Calculate the theoretical yield for the bromination of bothstilbenes and cinnamic acid, assuming the presence of excesspyridinium tribromide.
cinnamic acid - 150 mg
cis-stilbene - 100 μL
trans- stilbene - 100 mg
pyridinium tribromide - 200-385 mg
Write the equation and follow these steps. The example given is NOT organic; however, that won't change a thing. Just follow the steps.
http://www.jiskha.com/science/chemistry/stoichiometry.html
To calculate the theoretical yield for the bromination of bothstilbenes and cinnamic acid, we need to consider the balanced equation for the reaction and use stoichiometry to determine the amount of product that can be formed.
The balanced equation for the bromination of bothstilbenes and cinnamic acid can be written as follows:
Stilbene or Cinnamic Acid + Pyridinium tribromide → Brominated product + Pyridinium bromide
We are assuming the presence of excess pyridinium tribromide, which means that all the stilbene or cinnamic acid will react completely and be converted into the brominated product. Therefore, the limiting reagent in this case is the stilbene or cinnamic acid.
Let's calculate the theoretical yield for each compound:
1. Cinnamic Acid:
Given: 150 mg of cinnamic acid
We need to calculate the molar amount of cinnamic acid:
Molar mass of cinnamic acid = 148.16 g/mol
Moles of cinnamic acid = (mass of cinnamic acid) / (molar mass of cinnamic acid)
= (150 mg) / (148.16 g/mol)
= 1.013 x 10^-3 mol
Since the stoichiometry of the reaction states that 1 mole of cinnamic acid reacts with 1 mole of pyridinium tribromide to form 1 mole of the brominated product, the theoretical yield of the brominated product (in moles) is also 1.013 x 10^-3 mol.
2. cis-Stilbene:
Given: 100 μL of cis-stilbene
We need to convert the volume of cis-stilbene to mass:
Density of cis-stilbene = 0.940 g/mL (approx.)
Mass of cis-stilbene = (volume of cis-stilbene) x (density of cis-stilbene)
= (100 μL) x (0.940 g/mL)
= 0.094 g
We can now calculate the molar amount of cis-stilbene:
Molar mass of cis-stilbene = 180.23 g/mol
Moles of cis-stilbene = (mass of cis-stilbene) / (molar mass of cis-stilbene)
= (0.094 g) / (180.23 g/mol)
= 5.22 x 10^-4 mol
The stoichiometry of the reaction still applies, so the theoretical yield of the brominated product (in moles) is 5.22 x 10^-4 mol.
3. trans-Stilbene:
Given: 100 mg of trans-stilbene
We can calculate the molar amount of trans-stilbene:
Molar mass of trans-stilbene = 180.23 g/mol
Moles of trans-stilbene = (mass of trans-stilbene) / (molar mass of trans-stilbene)
= (100 mg) / (180.23 g/mol)
= 5.54 x 10^-4 mol
Similarly, the theoretical yield of the brominated product (in moles) is 5.54 x 10^-4 mol.
Keep in mind that these calculations assume 100% yield, which is not always achievable in practice. The actual yield may be lower due to side reactions, incomplete reactions, or other factors.
To calculate the theoretical yield for the bromination of both stilbenes and cinnamic acid, we need to determine the limiting reactant. The reactant that produces the least amount of product is the limiting reactant.
1. For cinnamic acid:
Given: cinnamic acid - 150 mg
We do not have the molar mass of cinnamic acid, so we cannot directly convert it to moles. However, we can assume that the density of cinnamic acid is approximately 1 g/mL, which means 150 mg is equal to 0.15 g.
2. For cis-stilbene:
Given: cis-stilbene - 100 μL, trans-stilbene - 100 mg
We need to convert the given volume (μL) of cis-stilbene to grams. However, we do not have the density of cis-stilbene. Let's assume the density of cis-stilbene is approximately 1 g/mL (similar to water). Therefore, 100 μL is equal to 0.1 mL, which is equal to 0.1 g.
3. Convert the amounts of pyridinium tribromide to grams:
Given: pyridinium tribromide - 200-385 mg
We'll consider both the minimum and maximum amounts provided:
- Minimum amount: 200 mg = 0.2 g
- Maximum amount: 385 mg = 0.385 g
4. Convert the amounts of reactants to moles:
To determine the limiting reactant, we need to convert the masses of each reactant to moles using their respective molar masses.
- Cinnamic acid: We'll assume a molar mass of 150 g/mol for cinnamic acid.
Moles of cinnamic acid = 0.15 g / 150 g/mol = 0.001 mol
- Cis-stilbene: We'll assume a molar mass of 180 g/mol for cis-stilbene.
Moles of cis-stilbene = 0.1 g / 180 g/mol ≈ 0.000556 mol
- Trans-stilbene: We'll assume a molar mass of 180 g/mol for trans-stilbene.
Moles of trans-stilbene = 0.1 g / 180 g/mol ≈ 0.000556 mol
5. Determine the limiting reactant:
Comparing the moles of each reactant, we see that cis-stilbene and trans-stilbene have the same number of moles, which is the lowest. Therefore, either cis-stilbene or trans-stilbene can be the limiting reactant.
6. Calculate the theoretical yield:
Since both cis-stilbene and trans-stilbene can be the limiting reactant, we will calculate the theoretical yield for both.
For cis-stilbene:
Using the balanced chemical equation for the bromination of cis-stilbene, we know that 1 mol of cis-stilbene reacts with 3 moles of pyridinium tribromide to produce 1 mol of product.
- Theoretical yield = 0.000556 mol (limiting reactant) * (1 mol product / 3 mol cis-stilbene) = 0.000185 mol
For trans-stilbene:
Using the balanced chemical equation for the bromination of trans-stilbene, we know that 1 mol of trans-stilbene reacts with 2 moles of pyridinium tribromide to produce 1 mol of product.
- Theoretical yield = 0.000556 mol (limiting reactant) * (1 mol product / 2 mol trans-stilbene) = 0.000278 mol
7. Convert the theoretical yields to the desired units:
- For cis-stilbene: Theoretical yield = 0.000185 mol * 104.15 g/mol (molar mass of product) ≈ 0.0193 g
- For trans-stilbene: Theoretical yield = 0.000278 mol * 104.15 g/mol (molar mass of product) ≈ 0.029 g
Therefore, the theoretical yield for the bromination of cis-stilbene is approximately 0.0193 g (19.3 mg) and for trans-stilbene is approximately 0.029 g (29 mg).