Question

Methyl salicylate (oil of wintergreen) is prepared by heating salicylic acid, , with methanol, . In an experiment, 1.60 g of salicylic acid is reacted with 15.0 g of methanol. The yield of methyl salicylate, , is 1.66 g. What is the percentage yield

Answer 1

Answer:

90.71%

Explanation:

From the equation of reaction:

$C_7H_6O_3 + CH_4O --> C_8H_8O_3$

1 mole of salicylic acid requires 1 mole of methanol in order to produce 1 mole of methyl salicylate.

mole = mass/molar mass

1.6 g of salicylic acid = 1.6/138.121 = 0.012 moles

15 g of methanol = 15/32.04 = 0.47 moles

Salicylic acid is the limiting reagent and hence, 0.012 moles of methyl salicylate would be produced according to the mole ratio above.

0.012 moles of methyl salicylate = 0.012 x 152.1494 = 1.83 g

The theoretical yield of methyl salicylate is 1.83 g while the actual yield is 1.66 g.

Percentage yield = actual yield/theoretical yield x 100%

                         1.66/1.83 x 100% = 90.71%