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19 September, 18:13

If 1.87 g‎ of acetic acid (ch3co2h) ‎ reacts with 2.31 g‎ of isopentylalcohol (c5h12o) ‎ to give‎ 3.20 g of isopentylacetate (c7h14o2) ‎, what is the percent yield of the reaction?

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  1. 19 September, 18:42
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    The reaction of acetic acid with isopentyl alcohol is

    CH3COOH + C5H12O - --> C7H14O2 + H2O

    Thus one mole of acetic acid reacts with one mole of isopentyl alcohol to give one mole isopentyl acetate

    Molar mass of acetic acid = 60.05 g / mole

    Moles of acetic acid used = Mass used / Molar mass = 1.87 / 60.05 = 0.0311

    Molar mass of isopentyl alcohol = 88.15 g / mole

    Moles of isopentyl alcohol = Mass / Molar mass = 2.31 / 88.15 = 0.0262

    Molar mass of isopentyl acetate = 130.19 g / mole

    Moles of isopentyl acetate = 3.20 / 130.19 = 0.0246

    As per equation 0.0262 should of isopentyl alcohol should react with 0.0262 moles of acetic acid to give 0.0262 isopentyl acetate

    Thus theoretical yield = 0.0262 moles

    Actual yield = 0.0246

    % yield = Actual yield X 100 / theoretical yield = 0.0246 X 100 / 0.0262 = 93.89%
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