Suppose a student started with 127 mg of trans-cinnamic acid and 0.520 mL of a 1

Question

Suppose a student started with 127 mg of trans-cinnamic acid and 0.520 mL of a 10% (v/v) bromime solution, and after the reaction and workup, ended up with 0.220 g of brominated product. Calculate the student's theoretical and percent yields. NOTE: The density of the 10% bromine solution is 3.103 g/mL The formula mass of bromine is 159.8 g/mol The formula mass of trans-cinnamic acid is 148.2 g/mol The formula mass of 2,3-bromo-3-phenylpropanoic acid product is 307.9 g/mol Please show me how to do this step by step. I do not know how to do it at all :(

Explanation / Answer

127 mg cinamic acid = (0.127/148.2) = 0.000857 0.22 brominated cinamic acid = (0.22/307.9) =0.0007145 moles 0.52 ml Br2 = 0.52 x3.103 =1.6135 gms = (1.6135/159.8) = 0.01 moles in general 1 mole cinamic acid+ 1mole Br2 ---> 1 mole brominated product based on calculated molar quantities expected brominated product =0.000857 moles = 0.000857 x307.9 = 0.26387 gms theoretical yield = 0.26387 gm % yield = (0.22/0.26387) x100 = 83.37 %

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