Question

In: Chemistry

1) A student recrystallizes 1.00 g of the unknown compound from 50 mL of water. After...

1) A student recrystallizes 1.00 g of the unknown compound from 50 mL of water. After allowing the mixture to cool to room temperature (25 °C), the student collects the recrystallized compound using vacuum filtration. What is the theoretical maximum percent recovery of the unknown compound for this recrystallization step?

2)A student recrystallizes 1.00 g of the unknown compound from 50 mL of water. After allowing the mixture to cool in an ice bath (0 °C), the student collects the recrystallized compound using vacuum filtration. What is the theoretical maximum percent recovery of the unknown for this recrystallization step?

Solutions

Expert Solution

Below two reasons for loss of compounds

1) In the filtration step, material was lost on filter paper: this is common at 25 o C and 0 o C

2) Material did not completely crystallize from solution: The percentage recovery depends on the final temperature of the solution after recrystallization, as at 0 o C the solute will most likely be less soluble than at room temperature

% of recovery = (amount pure product recovered/amount crude product used) X100

If you know the solubility of your unknown compound in water then we can calculate the exact % of recovery.

You will get maximum recovery filtration at 0 o C than room temperature filtration


Related Solutions

Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water....
Suppose that 0.323 g of an unknown sulfate salt is dissolved in 50 mL of water. The solution is acidified with 6M HCl, heated, and an excess of aqueous BaCl2 is slowly added to the mixture resulting in the formation of a white precipitate. 1. Assuming that 0.433 g of precipitate is recovered calculate the percent by mass of SO4 (2-) in the unknown salt. 2. If it is assumed that the salt is an alkali sulfate determine the identity...
A solution contains 12.00 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume...
A solution contains 12.00 g of unknown compound (non-electrolyte) dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -4.95 ∘C. The mass percent composition of the compound is 53.31% C, 11.19% H, and the rest is O. Part A What is the molecular formula of the compound? Express your answer as a molecular formula.
Weight of Unknown #4( in g) Volume of water (in mL) Unknown #4 solution (in g/L)...
Weight of Unknown #4( in g) Volume of water (in mL) Unknown #4 solution (in g/L) 0.118 50.00 2.36 0.127 50.00 2.54 0.112 50.00 2.24 Volume of NaOH used = Final concentration – Initial concentration                                           = 16.49 mL – 4.90 mL = 11.49mL NaOH= 0.01149 L NaOH Exp. # Concentration of NaOH Initial volume of NaOH Final volume of NaOH Volume of NaOH used for titration 1 0.1030M 4.90mL 16.49mL 11.59mL 2 0.1030M 16.49mL 26.01mL 9.52mL 3 0.1030M...
An unknown compound (152 mg) was dissolved in water to make 75.0 mL of solution. The...
An unknown compound (152 mg) was dissolved in water to make 75.0 mL of solution. The solution did not conduct electricity and had an osmotic pressure of 0.328 atm at 27°C. Elemental analysis revealed the substance to be 78.90% C, 10.59% H, and 10.51% O. Determine the molecular formula of this compound by adding the missing subscripts.
A solution contains 10.75g of an unknown compound dissolved in 50.0 mL of water. (Assume a...
A solution contains 10.75g of an unknown compound dissolved in 50.0 mL of water. (Assume a density of 1.00 g/mL for water.) The freezing point of the solution is -3.38 C. The mass percent composition of the compound is 60.98% C, 11.94% H, and the rest is O. -What is the molecular formula of the compound? Express your answer as a molecular formula..
What is the molality of a solution containing 50.6 mL of water (d= 1.00 g/mL) and...
What is the molality of a solution containing 50.6 mL of water (d= 1.00 g/mL) and 0.122 g of 2-pentanol?
A student conducts a titration of 50 mL of HCl with 1.00 M NaOH. The pH...
A student conducts a titration of 50 mL of HCl with 1.00 M NaOH. The pH at the equivalence point is Basic Neutral Acid
Compound Y has a solubility of 2.0 g/100 mL in water and 20.0 g/100 mL of...
Compound Y has a solubility of 2.0 g/100 mL in water and 20.0 g/100 mL of ether. What mass of compound Y would be removed from a solution of 1.8 g of Y in 100 mL of water by a single extraction with 100 mL of ether? What mass of compound Y would be removed from the original water solution by two extractions using 50 mL of ether each time?
A student dissolved 10.5 g of NH4Cl in 40 mL of water.  The temperature of the water...
A student dissolved 10.5 g of NH4Cl in 40 mL of water.  The temperature of the water changed from 22.5oC to 5.1oC.  Calculate the enthalpy of hydration of ammonium chloride in J/mol. 1.  A student collected the data below while performing a kinetics experiment. Trial Volume of 2.0 M Reactant A Volume of 0.5 M Reactant B Volume of H2O Time for B to react 1 10 mL 10 mL 30 mL 135 sec 2 20 mL 10 mL 20 mL 69 sec...
A 0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water...
A 0.1276 g sample of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0633 M NaOH solution. The volume of base required to bring the solution to the equivalence point was 18.4 mL. (a) Calculate the molar mass of the acid. (b) After 10.0 mL of base had been added during the titration, the pH was determined to be 5.87. What is the Ka of the unknown acid? (10 points)
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT