4. Reaction Scheme- Fil in the Blanks (3 points): 2 OMe 10% Pd/C OMe MeOH Pd/C M

Question

4. Reaction Scheme- Fil in the Blanks (3 points): 2 OMe 10% Pd/C OMe MeOH Pd/C MeOH MW or Conc.296.49 2.02 10% 3.0 mL volume density mass mmole equivalents 0.7 mL2 L Theoretical Yield 350 mg 15 mg Theoretical Yield catalyst solvent *The methyl oleate used in this experiment is only 70% pure (0.5 g of liquid only contains 0.35 g of methyl oleate) **22.4 L/mole gas @ STP. Assume STP 5. a) Show your calculations for determining the maximum possible amount maximum theoretical yield) of methyl stearate that could be formed in the reaction above. (1 point) b) What is the percentage yield if 237 mg of methyl stearate is produced from the reaction above (Show calculations) (1 point)

Explanation / Answer

The purity of the methyl oleate is 70 % only

So 350 mg contain = 70 / 100 x 350 = 245 mg.

So the actual methyl oleate going to react is 245 mg.

one molecule of methyl oleate will give one molecule of methyl stearate.

that means, one mole of methyl oleate will give one mole of methyl stearate.

to calculate the number of moles of methyl oleate in this reaction is

= weight in gram / molecular weight = 0.245 / 296.49 = 0.000826335 mols.

0.000826335 mols of methyl oleate will give 0.000826335 mols of methyl stearate.

To calculate the theoratical yield of this reaction = molecular weight of the product x Moles of product

Molecular weight of methyl stearate = 298.511

So Theoratical yield = 298.511 x 0.000826335 = 0.24667 g = 246.67 mg.

To calculate the percetage yield for 237 mg is

= 237 / 246. 67 x 100 = 96.08 %

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