Need to input the values into the below paragraph with calculated results where

Question

Need to input the values into the below paragraph with calculated results where needed. The unknown is propyl isobutyrate (C7H14O2) and its wieght is 1 gram.

Saponification of unknown: In a 100mL singleneck RBF, water (50mL), the unknown (Xg, Ymmol) and a stoichiometric amount of potassium hydroxide (Xg, Ymmol) were combined. The reaction was refluxed for 2 hours, and cooled to room temperature. The water mixture was then extracted with diethyl ether (2x25mL). The ether was discarded, and then the water layer was acidified using excess concentrated HCl (5mL). The water layer was then extracted a second time with diethyl ether (2x25mL), and the ether extracts dried over using magnesium sulfate. Removal of the volatile solvent, yielded the desired product (Xg, 75.9%).

Explanation / Answer

Ans is 2.138mgms,

reactants-------->products

According to chemical conservation law

n1v1+n2v2+n3v3=X

given mass of unknown is 1g;n=mass/molecular weight=1/130=7.69mmol of unknown=Y;Diethyl Ether moles=(2/74.12)*1000=26.98mmol

Therefore 7.69*50(unknown is diisolved in water)+7.69*50(KOH prepared in water taken same values as X,Y are same for unknown and KOH)+Diethylether is taken to seperate water 26.98mmol*50mL+excess of Conc.HCl is taken i.e 4mmol*5mL=X

X=384.5+384.5+1349+20

Therefore X=2138milli gms yield

2138/1000=2.138gms

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