Experiment 5_ Extraction of Contaminants from Water

docx

School

University of Michigan *

*We aren’t endorsed by this school

Course

211

Subject

Chemistry

Date

Feb 20, 2024

Type

docx

Pages

10

Uploaded by MagistratePigeon2614

Report
1 10/10/21 Experiment 5: Extraction of Contaminants from Water Using liquid-liquid extraction (LLE) to separate compounds based on differences in their solubility in organic vs. aqueous solvents. Goals : Using LLE to extract a contaminant from a polluted water sample. Also, you will use TLC to monitor which LLE layer contains the contaminant and to confirm that your LLE is successful before evaporating the solvent and calculating the yield. Lastly, you will explore how NMR can be used to assess your extraction success. Reagent Table: Compound Molecular Weight (g/mol) Density (g/mL) Used Safety Dichloromethane (DCM) ( CH 2 Cl 2 ) 84.93 1.33 60 mL Health hazard Hexane (C H ) ₁₄ 86.18 0.659 60 mL Irritant, flammable Ethyl Acetate ( C 4 H 8 O 2 ) 88.11 0.902 60 mL Irritant, flammable 2- phenylphenol (C H O) ₁₂ ₁₀ 170.2 1.21 Less than 1 mL (used for TLC spotting) Irritant, environmental hazard
2 Sodium Hydroxide (NaOH) 39.997 2.13 1 mL Corrosive, causes severe skin burns and eye damage Magnesium sulfate (MgSO ) 120.37 2.66 A few scoops Irritant (Polluted) Water (H O) 18.02 1.00 50 mL N/A Nitrogen (N ) 28.013 0/81 Continual gas Corrosive, acute toxic
3 Procedure : Class Group Work Procedure : 1. Using the commercially obtained 2-phenylphenol sample (solid in reagent jar), perform a TLC characterization to test different developing solvents (different ratios of hexane: ethyl acetate) Developing solvent my group tested: 75% hexane, 25% ethyl acetate Rf = 4.1 / 6.7 = 0.612 Visualization: UV 2. As a class, decide which developing solvent condition(s) will be the best (the Rf of 2- phenylphenol is between 0.2-0.8). Use these TLC conditions for the rest of the lab. Best developing solvent: 75% hexane, 25% ethyl acetate LLE Procedure : 1. Measure 50 mL of the polluted water sample using a graduated cylinder. Pour the 50 mL into a 125 mL separatory funnel. 2. Add 1 mL of acid (1 M HCl) to your polluted water in the separatory funnel.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
4 3. Working in the fume hood, add organic solvent (chosen volume) to the sep funnel. Agitate the mixture by swirling and venting. Repeat the swirling and venting process twice. Organic solvent: DCM Chosen volume: 1 x 60 mL 4. Allow the two layers to separate. The density of the organic solvent determines whether it is the top or bottom layer. (Least dense solvent will float) If an emulsion occurs : 1. Gently poke the suspended drops using a glass stirring rod to help break the emulsion. 2. If there is still an emulsion, perform a chemical wash: a. Separate the layers and put the organic layer in the sep funnel b. Add 5 mL of deionized water. Two layers will form. c. If needed, allow 10 minutes for the layers to fully separate. 5. Separate the organic and aqueous layers If extracted with less than 60 mL organic solvent : 1. Run a TLC plate (using the best developing solvent) of the layers to determine the success of the extraction at that point. Include a commercially obtained sample of 2- phenylphenol on your TLC plate. Visualize using UV light and circle all of the spots gently with a pencil. a. If you choose to do additional extractions at this point , return the aqueous layer to the sep funnel, add additional organic solvent, and repeat swirling and separating for each extraction. Your total volume of organic solvent should not exceed 60 mL
5 6. When you have completed your total number of extractions, transfer and combine all organic layers in an Erlenmeyer flask. 7. Add a few scoops of anhydrous magnesium sulfate (MgSO ) to the flask of organic layers. Swirl the flask. If the MgSO remains in the single clump, add another scoop. Repeat until the MgSO does not remain clumped. Look for a small amount of loose salt crystals as you swirl. 8. Pre-weigh a large vial/beaker. Record this weight in the table below. 9. Pour the liquid only into the pre-weighed vial. Do not pour the drying agent into the container. 10. Run a TLC of your dried organic layer and final aqueous layer along with a commercially obtained sample of the 2-phenylphenol. Visualize using UV light and circle all of the spots gently with a pencil. 11. Using the pressurized air or N in the fume hood, evaporate the organic solvent in the organic layer beaker. You might keep the beaker in a bath of tap water during evaporation, but if there is only a small amount of solvent, this is not necessarily required. 12. Record the weight of your beaker with the dried 2-phenylphenol. If time remains, run an IR and a NMR. 13. Calculate the mass of the recovered material. Weight (g) Pre-weighed empty beaker 184.79 Beaker and extracted material 185.17 Extracted material 0.38 14. Based on the weight of the extracted material, determine the concentration of the 2- phenyl phenol in your 50 mL sample (in the data section)
