# chem help

Watch
Announcements
#1
0
2 years ago
#2
Sorry you've not had any responses about this. Are you sure you've posted in the right place? Here's a link to our subject forum which should help get you more responses if you post there. 0
2 years ago
#3
Calculate the moles of phenol : 100/ (Mr of phenol)
Calculate the theoretical moles of the product using the 1:1 ratio of the equation
Calculate the theoretical mass of the product by doing moles x Mr of product
Multiply this mass by 0.27 to find the mass produced if the yield is 27%

Posted from TSR Mobile
0
#4
(Original post by Lilith2400)
Calculate the moles of phenol : 100/ (Mr of phenol)
Calculate the theoretical moles of the product using the 1:1 ratio of the equation
Calculate the theoretical mass of the product by doing moles x Mr of product
Multiply this mass by 0.27 to find the mass produced if the yield is 27%

Posted from TSR Mobile
http://pmt.physicsandmathstutor.com/...0Compounds.pdf

Question 5b

1g=1000mg
250x3x7= 5.25 g

moles chloral hydrate= 5.25/165.5 = 0.0317 mol
mass trichloro= mols x Mr = 0.0317 x 147.5 = 4.6789 g

percentage yield= actual/theoretical = 60%
Actual = theoretical x 0.60 = 4.6789 x 0.60 = 2.81 g

0
2 years ago
#5
(Original post by chemquestion)
http://pmt.physicsandmathstutor.com/...0Compounds.pdf

Question 5b

1g=1000mg
250x3x7= 5.25 g

moles chloral hydrate= 5.25/165.5 = 0.0317 mol
mass trichloro= mols x Mr = 0.0317 x 147.5 = 4.6789 g

percentage yield= actual/theoretical = 60%
Actual = theoretical x 0.60 = 4.6789 x 0.60 = 2.81 g

5.25 g is what you get for 60% yield.
(5.25 / 60 ) ×100
Then divide this new mass by the Mr
Use 1:1 ratio and multiply by reactant Mr which gives you 7.8

Posted from TSR Mobile
0
#6
(Original post by Lilith2400)
5.25 g is what you get for 60% yield.
(5.25 / 60 ) ×100
Then divide this new mass by the Mr
Use 1:1 ratio and multiply by reactant Mr which gives you 7.8

Posted from TSR Mobile
Thank you
As I can see this question requires a slightly different method than the one above...
what's the difference between this question and the one I asked before?
0
2 years ago
#7
(Original post by chemquestion)
Thank you
As I can see this question requires a slightly different method than the one above...
what's the difference between this question and the one I asked before?
The first one is finding the mass of a product formed from a certain mass of reactant given and the second is finding the mass of a reactant required to form a certain amount of product given.
The second is the same method but reversed - you're just working backwards.
0
#8
(Original post by Lilith2400)
The first one is finding the mass of a product formed from a certain mass of reactant given and the second is finding the mass of a reactant required to form a certain amount of product given.
The second is the same method but reversed - you're just working backwards.
I meant in terms of the yield calculation part
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### How are you finding researching unis for 2021 entry?

I have been able to get all the information I need from online research (66)
19.41%
I have tried virtual events and found them useful (73)
21.47%
I have tried virtual events and did not find them useful (63)
18.53%
I would be interested in trying socially distanced or scaled down in person events (70)
20.59%
I want to but don't know where to start with researching unis for 2021 entry (34)
10%
I haven't started researching yet (34)
10%