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Morphine has the formula c17h19no3. it is a base and accepts one proton per molecule. it is isolated from opium. a 0.685 −g sample of opium is found to require 8.91 ml of a 1.16×10−2 m solution of sulfuric acid for neutralization. part a assuming that morphine is the only acid or base present in opium, calculate the percent morphine in the sample of opium.

Respuesta :

ChemistryHelper2024
2 C₁₇H₁₉NO₃ + H₂SO₄ → Product
Moles of H₂SO₄ = M x V(liters) = 0.0116 x 8.91/1000 = 1.033 x 10⁻⁴ mole 
moles of morphine = 2 x moles of H₂SO₄ = 2.066 x 10⁻⁴
Mass of morphine = moles x molar mass of morphine = 2.066 x 10⁻⁴ x 285.34 
                             = 0.059 g
percent morphine = [tex] \frac{mass of morphine}{mass of opium} x 100[/tex]
                             = [tex] \frac{0.059}{0.685} x 100 [/tex] = 8.6 %   

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