Suppose a student started with 133 mg of trans-cinnamic acid and 0.50 mL of a 10% (v/v) bromime solution, and after the reaction and workup, ended up with 0.236 g of brominated product. Calculate the student's theoretical and percent yields.
To calculate the theoretical yield and percent yield, we need to first understand the reaction and determine the limiting reagent. Let's break down the steps:
1. Calculate the moles of trans-cinnamic acid:
- Given: mass of trans-cinnamic acid = 133 mg = 0.133 g
- To calculate moles, divide mass by the molar mass.
- The molar mass of trans-cinnamic acid (C9H8O) is approximately 148.17 g/mol.
- Moles of trans-cinnamic acid = 0.133 g / 148.17 g/mol
2. Calculate the moles of bromine:
- Given: volume of bromine solution = 0.50 mL
- To calculate moles, we need to convert the volume to volume percent.
- Given: 10% (v/v) bromine solution means 10 mL of bromine per 100 mL of solution.
- Moles of bromine = (volume of bromine solution / 100 mL) * 10 mL * density of bromine * molar mass of bromine
3. Determine the limiting reagent:
- The reactant that produces the least amount of product is the limiting reagent.
- Compare the moles of trans-cinnamic acid and bromine to determine the limiting reagent.
4. Calculate the theoretical yield:
- Once we know the limiting reagent, we can calculate the moles of product formed using the stoichiometry of the balanced equation:
Cinnamic acid + Bromine → Brominated product
1 mole + 1 mole → 1 mole
5. Calculate the percent yield:
- The percent yield is the actual yield divided by the theoretical yield, multiplied by 100%.
- Given: mass of brominated product = 0.236 g
Now, let's proceed with the calculations:
1. Moles of trans-cinnamic acid = 0.133 g / 148.17 g/mol = 0.000899 moles
2. Moles of bromine = (0.50 mL / 100 mL) * 10 mL * density of bromine * molar mass of bromine
Let's assume the density of bromine is 3.12 g/mL and the molar mass of bromine is 79.90 g/mol:
Moles of bromine = (0.50 / 100) * 10 * 3.12 g/mL / 79.90 g/mol = 0.0196 moles
3. From step 2, we can see that there is an excess of bromine (0.0196 moles) compared to trans-cinnamic acid (0.000899 moles). Therefore, bromine is the limiting reagent.
4. The stoichiometry of the reaction tells us that 1 mole of bromine corresponds to 1 mole of the brominated product. Therefore, the theoretical yield is 0.0196 moles.
5. Percent yield = (actual yield / theoretical yield) * 100%
Actual yield = 0.236 g
Theoretical yield = 0.0196 moles * molar mass of the brominated product
After obtaining the molar mass of the brominated product, you can calculate the theoretical yield and percent yield using the values mentioned above.
Note: I couldn't provide the molar mass of the brominated product as it's not given in the question. Please find the molar mass of the brominated product and substitute it into the calculations to obtain the final answer.