if you could explain how you got your answer, that will be very much appreciated

Question

if you could explain how you got your answer, that will be very much appreciated thanks!

Suppose a student started with 129 mg of trans-cinnamic acid and 0.500 ml of a 10% (v/v) bromime solution, and after the reaction and workup, ended up with 0.191 g of brominated product. Calculate the student's theoretical and percent yields. 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

Explanation / Answer

moles of trans cinamic acid = 0.129/148.2 =0.00087 bromine mas = 0.5 x3.103 =1.5515 gms moles of bromine = 1.5515/159.8 = 0.0097 moles of product = 0.191/307.9 = 0.00062 1 mole trans cn acid reacts with 1 mole bromine to give 1 mole 2,3 dibromo product since trans cinamic acid is present =0.00087 moles expected moles of product = 0.00087 mass of expected (theoretical) = 0.00087 x307.9 = 0.2678 gm theoretical yield = 0.2678 gm % yield = (actualmass/theoretical) x100 = (0.191/0.2678) x100 = 71.32

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