In an experiment, 13g of methanol and 11 g of carbon monoxide were placed in a reaction vessel. Given the following reaction
C:12.0 amu O:16.0 amu H:1.0 amu
CH3OH + CO --> HC2H3O2
1 What is the theoretical yield of acetic acid?
2 If the actual yield of acetic acid is 19.1 g, what is the percent yield?
See your previous problem. This is not a LR problem BUT in the previous problem you calculated TWO problems like this while this is just one of those problems. The theoretical yield is the mols produced by that one calculation. We'll call that TY. The actual yield (AY) is 19.2g.
% yield = (AY/TY)*100 = ?
To find the theoretical yield of acetic acid (HC2H3O2), we need to calculate the molar masses of methanol (CH3OH) and carbon monoxide (CO), determine the limiting reactant, and use stoichiometry to find the maximum amount of acetic acid that can be produced.
1. Calculating Molar Mass:
- Methanol (CH3OH):
- C: 12.0 g/mol
- H: 1.0 g/mol x 4 = 4.0 g/mol
- O: 16.0 g/mol + 1.0 g/mol = 17.0 g/mol (1 oxygen in alcohol group)
Total molar mass of CH3OH = 12.0 g/mol + 4.0 g/mol + 17.0 g/mol = 33.0 g/mol
- Carbon Monoxide (CO):
- C: 12.0 g/mol
- O: 16.0 g/mol
Total molar mass of CO = 12.0 g/mol + 16.0 g/mol = 28.0 g/mol
2. Determine the Limiting Reactant:
To determine the limiting reactant, we compare the number of moles of each reactant to the stoichiometric ratio in the balanced equation. The reactant that produces fewer moles of acetic acid is the limiting reactant.
- Moles of Methanol (CH3OH):
Moles of CH3OH = Mass of CH3OH / Molar mass of CH3OH
Moles of CH3OH = 13g / 33.0 g/mol ≈ 0.394 mol
- Moles of Carbon Monoxide (CO):
Moles of CO = Mass of CO / Molar mass of CO
Moles of CO = 11g / 28.0 g/mol ≈ 0.393 mol
Since CO has the fewer moles, it is the limiting reactant.
3. Stoichiometry:
We will use the mole ratio of CO and acetic acid (from the balanced equation) to find the maximum moles of acetic acid that can be produced.
From the balanced equation: CH3OH + CO → HC2H3O2
The molar ratio of CO to HC2H3O2 is 1:1.
Maximum moles of HC2H3O2 that can be produced = Moles of CO ≈ 0.393 mol
4. Calculate Theoretical Yield:
To find the theoretical yield in grams, we multiply the moles of acetic acid by its molar mass.
Theoretical Yield of HC2H3O2 = Moles of HC2H3O2 × Molar mass of HC2H3O2
Theoretical Yield of HC2H3O2 = 0.393 mol × (12.0 g/mol + 2(1.0 g/mol) + 3(16.0 g/mol))
Theoretical Yield of HC2H3O2 = 0.393 mol × 60.0 g/mol ≈ 23.6 g
Therefore, the theoretical yield of acetic acid (HC2H3O2) is approximately 23.6 grams.
Now, let's move on to the second part of the question.
2. Percent Yield:
The percent yield can be calculated using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) × 100
Given that the actual yield of acetic acid is 19.1 grams, we can substitute the values in the formula:
Percent Yield = (19.1 g / 23.6 g) × 100
Percent Yield ≈ 80.93%
Therefore, the percent yield of acetic acid is approximately 80.93%.