Suppose a student started with 128 mg of trans-cinnamic acid and 0.52 mL of a 10% (v/v) bromine solution, and after the reaction and workup, ended up with 0.184 g of brominated product. Calculate the student's theoretical and percent yields.

what are the equations and do i need to convert anything?

Here is a link showing cinnamic acid. Br2 adds across the double bond.

http://en.wikipedia.org/wiki/Cinnamic_acid

This is a limiting reagent problem; I now because values are given for BOTH reactants.
1. Convert 128 mg cinnamic acid to mols. mols = grams/molar mass
2. Convert the bromine to mols.

3a. Using the coefficients in the balanced equation, convert mols cinnamic acid to mols of the poduct.
3b. Do the same for mols Br2.
3c.It is likely that the mols for 3a and 3b will be different which means one of them is wrong.
3d. The correct value in limiting regent problems is ALWAYS the smaller value and the reagent providing that value is the limiting reagent.

4. g product = mols x molar mass. This is the theoretical yield (TY)

5. %yield = (actual yield/TY)*100 = ?

how do i find the limiting reagent without a reaction? is everything 1 mol to 1 mol?

i found 0.0008693 mols for the cinnamic acid
and 0.0023 mols for Br
that means the cinnamic is the limiting reagent

so i converted and got 0.069 g Br as the theoretical but that doesnt make any sense because shouldnt the theoretical be higher than the actual? (the actual is 0.184 g Br)

i dono why im having so much trouble, thanks for the help

The product isn't bromine, it's 2 3-dibromo-3-phenylpropanoic acid. The MM for the product should be 307.97g.

To calculate the theoretical and percent yields, we first need to understand the chemical equation for the reaction and determine any necessary conversions.

The chemical equation for the reaction between trans-cinnamic acid (C9H8O2) and bromine (Br2) can be represented as follows:

C9H8O2 + Br2 → brominated product

Based on the given information, we have 128 mg (or 0.128 g) of trans-cinnamic acid and 0.52 mL of a 10% (v/v) bromine solution.

Now, let's address the conversions:
1. Convert the mass of trans-cinnamic acid from mg to g.
128 mg = 0.128 g

2. For the bromine solution, we need to determine the mass of bromine present in 0.52 mL of the solution.
Since the bromine solution is given as a volume/volume percentage, it means that 10% of the solution is bromine by volume.

To calculate the mass of bromine present in 0.52 mL of the solution:
First, convert the volume (0.52 mL) to mass by using the density of the solution (density is essential as we are converting from volume to mass).

Let's assume the density of the bromine solution is known. For example, if the density of the solution is 2.0 g/mL, then:
0.52 mL * 2.0 g/mL = 1.04 g

Therefore, we have 1.04 g of the bromine solution, which contains 10% bromine. So, we need to calculate 10% of 1.04 g to obtain the mass of bromine in the solution:
10% of 1.04 g = 0.104 g

Now, we have the following information:
Mass of trans-cinnamic acid = 0.128 g
Mass of bromine = 0.104 g
Mass of brominated product = 0.184 g

To calculate the theoretical yield, we need to compare the ratio of the coefficients in the balanced chemical equation to determine the limiting reactant. However, since the reaction is not fully defined, it is not possible to calculate the theoretical yield accurately. Therefore, we will assume that trans-cinnamic acid is the limiting reactant.

The molar mass of trans-cinnamic acid (C9H8O2) is calculated as follows:
C (12.01 g/mol) + H (1.01 g/mol) * 8 + O (16.00 g/mol) * 2 = 148.16 g/mol

Now, we can calculate the theoretical yield based on the assumption that all the trans-cinnamic acid reacted to form the brominated product. To do this, we use the molar mass of trans-cinnamic acid:

Theoretical yield = (0.128 g)/(148.16 g/mol) * (molar mass of brominated product)
Please provide the molar mass of the brominated product for a more accurate calculation.

To calculate the percent yield, we use the formula:

Percent yield = (actual yield/theoretical yield) * 100

Note: Without additional information about the molar mass of the brominated product or any other related details, it is not possible to provide a precise answer.