo) / (Waq/Vaq)
For first extraction:
Vo = 30mL
Vaq = 100mL
Wo + Waq = 2.7
K = 15
Therefore, 15 = (Wo/30) / (Waq/100)
15 = Wo x 100 / Waq x 30
(15 x 30) /100= Wo / Waq
Wo/Waq = 4.5
Wo = 4.5 Waq
Wo = 4.5 (2.7 – Wo)
Wo = 12.15 – 4.5Wo
Wo + 4.5Wo = 12.15
5.5Wo = 12.15
Wo = 12.15/5.5
Wo = 2.20
Hence, Waq = 2.7-2.2 = 0.5 g
Thus, after first extraction, amount of solute in organic phase is 2.2g and that in aqueous phase is 0.5g
For second extraction:
Vo = 30mL
Vaq = 100mL
Wo + Waq = 0.5
K = 15
Therefore, 15 = (Wo/30) / (Waq/100)
15 = Wo x 100 / Waq x 30
(15 x 30) /100= Wo / Waq
Wo/Waq = 4.5
Wo = 4.5 Waq
Wo = 4.5 (0.5 – Wo)
Wo = 2.25 – 4.5Wo
Wo + 4.5Wo = 2.25
5.5Wo = 2.25
Wo = 2.25/5.5
Wo = 0.409
Hence, Waq = 0.5 – 0.409 = 0.091 g
Thus, in second extraction, amount of solute in organic phase is 0.409g and that in aqueous phase is 0.091g
Thus, after two extractions of 30mL each, total solute in organic phase is 2.20 + 0.409 = 2.609g
Therefore, % of solute recovered = (2.609 / 2.7) x 100 = 96.63%
Hence, percentage of organic compound that can be recovered after two extractions is 96.63%
2. What volume of an organic solvent must be used to effect 90% separation in one...
Extractions rely on differences in solubility of the components that are being separated. The amount of a material that will transfer into the solvent in an extraction can be calculated using a partition coefficient. This is a ratio of the solubility of the material in each solvent that can be used to predict outcomes of extractions. A certain organic compound, Z, has the distribution coefficient (partition coefficient), Kp = 10.45 when partitioned between ether and water. Therefore, the equation used...
Provide a flow chart detailing the acid/base extraction/separation of the compounds show below. Your answer must employ the following reagents: methylene chloride, hydrochloric acid (1M & 6M), sodium hydroxide (1M & 6M), 10% sodium bicarbonate (aq). Clearly indicate the product(s) and layers formed following each step of your separation scheme. IMPORTANT: p-cresol is soluble in sodium hydroxide solution but insoluble in neutral water or sodium bicarbonate solution. 2-ethylbenzoic acid is soluble in both sodium hydroxide and sodium bicarbonate solutions. Use...
Separation techniques /LLE Do all parts Due: Wednesday Feb 27th Assignment #4-Liquid-Liquid Extraction Reading/References: Ch 8 [Seader, Henley, & Roper (SHR) Textbook] 1) A solution containing 10 g/L of a valuable protein and 2 g/L of a protein impurity is extracted in a stirred vessel using an organic solvent. Distribution coefficients are K = 6.5 for the valuable protein and 1.5 for the impurity. The initial volume is 100 L, and 400 L of solvent are used for the extraction....
Acid/Base Extraction 1. Provide a flow chart detailing the acid/base extraction/separation of the compounds shown below. Your answer must employ the following reagents: methylene chloride, hydrochloric acid (1M & 6M), sodium hydroxide (1M & 6M), 10% sodium bicarbonate (aq). Clearly indicate the product(s) and layers formed following each step of your separation scheme. IMPORTANT: p-cresol is soluble in sodium hydroxide solution but insoluble in neutral water or sodium bicarbonate solution. 2- ethylbenzoic acid is soluble in both sodium hydroxide and...
Question 1 p Flag question Not yet answered Marked out of 1 What is an extraction? Select one: a. Separation of compounds in a mixture based on their solubility in miscible solvents. b. Separation of compounds in a mixture based on their solubility in immiscible solvents. Question 2 Not yet answered Marked out oft Flag question In a liquid-liquid extraction the compound (or solute) prefers to be dissolved in one liquid versus the other liquid. Select one: True False Question...
Introduction: The technique used to separate an organic compound from a mixture of compounds is called Extraction. Extraction process selectively dissolves one or more of the mixture compounds into a suitable solvent. The solution of these dissolved compounds is referred to as the Extract. Here the organic solvent dichloromethane is used to extract caffeine from an aqueous extract of tea leaves because caffeine is more soluble in dichloromethane (140 mg/ml) than it is in water (22 mg/ml). However, there are...
Exercise 2 Separation of a Mixture Based on Acid-Base Properties One purpose of this exercise is to learn how to use a separatory funnel to extract a single component away from other compounds in solution. To do so, we will apply the principles of solubility and acid-base behavior you’re seeing in class. One of the compounds is neutral in the acid-base sense. It has no ability to either donate or accept a proton from an aqueous solution, and will remain...
I dont understand the process for this type of question. It has to do with determining the solubility in mL to achieve Purification of an Organic Compound. removing impurities to get compound A in pure form. If someone could show me the process step by step that would be great thank you. A= 10g impurity 1= 1.0g partially soluble impurity 2= 1.0g insoluble the question is really how to seperate, and the next part of the question is how to...
Part A 1. If you had used more methylene chloride in each step, you could have extracted more caffeine. Explain why you did not. Hint - What step would have taken longer (Hint: not drying). PART A Mass of Beaker & Caffeine Mass of Beaker 167.750g 67.6809 Mass of Caffeine 0.0 75 Calculations. Show your work and circle the answers. Mass of caffeine recovered: 012919 b) Actual:.07 % caffeine recovered: a) Predicted: 9 7:213% b) Actual: 92.105% PART B Unknown...
need help with number 2 HW #3 for Chem 318 Due: April 29, 2019 1. Solute A has a K 2.5 for an extraction between water (phase 1) and chloroform (phase 2). If 100 mL of a 0.01 fraction will be extracted? For the same solute and, what fraction will be extracted if 5 extractions wi 80 mL chloroform each are used (instead of one 400 ml extraction)? M solution containing solute A in water is extracted one time with...