Answer:
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%
