In: Chemistry
a) Calculate the theoretical yield of your product. The theoretical yield should be quoted as a mass, not a number of moles. Units must be shown on all numbers that need them. Calculations that are done with an incorrect number of significant figures will be counted as incorrect. Your work must be shown.
b) Calculate the percent yield for your product. This calculation should be done with masses of your product and the theoretical yield, not the number of moles. Calculations that are done with an incorrect number of significant figures will be counted as incorrect.) Your work must be shown.
c) Calculate the atom economy for the epoxidation. (Calculations that are done with an incorrect number of significant figures will be counted as incorrect). Your work must be shown. The equation is:
Carvone + HOOH product + H2O (the hydroxide is catalytic).
carvone mass: .709 g
Mass of organic product: .519g
I don't think you need these numbers: mass of flask: 37.935, mass with carvone in flask: 38.644, mass of empty beaker product was placed into: 49.816, mass of organic product in beaker:50.335
Given reaction is Carvone + HOOH -----------> carvone epoxide + H2O
Molar mass of carvone = 150.2 g/mol
Molar mass of carvone epoxide = 166.2 g/mol
a)
Carvone + HOOH -----------> carvone epoxide + H2O
1 mol 1 mol
150.2 g 166.2 g
0.709 g ?
Then,
? = ( 0.709 g/ 150.2 g ) x 166.2 g of product
= 0.784 g of product
This is the theoretical yield of the product.
Hence, theoretical yield of the product = 0.784 g
b) Given that actual yield of the product = 0.519 g
Hence,
percent yield = ( actual yield of the product / theoretical yield of the product) x 100
= 66.2 %
Therefore,
percent yield of the product = 66.2 %
c)
atom economy for the epoxidation = ( atomic mass of the product/ sum of atomic masses of reactants) x 100
= [ atomic mass of carvone epoxide / atomic mass of (carvone + HOOH)] x 100
= [ 166.2 / (150.2 + 34) ] x 100
= 90.22 %