Question

Methyl salicylate (oil of wintergreen) is prepared by heating salicylic acid,C7H6O3 , with methanol,CH3OH .

C7H6O3+CH3OH---->C8H8O3+H2O

In an experiment, 1.50 g of salicylic acid is reacted with 11.8 g of methanol. The yield of methyl salicylate,C8H8O3 , is 1.45 g. What is the percentage yield?

Percentage yield = %

Answer 1

Mass of salicylic acid = 1.50 g

Molar mass of salicylic acid = 138.121 g/mol

Moles of  salicylic acid = Mass/ Molar mass = 1.50 g/(138.121 g/mol) = 0.01086 moles

Mass of Methanol = 11.8 g

Molar mass of methanol = 32.04 g/mol

Moles of methanol = 11.8 g/(32.04 g/mol) = 0.36829 mol

1 mol of C₇H₆O₃ reacts with 1 mol of CH₃OH

SO, 0.01086 moles of C₇H₆O₃ reacts with 0.01086 mol of CH₃OH

Hence  C₇H₆O₃ is a limiting reagent

1 mol of  C₇H₆O₃ produces 1 mol of C₈H₈O₃

So,  0.01086 moles of C₇H₆O₃ produces 0.01086 mol of C₈H₈O₃

Molar mass of C₈H₈O₃ = 152.15 g/mol

Mass of C₈H₈O₃ = Moles *MOlar mass = 0.01086 mol*152.15 g/mol = 1.65 g

Thus, Perecntage yield = (Experimental yield/ Theoretical yield) *100%

= (1.45 g / 1.65 g) *100% = 87.88%