Calculate the percent yield if 104 g of CH3CO2CH2CH3 is produced by the reaction of 111 g of CH3CO2H according to the following equation:
CH3CO2H(l) + CH3CH2OH(l) �¨ CH3CO2CH2CH3(l) + H2O(l)
Here is a worked example of a stoichiometry problem.
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To calculate the percent yield, you need to have both the actual yield and the theoretical yield of the reaction.
1. Calculate the molar mass of CH3CO2H (acetic acid) and CH3CO2CH2CH3 (ethyl acetate).
- Molar mass of CH3CO2H:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of CH3CO2H = (2 x C) + (4 x H) + (2 x O) = (2 x 12.01) + (4 x 1.01) + (2 x 16.00) = 60.05 g/mol
- Molar mass of CH3CO2CH2CH3:
C = 12.01 g/mol
H = 1.01 g/mol
O = 16.00 g/mol
Molar mass of CH3CO2CH2CH3 = (6 x C) + (12 x H) + (2 x O) = (6 x 12.01) + (12 x 1.01) + (2 x 16.00) = 88.15 g/mol
2. Calculate the theoretical yield.
Assuming that the reaction goes to completion, the theoretical yield is based on the stoichiometry of the balanced equation.
According to the balanced equation:
1 mole of CH3CO2H produces 1 mole of CH3CO2CH2CH3.
So, the theoretical yield of CH3CO2CH2CH3 is equal to the number of moles of CH3CO2H used.
To calculate the number of moles of CH3CO2H:
moles = mass / molar mass
moles = 111 g / 60.05 g/mol = 1.849 mol
Therefore, the theoretical yield of CH3CO2CH2CH3 is 1.849 mol.
To calculate the theoretical yield in grams:
theoretical yield = moles x molar mass
theoretical yield = 1.849 mol x 88.15 g/mol = 162.86 g
3. Calculate the percent yield.
Percent yield = (actual yield / theoretical yield) x 100
Given that 104 g of CH3CO2CH2CH3 is obtained, the actual yield is 104 g.
Percent yield = (104 g / 162.86 g) x 100 = 63.78%
Therefore, the percent yield of CH3CO2CH2CH3 in this reaction is approximately 63.78%.
To calculate the percent yield, we need to determine the theoretical yield and the actual yield. The theoretical yield is the maximum amount of product that can be obtained according to the balanced chemical equation, while the actual yield is the amount of product obtained in the experiment.
First, we need to determine the limiting reactant. The limiting reactant is the reactant that is completely consumed and determines the amount of product that can be formed. To find the limiting reactant, we need to compare the amount of product that can be formed from each reactant.
The balanced chemical equation shows that 1 mole of CH3CO2H reacts with 1 mole of CH3CH2OH to produce 1 mole of CH3CO2CH2CH3. Therefore, the molar ratio between CH3CO2H and CH3CO2CH2CH3 is 1:1.
1. Calculate the number of moles of CH3CO2H using its molar mass:
Moles of CH3CO2H = mass of CH3CO2H (g) / molar mass of CH3CO2H
Molar mass of CH3CO2H = 60.05 g/mol
Moles of CH3CO2H = 111 g / 60.05 g/mol = 1.84 mol (rounded to 2 decimal places)
2. Calculate the number of moles of CH3CH2OH using its molar mass:
Moles of CH3CH2OH = mass of CH3CH2OH (g) / molar mass of CH3CH2OH
Molar mass of CH3CH2OH = 46.07 g/mol
Moles of CH3CH2OH = 111 g / 46.07 g/mol = 2.41 mol (rounded to 2 decimal places)
Since the molar ratio is 1:1, the limiting reactant is CH3CO2H, as it has fewer moles available for the reaction compared to CH3CH2OH.
3. Calculate the theoretical yield.
Theoretical yield = moles of limiting reactant * molar mass of CH3CO2CH2CH3
Molar mass of CH3CO2CH2CH3 = 88.11 g/mol
Theoretical yield = 1.84 mol * 88.11 g/mol = 162 g (rounded to 3 decimal places)
4. Calculate the percent yield.
Percent yield = (actual yield / theoretical yield) * 100
Actual yield = 104 g (given)
Percent yield = (104 g / 162 g) * 100 = 64.2% (2 decimal places)
Therefore, the percent yield of CH3CO2CH2CH3 in this reaction is 64.2%.