Goil of wintergreen, c8h8o3 is prepared when methanol, ch3oh, is reacted with salicylic acid, c7h6o3. what is the percent yield of a reaction in which 15.3 g of oil of wintergreen is collected after 16.3 g salicylic acid is reacted with an excess of methanol? ch3oh + c7h6o3 c8h8o3 + h2o

85.2% yield. Determine the molar masses of salicylic acid and oil of wintergreen, by first 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 C7H6O3 = 7 * 12.0107 + 6 * 1.00794 + 3 * 15.999 = 138.11954 g/mol Molar mass C8H8O3 = 8 * 12.0107 + 8 * 1.00794 + 3 * 15.999 = 152.14612 g/mol Now determine number of moles of each. Moles C7H6O3 = 16.3 g / 138.11954 g/mol = 0.118013715 mol Moles C8H8O3 = 15.3 g / 152.14612 g/mol = 0.100561224 mol Looking at the balanced equation CH3OH + C7H6O3 ==> C8H8O3 + H2O it's obvious that for every mole of C7H6O3 used, one mole of C8H8O3 should be produced. But the actual production is lower. To figure the percent yield, just divide the moles produced by the expected production. So 0.100561224 / 0.118013715 = 0.852114721 = 85.2114721% Since we have 3 significant figures, round to 85.2%

The answer would be solvent

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