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Methyl salicylate (oil of wintergreen), C₈H₈O₃, is prepared by heating salicylic acid, C₇H₆O₃ (MW = 138.1 amu), with methanol, CH₃OH, (MW = 32.04 amu). C7H6O₃ + CH₃OH ----> C₈H₈O₃, + H₂O In an experiment, 1.50 g of salicylic acid is reacted with 11.20 g of methanol. The yield of methyl salicylate is 1.11 g. What is the percentage yield of the reaction? Express your answer to one decimal place.

User Jon Hess
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Final answer:

The percentage yield of the reaction producing methyl salicylate is calculated as (actual yield / theoretical yield) × 100%, resulting in a yield of 66.9%.

Step-by-step explanation:

The percentage yield of a chemical reaction indicates the efficiency of the reaction. To calculate it, you need to know the theoretical yield, which is the amount of product that would be formed if the reaction were perfect and all reactants were converted into products. Then you compare it with the actual yield, which is the amount of product actually obtained from the reaction.

In this case, to determine the theoretical yield, we need to find out which reactant is the limiting reagent. Since salicylic acid has a molecular weight (MW) of 138.1 amu, 1.50 g corresponds to 0.0109 moles of salicylic acid. Methanol, with a MW of 32.04 amu, equates to 0.349 moles for 11.20 g. The reaction shows a 1:1 mole ratio between salicylic acid and methanol, so salicylic acid is the limiting reagent. Therefore, the theoretical yield of methyl salicylate would be the same number of moles as salicylic acid, which is 0.0109 moles, taking into account its molar mass is approximately 152.15 g/mol, gives us a theoretical yield of approximately 1.66 g.

To find the percentage yield, use the formula: (actual yield / theoretical yield) × 100%. So, the percentage yield is (1.11 g / 1.66 g) × 100% = 66.9% when expressed to one decimal place.

User Locks
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