Please help!! This is a question about an ochem lab multipstep synthesis involvi

Question

Please help!! This is a question about an ochem lab multipstep synthesis involving the preparation of 4,4-diphenyl-3-buten-2-one. The first step created a product of ethyl acetoacetate ethylene ketal with toluene as one of the reagents. In step 2, this product is used as a reactant and we need to calculate how much of this ethyl acetoacetate ethylene ketal needs to be used if we need 0.07 moles of it. However, our ethyl acetoacetate ethylene ketal had 4.06% leftover toluene in it. Can you please help with how to calculate how much ethyl acetoacetate ethylene ketal we need in mL for step 2 taking into account the leftover toluene???

so....

-Step 1 actual yield of ethyl acetoacetate ethylene ketal product: 21.24 g.

-this contains 4.06% leftover toluene (this was found via a calculation of a mole fraction (percent))(found from an NMR analysis)

-need 0.07 moles of ethyl acetoacetate ethylene ketal product

-how many mL does this come out to???

thank you!!!!

Explanation / Answer

Molecular wegith of Ethyl acetoacetate Etylene ketal = 174.2 gm/ mol

Required quantity is 0.07mol, so the actual amount of Ketal required - 174.2 X 0.07 = 12.19 gm

Now, You have 4.06% leftover toluene. Therefore

1gm step 1 material = 1-0.0406 gm = 0.9594 gm Ketal

Thus 1 gm step 1 material = 0.9594 gm Ketal

x gm step 1 material = 12.19 gm Ketal

Solving this, x= 12.19 / 0.9594 = 12.70 gm step 1 material must be taken

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects