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

See explanation below

Explanation:

First, we need to know the formula of the carvone, which is C₁₀H₁₄O (MM = 150.2 g/mol) and the product which is C₁₀H₁₄O₂ (MM = 166.2 g/mol).

a) We have the initial mass of carvone which is 0.709 g. With this mass, and assuming a mole ratio of 1:1, we can calculate the theorical moles of the product:

C₁₀H₁₄O + HOOH > C₁₀H₁₄O₂ + H₂O

Let's calculate the moles of carvone:

n = m/MM

n = 0.709/150.2 = 4.72x10⁻³ moles

As we stated before, we have a 1:1 mole ratio, so the moles of carvone will be the mole of the products, so the moles of the carvone epoxide are 4.72x10⁻³ moles. To get the theorical yield, we just use the molecular mass of the carvone epoxide:

m = 4.72x10⁻³ * 166.2 = 0.784 g

This would be the theorical yield of the product

b) To get the percent of yield, we use the mass of the theorical yield and the actual mass obtained in the experiment:

%yield = exp yield / theo yield * 100

Replacing:

%yield = 0.519/0.784 * 100

%yield = 66.2 %

c) to get the atom economy, we just apply the following expression:

%AE = MM of desired product / MM of reactants * 100

In this case we already have the molecular mass of the product and one reactant. We only need to know the molecular mass of the HOOH which is 34 g/mol. Applying the formula we have:

%AE = 166.2 / (150.2+34) * 100

%AE = 90.2%

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