Data: (part of 15 point lab report grade) ter of alcohol unknown used: 5 ass (g)
ID: 954981 • Letter: D
Question
Data: (part of 15 point lab report grade) ter of alcohol unknown used: 5 ass (g) of alcohol used: 3.sblo grams ass (g) of purified ester obtained: 0143 bserved boiling point range of ester: 105'C escribe odor of ester: Nullnkin Remaal rinciple peaks in the IR spectrum of the alcohol unctional groups stretches or bends correspond to rinciple peaks in the IR spectrum of the product e Peaks in the NMR spectrum for the product este hemical shift, the integration (relative intensity) riplet . ..) for each unique peak set.Explanation / Answer
solution:
Given that weight of alcohol(2-pentanol)=3.866g
number of moles of alcohol(2-pentanol)=weightof alcohol/molecular weight of alcohol
=3.866/88.148(since molecular weight of 2-pentanol-88.148g/mol)
=0.0838580moles
0.0838580moles to give 0.0838580moles of ester.
calculate the mass of 0.0838580moles of ester(that is multiplication of molecular weight of eaterxnumber of moles),you will get theoritical yield.
percent yield=(Actual yield/theoritical yield)*100 (Actual yield is the yield of product in grams)
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.