Suppose that in the synthesis of isoamyl acetate, a student began with 6.289 grams of acetic acid and 3.75 grams of isoamyl alcohol. If the mass of isoamyl acetate obtained were 4.828, what was the percent yield?

The percent yield of the reaction is 87.24 %.

Explanation:

To calculate the number of moles, we use the equation:

    .....(1)

Given mass of acetic acid = 6.289 g

Molar mass of acetic acid = 60 g/mol

Putting values in equation 1, we get:

Given mass of isoamyl alcohol = 3.75 g

Molar mass of isoamyl alcohol = 88.15 g/mol

Putting values in equation 1, we get:

The chemical equation for the reaction of acetic acid and isoamyl alcohol follows:

By Stoichiometry of the reaction:

1 mole of isoamyl alcohol reacts with 1 mole of acetic acid

So, 0.0425 moles of isoamyl alcohol will react with = of acetic acid

As, given amount of acetic acid is more than the required amount. So, it is considered as an excess reagent.

Thus, isoamyl alcohol is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of isoamyl alcohol produces 1 mole of isoamyl acetate

So, 0.0425 moles of isoamyl alcohol will produce = of isoamyl acetate

Now, calculating the mass of isoamyl acetate from equation 1, we get:

Molar mass of isoamyl acetate = 130.2 g/mol

Moles of isoamyl acetate = 0.0425 moles

Putting values in equation 1, we get:

Experimental yield of isoamyl acetate = 4.828 g

Theoretical yield of isoamyl acetate = 5.534 g

Putting values in above equation, we get:

Hence, the percent yield of the reaction is 87.24 %.

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