experiment 1
Post on 03-Sep-2014
187 Views
Preview:
TRANSCRIPT
Espiritu, Ma. Sharmaine D. & Sy, Diane Nicole L. Superman & Wonderwoman
Espiritu, Ma. Sharmaine D. & Sy, Diane Nicole L. Ma. Sharmaine D. Espiritu & Diane Nicole L. Sy
• The products of a chemical reaction are not necessarily in its purest form.
• Crystallization used in order to extract pure compound.
• Addition of water to sample and boiling until
it dissolves • Cooling of the solution, addition of activated
carbon, boiling • Filtration using a syringe • Cooling • Paper filtration, drying • Weighing
• Varying solubilities • Increase and decrease in solubility • Saturation • Crystal formation • Adhesion
Weight of impure sample: 0.1 g Weight of pure crystals: 0.05 g Percent Recovery: 50% Sources of error: incomplete dissolution, excessive activated carbon, slow filtration, remaining liquid in syringe, rapid cooling, wet filter paper
1. Properties of an ideal solvent for purification by crystallization
• Concept of “like dissolves like” and polarity of substances
• Solubility • of compound • vs. impuri:es
• Boiling point of the solvent should also be lower than the mel:ng point of the compound
2. Should cooling be slow or rapid? Explain.
• Slow
3. Advantage of water as solvent
• Nonflammable • Nontoxic • ‘Universal’ solvent
4. The solubility of benzoic acid in water is 0.21 g per 100 mL of water at 10°C, 0.27 g per 100 mL at 18°C, 2.75 g per 100 mL at 80°C and 6.80 g per 100 mL at 95°C. Two students crystallized 10 g samples of benzoic acid from water, the first dissolving benzoic acid at 80°C and filtering at 10°C, the second dissolving the acid at 95°C and filtering at 18°C. Calculate the quantity of water each student was required to use and the maximum recovery of benzoic acid possible in each case.
A: At 80°C Amount of water = 100x10/2.75 = 363.64 mL At 10°C Max. Recovery = 10 – (363.64x0.21/100) = 9.24 g
B: At 95°C Amount of water = 100x10/6.80 = 147.06 mL At 18°C Max. Recovery = 10 – (147.06x0.27/100) = 9.60 g
4. A solid (X) is soluble in water to the extent of 1 g per 100 g of water at room temperature and 10 g per 100 g of water at the boiling point. How would you purify X from a mixture of 10 g of X with 0.1 g of impurity Y, which is completely insoluble in water, and 1 g of impurity Z, having the same solubility characteristics in water as X?
Dissolve in water and boil. Cool to room temperature, then filter (Residue: Y) Add water, heat un:l all crystals dissolve. Slowly cool and then filter. (Residue: X)
How much pure X could be obtained after one recrystallization from water? 10 G−1 G(HIIJKLMNHOP QHRSP KT U)/10 G L 100%=90% 90 % pure X
How much pure X could be obtained after one recrystallization from a mixture of 10 g of X with 9 g of Z? At boiling point At room temperature
X recovered = 1 g
Based on the result obtained, what is suggested about the use of crystallization as a purification technique?
• Inaccurate • Repeated crystalliza:ons • Change in solvent used
• Bansal, R. (2003). A textbook of organic chemistry. (4th ed., pp. 116-117). New Delhi, India: New Age Internation (P) Limited.
• Welton, T. & Reichardt. C. (2011). Solvents and solvent effects in organic chemistry (4th ed.). Weinheim, Germany: Wiley-VCH Verlag & Co.
• Gilman, J. (1963). The art and science of growing crystals. Weinheim, Germany: John Wiley & Sons Inc.
• Williamson, K. & Masters, K. (2011). Macroscale and microscale organic experiments (6th ed.). Belmont, CA: Brooks/Cole.
top related