With 0.304g of Methyl salicylate, 0.22g of salicylic acid wasproduced. C 8 H 8 O

Question

With 0.304g of Methyl salicylate, 0.22g of salicylic acid wasproduced. C8H8O3 (Methylsalicylate) + 2NaOH --> Disodium salicylate + CH3OH +H2O Disodium salicylate + H2SO4 -->C7H6O3(salicylic acid) +Na2SO4 What is the percent yield? (Molar mass of Methyl salicylate:152.15 g/mol, Molar mass of salicylic acid: 138.12 g/mol) With 0.304g of Methyl salicylate, 0.22g of salicylic acid wasproduced. C8H8O3 (Methylsalicylate) + 2NaOH --> Disodium salicylate + CH3OH +H2O Disodium salicylate + H2SO4 -->C7H6O3(salicylic acid) +Na2SO4 What is the percent yield? (Molar mass of Methyl salicylate:152.15 g/mol, Molar mass of salicylic acid: 138.12 g/mol) What is the percent yield? (Molar mass of Methyl salicylate:152.15 g/mol, Molar mass of salicylic acid: 138.12 g/mol)

Explanation / Answer

percent yield = actual yield/ theoretical yeild x 100% actual yield = 0.22g theoretical yield: 0.304g x mol/ 152.15g = 0.001998moles of methly salicylatereacts with sodium hydroxide. therefore 0.001998 moles of disodium salicylate is produced.0.001998 moles of sodium salicylate reacts withH2SO4 to produce 0.00198 moles of salicylicacid. to find the theoretical yield. find the mass in grams ofsalicylic acid. 0.001998moles x 138.1g/mol = 0.276grams so % yield = 0.22/0.276 x 100% = 79.7% hope that helps

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