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.