Question

Methyl salicylate (oil of wintergreen) is prepared by heating salicylic acid (C7H6O3) with methanol

(CH3OH) C7H6O3 + CH3OH--> C8H8O3 + H2O

What is the percent yield if reaction of 5.172 g of salicylic acid with an excess of methanol produced a measured yield of 3.733 g of oil of wintergreen?

Answer 1

Molar mass of C7H6O3,
MM = 7*MM(C) + 6*MM(H) + 3*MM(O)
= 7*12.01 + 6*1.008 + 3*16.0
= 138.118 g/mol

mass of C7H6O3 = 5.172 g
mol of C7H6O3 = (mass)/(molar mass)
= 5.172/138.118
= 0.0374 mol

According to balanced equation
mol of C8H8O3 formed = moles of C7H6O3
= 0.0374 mol


Molar mass of C8H8O3,
MM = 8*MM(C) + 8*MM(H) + 3*MM(O)
= 8*12.01 + 8*1.008 + 3*16.0
= 152.144 g/mol


mass of C8H8O3 = number of mol * molar mass
= 0.0374*152.144
= 5.6972 g

% yield = actual mass*100/theoretical mass
= 3.733*100/5.6972
= 65.5 %
Answer: 65.5 %