Question

What is the theoretical yield of p-bromotoluene (C7H7Br) from reacting 1.5 grams of toluene (C7H8) with an excess of bromine and an iron catalyst?

Please help

Answer 1

Write the equation.
Convert 1.5 g toluene to moles. # moles = g/molar mass.
Using the coefficients in the balanced equation, convert moles toluene to moles of the bromotoluene.
Convert moles bromotoluene to grams. # g = moles x molar mass.
Post your work if you get stuck.

Answer 2

The reaction is:

C7H8 + Br2 --> C7H7Br + HBr

Find the molar masses of C7H8 and of C7H7Br

The theoretical yield is:

(1.5g C7H8)(mol.mass C7H7Br / mol.mass C7H8)

Answer 3

I don't understand the part where you convert moles toluene to moles of the bromotoluene.

Answer 4

Use the equation GK wrote for you.
1 mole C7H8 give you 1 mole C7H7Br, so
moles C7H7Br = moles toluene x (1 mole C7H7Br/1 mole toluene) = moles C7H7Br.

Answer 5

# moles of C7H8= .01628

.01628 mol of toluene*(1 mol C7H7Br/1 mole toluene)=.01628 mol C7H7Br

?????

Answer 6

That's right. Since the ratio of C7H8 to moles C7H7Br is 1:1, then the moles of one equal moles of the other.
But you will see that the units of C7H8 cancel(which is what you want) and leaves the answer in units of moles C7H7Br (which is what you want). I'm sure you have seen this done in your classroom. This is dimensional analysis.

Answer 7

I got 100% yield percent
It doesn't sound right

Answer 8

I think you are confused. The problem, as you posted it, is to determine the theoretical yield starting with a 1.5 g sample. So the problem wants to know, in grams, how much of the C7H7Br is formed IF it is 100% yield. Give me a few minutes and I'll work the problem and see how much you should have. In the meantime, post how much you have and we can compare numbers.

Answer 9

Check my work. I obtained 2.79 g if I didn't make an arithmetic error.

Answer 10

moles C7H8 = 1.5/92 = 0.0163
moles C7H7Br = 0.0163 x (1 mole C7H7Br/1 mole C7H8) = 0.0163 x (1/1) = 0.0163 moles C7H7Br.
grams C7H7Br = moles x molar mass = 0.0163 x 171.04 = 2.79 g C7H7Br and I would round that to 2.8 g to agree with the number of significant figures in the 1.5. I hope you understand what I've done.