Suppose a student started with 135 mg of trans-cinnamic acid and 0.52 mL of a 10% (v/v) bromime solution, and after the reaction and workup, ended up with 0.202 g of brominated product. Calculate the student's theoretical and percent yields.
I gave instructions for solving this last night (late). Same problem but different numbers. Just change the numbers.
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To calculate the theoretical yield and percent yield, we first need to determine the limiting reactant for the reaction. The limiting reactant is the one that will be completely consumed and thus determines the maximum amount of product that can be obtained.
In this case, we have two reactants: trans-cinnamic acid and bromine solution. We will compare the moles of each reactant to determine the limiting reactant.
1. Convert the mass of trans-cinnamic acid to moles:
To do this, we need the molar mass of trans-cinnamic acid, which is 148.16 g/mol.
Moles = Mass / Molar mass = 0.135 g / 148.16 g/mol = 0.000912 mol
2. Convert the volume of bromine solution to moles:
Since the bromine solution is given as a percentage, we need to determine the mass of bromine in the solution first.
Mass of bromine = Volume of solution (mL) * Density of solution (g/mL) * Concentration of bromine (%)
Let's calculate the mass of bromine:
Mass of bromine = 0.52 mL * 1 g/mL * 0.10 = 0.052 g
Next, we need the molar mass of bromine, which is 79.90 g/mol.
Moles = Mass / Molar mass = 0.052 g / 79.90 g/mol = 0.000651 mol
3. Compare the moles of each reactant to determine the limiting reactant:
The ratio of trans-cinnamic acid to bromine is 1:1, so the reactant with the lower number of moles is the limiting reactant.
In this case, trans-cinnamic acid has 0.000912 mol and bromine has 0.000651 mol.
Since bromine has fewer moles, it is the limiting reactant.
4. Calculate the theoretical yield:
The molar ratio between bromine and the brominated product is also 1:1.
Therefore, the moles of the brominated product will be the same as the moles of bromine.
Moles of brominated product = 0.000651 mol
Finally, we need to convert moles of the brominated product to grams using its molar mass.
The molar mass of the brominated product is unknown, but we can calculate it using its mass and the given mass of brominated product.
The mass of the brominated product is 0.202 g.
Molar mass = Mass / Moles = 0.202 g / 0.000651 mol = 310.76 g/mol
The theoretical yield is therefore 0.202 g.
5. Calculate the percent yield:
Percent yield = (Actual yield / Theoretical yield) * 100
In this case, the actual yield is given as 0.202 g.
Percent yield = (0.202 g / 0.202 g) * 100 = 100%
Therefore, the student's theoretical yield is 0.202 g, and the percent yield is 100%.