6 Data: My TLC plate: 1x60 mL Other organic solvent plates: Ethyl Acetate 1x60mL Visualization: UV D.S.: 75% hexane, 25% ethyl acetate Rf (P) = 2.5 / 5.0 = 0.50 Rf (AQ) = N/A Rf (O) = 2.2 / 5.0 = 0.44 Mass of recovered material: 0.13 g P = 2-phenylphenol AQ = Aqueous layer O = Ethyl acetate
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
7 Hexane 1x60mL Visualization: UV D.S.: 75% hexane, 25% ethyl acetate Rf (S) = 3.1 / 4.3 = 0.72 Rf (O) = 2.6 / 4.3 = 0.60 Rf (A) = 2.5 / 4.3 = 0.58 Mass of recovered material: 0.15 g S = 2-phenylphenol O = Hexane A = Aqueous layer Concentration of 2-phenylphenol calculation (my compound - DCM ):
8 Analysis and Conclusion : By examining the TLC plates, the organic solvent that was able to extract the most 2- phenylphenol was ethyl acetate. On the TLC plate for ethyl acetate, there was no appearance of a spot for the aqueous solution. This demonstrates that the ethyl acetate extracted 2- phenylphenol completely. DCM is very close second because the pigmentation of the aqueous solution spot was very minimal. And lastly, hexane extracted the least amount of 2- phenylphenol because the aqueous solution spot on the TLC plate was evident and had an Rf value near the Rf for the 2-phenylphenol. This is because hexane is a nonpolar solvent and 2- phenylphenol is a polar molecule, making hexane the least fit to extract 2-phenylphenol completely. As with the mass of recovered material, ethyl acetate extracted the least amount with 0.13 g, hexane was 0.15 g and DCM was 0.38 g. While this trend indicates that DCM has the highest concentration of 2-phenylphenol, ethyl acetate is still shown to contain the least concentration. Because we are only examining one data point for each of the organic compounds, the mass of recovered materials could have been inaccurate (higher or lower than expected) due to human error and possibly the difficulty to break the emulsions. When you increase the amount of organic solvent, the amount of extracted 2- phenylphenol will increase as well. This is because the solvent can become more saturated with the 2-phenylphenol before it is removed from the sep funnel. Also, as the number of extractions increases, the extracted amount of the 2-phenylphenol increases to a certain point until the organic solvent can no longer take out any more 2-phenylphenol from the polluted water. Factors that impact the success of extraction : polarity, pH, human error Polarity : Ethyl acetate and DCM are polar, meaning that they can form hydrogen bonds with the OH group on 2-phenylphenol. On the other hand, hexane is a nonpolar solvent so it is not attracted to the 2-phenylphenol and is unable to accept hydrogen bonds. pH : The lower the pH (the more acidic), the more likely that no deprotonated contaminant exists. This means that it will be easier to get the contaminant to associate more with the organic layer with a lower pH. Human error : Through manual mixing and venting, along with separating the organic solvent and the aqueous solution layers from the sep flask, human error plays a large factor in what could be impacting the conditions and measurement of the extractions.
9 Using H-NMR and IR spectra, one can evaluate the purity of the extracted compound. For H-NMR, if the number of sharp signals at the anticipated position matches with the number of protons with the correct splitting pattern, one can say that the compound is pure. Similarly with IR, if the anticipated functional groups of the compound are displayed on the spectra, then the compound is also deemed as pure. The combination of using both H-NMR and IR to determine the extent of purity of the compound would produce accurate conclusions. NMRDB C-NMR and H-NMR for 2-phenylphenol : (Did not collect in the lab)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
10 All H’s in rings are shown in this area on the H-NMR spectra. The proton attached to the O does not show up on the spectra.