Suppose that you obtain 1.0 g of isopentyl acetate from reacting 1.0 ml of acetic acid with 1.0 ml of isopentyl alcohol. calculate a percent yield. (see technique 3.2)

83.72%

Explanation:

The equation of of Reaction is given below;

CH3COOH + C5H11OH > CH3COOC5H11 + H2O. (1).

The equation (1) shows that one mole of acetic acid react with one mole of isopentyl alcohol to give one mole of isopentyl acetate and one mole of water.

RATIO===> 1 : 1 > 1 : 1.

We are given 1.0 mL of acetic acid and 1.0 mL of isopentyl alcohol.

===> Things to know: The density of acetic acid = 1.05 g/cm^3 and the density of isopentyl alcohol= 0.809 g/cm^3, 1 mL = 1 cm^3, molar mass of isopentyl alcohol = 88.179 g/mol.

So, the density of acetic acid = 1.05 g/mL and the density of isopentyl alcohol= 0.809 g/mL.

Therefore, mass of acetic acid;

= 1.0 mL acetic acid ×( 1.05 g/ mL).

= 1.05 g acetic acid.

Mass of isopentyl alcohol= 1.0 mL × 0.809 g/mL = 0.809 grams isopentyl alcohol.

Since one mole of each reactants formed the product, we have to find the limiting reagent. This is done by converting the mass into moles.

Hence, for acetic acid; the number of moles = mass/molar mass.

===> 1.05 grams of acetic acid/ 60.052 g/mol.

===> 0.017485 moles acetic acid.

For isopentyl alcohol; 0.809/ 88.179 grams per mol.

==> 0.0091745 mol isopentyl alcohol.

We can see that isopentyl alcohol is the limiting reagent. So, 0.0091745 mole isopentyl alcohol will produce 0.0091745 mol isopentyl acetate.

Hence, the mass of isopentyl acetate= number of moles × molar mass= 0.0091745 × 130.19 grams per mol.

==> 1.19 grams isopentyl acetate.

FINAL STEP: calculating the percent yield. The actual yield is 1.0 g, theoretical yield = 1.19 grams

Percent yield = actual yield/ theoretical yield × 100.

Percent yied;

= 1.0 grams/ 1.19 grams × 100.

= 83.72 %.

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