Solve Recrystalization Lab Homework - Comp. & Yield of Crystals, Mother liquor

  • Thread starter Thread starter mirandasatterley
  • Start date Start date
  • Tags Tags
    Lab
Click For Summary
SUMMARY

The discussion centers on a crystallization homework problem involving two compounds, A and B, with solubility of 1g/100mL at 20°C and 12g/100mL at 100°C. The solution, containing 1.5g of A and 10g of B, dissolves completely at 100°C. Upon cooling to 20°C, the crystallization yields 0.125g of A and 0.833g of B, resulting in a crystal composition of 13.05% A and 86.95% B. The yield of the process is implicitly calculated through these percentages, while the composition of the mother liquor remains unclear to the participants.

PREREQUISITES
  • Understanding of solubility principles at varying temperatures
  • Knowledge of crystallization processes
  • Ability to perform mass and percentage calculations
  • Familiarity with the concept of mother liquor in crystallization
NEXT STEPS
  • Research the calculation of yield in crystallization processes
  • Learn about the composition and analysis of mother liquor
  • Study solubility curves and their applications in crystallization
  • Explore practical examples of crystallization in laboratory settings
USEFUL FOR

Chemistry students, laboratory technicians, and anyone involved in crystallization processes or studying separation techniques in chemistry.

mirandasatterley
Messages
61
Reaction score
0

Homework Statement



You are given a mixture containing two compounds, A and B. Both compounds have a solubility of 1g/100mL of solvent at 20 degree celsius and 12g/100mL of solvent at 100 degree celsius. The sample is composed of 1.5g of A and 10g of B. At 100 degree celsius all of the sample just dissolves in a minimum amount of solvent. The solution is cooled to 20 degree celsius and crystals are collected. Calculate the composition of the crystals and the yeild of the process. What is the composition of the mother liquor?

Homework Equations



No equations.

The Attempt at a Solution



Compound A: 1.5g x 100mL/12g = 12.5mL
Compound B: 10g x 100mL/12g = 83.3mL
So from this, of the solution (solvent that just dissolves A and B), 12.5mL dissolves A and 83.8mL dissolves B.

When cooled, the solubility is only 1g/100mL, so
Compound A: 12.5mL x 1g/100mL = 0.125g
Compound B: 83.3mL x 1g/100mL = 0.833g
Therefore the crystals collected have composition:
Compound A = 0.125g/(0.125 +0.833)g x 100% = 13.05%
Compound B = 0.833g/(0.125 +0.833)g x 100% = 86.95%

Does any of this make sense? If not, where did I go wrong? I'm also unsure of how to calculate the yeild of the process and the composition of the mother liquor.

Any help is greatly appreciated. Thanks.
 
Physics news on Phys.org
You already calculated the yield when you figured out the %'s of compound A and B...as for the composition of mother liquor, I have no clue what its asking for.
 

Similar threads

Replies
4
Views
23K