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Unformatted text preview: Sheet1 Page 1 (26 points) The local anesthetic piperocaine was awarded U.S. patent 1,784,903 in 1930. In preparing for the patent applica Determine the empirical formula of piperocaine from the data. To simplify the calculations, work with 100 g of the drug. Any amount will work as an empirical formula gives a ratio of nu 67.17 g C /12.011 g/mol = 5.592 mol C 8.86 g H/1.00794 g/mol = 8.79 mol H 11.19 g N/14.0067 g/mole = 0.799 mol N 12.78 g O/15.9994 g/mol = 0.799 mol O The ratio is 7:11:1:1 and the empirical formula is C7H11NO. In a separate determination, the molecular weight of piperocaine was found to be 2.50 x 102 g/mole. Determine the mole Calculate the molecular weight based on the empirical formula. MW = (7 x 12.011) + (11 x 1.00794) + (1 x 14.00674) + (1 x 15.9995) = 125.17 g/mol 125.17n = 250 With today's instrumentation, the molecular weight of piperocaine can be easily determined with a sample size of 250 ng 250 ng = 250 x 10-9 g. The number of moles of the drug equals 2.5 x 10-7/2.50 x 102 g/mole = 1.0 x 10-9 mole. Multiply (26 points) A medicinal chemist working for the competition prepared a related local anesthetic dibucaine hydrochloride. Th C16H21N2O3 + 2 C2H5Cl = C20H30ClN2O3 + HCl The chemist performing the synthesis decided to use stoichiometric amounts of the two reactants, i.e. neither reactant is in Calculate the mass of C2H5Cl, required for the synthesis....
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This note was uploaded on 07/21/2011 for the course CHEM 111 taught by Professor Sawyers during the Spring '11 term at Virginia College.
- Spring '11