Question

Given 7.75 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100% yield? express your answer in grams to three significant fi

Answer 1

10.2 grams of ethyl butyrate synthesized. The balanced equation for the reaction of butanoic acid (C4H8O2) with ethanol (C2H6O) to produce ethyl butyrate (C6H12O2) is: C4H8O2 + C2H6O ==> C6H12O2 + H2O So for each mole of C4H8O2 used, 1 mole of C6H12O2 will be produced. So let's calculate the reactant and product molar masses. Start by looking up the atomic weights of the involved elements: Atomic weight carbon = 12.0107 Atomic weight hydrogen = 1.00794 Atomic weight oxygen = 15.999 Molar mass C4H8O2 = 4 * 12.0107 + 8 * 1.00794 + 2 * 15.999 = 88.10432 g/mol Molar mass C6H12O2 = 6 * 12.0107 + 12 * 1.00794 + 2 * 15.999 = 116.15748 g/mol Moles C4H8O2 = 7.75 g / 88.10432 g/mol = 0.087963905 mol Mass C6H12O2 = 0.087963905 mol * 116.15748 g/mol = 10.21766549 g Rounding to 3 significant figures gives 10.2 grams of ethyl butyrate synthesized.