Morphine has theformula C15H12NO3. It is a base and accepts one proton permolecu
ID: 76547 • Letter: M
Question
Morphine has theformula C15H12NO3. It is a base and accepts one proton permolecule. It is isolated from opium. A 0.681g sample of opium isfound to require 8.93mL of a 1.17×102 Msolution of sulfuric acid forneutralization. assuming the morphine is the only acid or basepresent, calculate the percent morphine inopium. Morphine has theformula C15H12NO3. It is a base and accepts one proton permolecule. It is isolated from opium. A 0.681g sample of opium isfound to require 8.93mL of a 1.17×102 Msolution of sulfuric acid forneutralization. assuming the morphine is the only acid or basepresent, calculate the percent morphine inopium.Explanation / Answer
Moleculer formula of Morphine is C15H12NO3 Molar mass ofMorphine = 15*12+12*1+14*1+16*3 = 180 + 12+ 14 +48 = 254 g Weight of opium = 0.681 g Morphine is isolated fromopium. Now opium is neutralized by 8.93 mL of1.17*10-2 M sulfuric acid solution. So thenumber of moles present in sulfuric acid solution n = molarity * Volume = 1.17*10-2 mol/L *8.93*10-3 L = 10.448*10-5 moles Number of moles of opium = number of moles of sulfuric acidsolution that is number of moles of opium = no of moles of morphine. So mass of morphine in opium = molar mass of Morphine *numberof moles = 254*10.448*10-5 =0.0265gmol The percentage of Morphine in opium is = (mass of Morphine in opium /total mass of opium)*100 =( 0.0265/0.681)*100 = 3.89 % . The percentage of Morphine in opium is 3.89 % . =0.0265gmol The percentage of Morphine in opium is = (mass of Morphine in opium /total mass of opium)*100 =( 0.0265/0.681)*100 = 3.89 % . The percentage of Morphine in opium is 3.89 % .Related Questions
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.