6. 11.00 g of compound X (solubility 1.0 g/100 ml. at 0 °C) is contaminated with

Question

6. 11.00 g of compound X (solubility 1.0 g/100 ml. at 0 °C) is contaminated with 1.00 g o (solubility 1.1 g/100 mL at o "C) and 1.00 g of compound Z (solubility 1.5 g/100 ml at impure mixture is recrystallized from 70 mL of hot solvent, then cooled (The collected crystals are called a crop). 0 C). The to 0 °C, followed by filtration. Showing your calculations, answer a-f below a. What is the composition, in grams, of X, Y, and Z in the first crop of crystals? b. What is the mass percentage of X in the first crop? lla X c. Has the purity of compound X in the first crop improved over that in the original mixture? Explain. 10mL x 70 mL 1.542 100 uL 103 x 100 M1.87- 100m. yes, puv The mother liquor (the liquid left after the step above-Hint: There is 70 ml of mother liquor.) was evaporated to 40 mL, then cooled to 0 °C and filtered to obtain a second crop of crystals: d. What is the composition, in grams, of the second erop? e. What is the mass percentage of X in the second crop? f. Is the second crop of compound X as pure as the first crop? Explain. g. Is the second crop of the compound X as pure as the original mixture? Explain.

Explanation / Answer

At zero degree the solubility of Z is 1.5g per 100ml and that of X and Y are 1g/100ml and 1.1g/100ml. Now if we cool 70ml of the solution to 0oC the substance with less solubility will crystallize completely. So the A will deposit first

Amount of A in the crystall = 11g in the sample - 0.7g in 70ml solution = 10.3 g in crystall

Amount of B in the crystall = 1g in the sample - 0.077g in 70ml solution = 0.923 in crystall

Amount of C in the crystall = 1g in the sample - 0.105 in 70ml solution = 0.895 in crystall

Now the % of A,B,C are already calculated in the worksheet. They are for crude sample (A-84.6%)

Now in the new sample the % of A is 10.3/(10.3+0.923+0.895)=85%. So the percentage of A is improved.

Now the solution. contains 0.7g of A, 0.77g of B and 0.105g of B. If it is reduced to 40ml and crystallised.

for A 0.7g in sample - 0.4g in 40ml = 0.3g in crystall

for B 0.077g in sample - 0.044g in 40ml = 0.033g in crystall

for C 0.105g in sample - 0.06g in 40ml = 0.045 in crystall

Now the % of A in the second sample is equal to 0.3/(0.3+0.033+0.045) = 79%. The yieldof A in the second crap is reduced

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