Theoretical Yield in Multi-Step Synthesis

In summary, the theoretical yield of methyl m-nitrobenzoate from 10.0 g of benzoic acid can be calculated by using the formulas given in the problem and converting between moles and grams. The final answer is 14.83 g of methyl m-nitrobenzoate.
  • #1
Soaring Crane
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Homework Statement



Exactly how do you find the theoretical yield of methyl m-nitrobenzoate from 10.0 g of benzoic acid?


Homework Equations



1) Ph-COOH + CH3OH + H2SO4 --> Ph-COOCH3
2) Ph-COOCH3 + HNO3 + H2SO4 --> methyl m-nitrobenzoate



The Attempt at a Solution



10.0 g benzoic acid*(1 mol benzoic acid/122.122 g benzoic acid)*(1 mol methyl benzoate/1 mol benzoic acid)*(136.149 g methyl benzoate/1 mol methyl benzoate) = 11.15 g methyl benzoate from FIRST STEP as theoretical yield

11.15 g methyl benzoate*(1 mol methyl benzoate/136.149 g methyl benzoate)*(1 mol methyl m-nitrobenzoate/1 mol methyl benzoate)*(181.147 g methyl m-nitrobenzoate/1 mol methyl m nitrobenzoate) = 14.83 g methyl m-nitrobenzoate from 10.0 g benzoic acid as theoretical yield?

Thanks.
 
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  • #2
That looks good if the formula wts. you give are accurate. I didn't check.
Also, you could have just multiplied the number of moles of benzoic acid by the formula wt. of m-nitromethylbenzoate.
 
  • #3


I would like to commend you for your attempt at solving this problem. Your approach is correct, and the equations you have used are appropriate for finding the theoretical yield of methyl m-nitrobenzoate from 10.0 g of benzoic acid. However, I would like to point out a few things that may help you improve your solution.

Firstly, when converting from grams to moles, it is important to use the molar mass of the specific compound, rather than the generic molar mass of the functional group. In this case, the molar mass of benzoic acid is 122.122 g/mol, while the molar mass of methyl benzoate is 136.149 g/mol. Therefore, the first step would be:

10.0 g benzoic acid*(1 mol benzoic acid/122.122 g benzoic acid) = 0.0819 mol benzoic acid

Similarly, for the second step, the correct conversion would be:

11.15 g methyl benzoate*(1 mol methyl benzoate/136.149 g methyl benzoate) = 0.0819 mol methyl benzoate

This is important because using the incorrect molar mass can lead to errors in the final result.

Secondly, when calculating the theoretical yield in a multi-step synthesis, it is important to consider the limiting reagent. In this case, the limiting reagent would be benzoic acid, as it is the starting material with the lowest number of moles. Therefore, the theoretical yield of methyl m-nitrobenzoate would be 11.15 g, as calculated in the first step.

Overall, your approach is correct, but it is important to use the correct molar masses and consider the limiting reagent when calculating the theoretical yield in a multi-step synthesis. Keep up the good work!
 

1. What is theoretical yield in multi-step synthesis?

Theoretical yield in multi-step synthesis refers to the maximum amount of a product that can be obtained from a chemical reaction, based on the stoichiometry of the reactants and assuming perfect conditions and complete conversion of all reactants into products.

2. How is theoretical yield calculated in multi-step synthesis?

Theoretical yield is calculated by multiplying the amount of the limiting reactant by the stoichiometric ratio between the limiting reactant and the desired product in each step of the synthesis. This is done for each step of the reaction and then the final theoretical yields are added together to get the overall theoretical yield.

3. What is the significance of theoretical yield in multi-step synthesis?

Theoretical yield serves as a benchmark for the actual yield obtained in a chemical reaction. It helps to determine the efficiency of the reaction and identify any sources of error. It also allows for the calculation of percent yield, which is a measure of the actual yield compared to the theoretical yield.

4. What factors can affect the theoretical yield in multi-step synthesis?

Theoretical yield can be affected by a variety of factors, including the purity and quality of the reactants, the reaction conditions (temperature, pressure, etc.), and the efficiency of the reaction steps. Any errors or variations in these factors can result in a lower or higher theoretical yield than expected.

5. Can the theoretical yield ever be achieved in real-life multi-step syntheses?

No, it is highly unlikely that the theoretical yield will ever be achieved in real-life multi-step syntheses. This is because there are always sources of error and inefficiencies in chemical reactions, and achieving a perfect conversion of all reactants into products is nearly impossible. However, the closer the actual yield is to the theoretical yield, the more efficient the reaction is considered to be.

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