Chemical Forums
1 Hour
1 Day
1 Week
1 Month
Forever
January 07, 2025, 09:40:59 AM
Forum Rules
: Read This Before Posting
Home
Help
Search
Login
Register
Chemical Forums
Chemistry Forums for Students
Organic Chemistry Forum
Theor./Percent Yield from by-product
« previous
next »
Print
Pages: [
1
]
Go Down
Topic: Theor./Percent Yield from by-product (Read 15233 times)
0 Members and 1 Guest are viewing this topic.
jrpestana
Guest
Theor./Percent Yield from by-product
«
on:
June 14, 2005, 12:27:55 AM »
In my preparation of n- butyl bromide experiment, I was able to calculate the following data.
Theoretical Yield
12.15 g -- 0.164 moles 1-butanol
22.47 g n-butyl bromide
Exp. Yield
10.71 g n-butyl bromide for a 47.66% yield.
Now suppose I isolated the di-n-butyl ether (and determined it weigh 0.8g)... how would I determine the percent yield of the ether and how would I be able to show the effect the maximun possible weight of n-butyl bromide?
You think I'm to calculate a new theoretical yield for the ether and then use the simple percent yield calculation? They do not give me any other figure for the ether so I wonder if I would just relate it to my currrent theoretical yield.
Thanks in advance for inputs...
Logged
movies
Organic Minion
Retired Staff
Sr. Member
Posts: 1973
Mole Snacks: +222/-21
Gender:
Better living through chemistry!
Re:Theor./Percent Yield from by-product
«
Reply #1 on:
June 14, 2005, 12:48:36 AM »
First write out the balanced equation for the formation of dibutyl ether. Then figure the theoretical maximum yield of dibutyl ether (assume that this would be the only reaction occuring). Then divide your experimental yield by this theoretical maximum.
That should do it.
Logged
jrpestana
Guest
Re:Theor./Percent Yield from by-product
«
Reply #2 on:
June 14, 2005, 04:00:16 PM »
(CH3CH2CH2CH2OH)2 ---> CH3CH2CH2CH2--O(+)--CH2CH2CH2CH3 ??
This is how I calculated my theoretical yield.
15 mL 1-butanol x 0.810 g 1-butanol / 1 mL 1-butanol =
12.15 g 1-butanol
12.15 g 1-butanol x 1 mole 1-butanol / 74.12 g 1-butanol =
0.164 moles 1-butanol
0.164 moles 1-butanol x
137.03 g n-butyl bromide / 1 mole n-butyl bromide
= 22.47 g n-butyl bromide
Percent Yield
8.4 mL n-butyl bromide x 1.275 g n-butyl bromide / 1 mole n-butyl bromide =
10.71 g n-butyl bromide
Experiment yield / Theoretical yield x 100%
10.71 g n-butyl bromide / 22.47 g n-butyl bromide x 100% = 47.66% yield
So you're saying place the MM of of di-n-butyl ether (which I don't have) in the place of the MM of n-butyl bromide?... and divide my exp yield by that number?... I'm still a bit lost... Thanks for the help
«
Last Edit: June 14, 2005, 04:01:40 PM by jrpestana
»
Logged
movies
Organic Minion
Retired Staff
Sr. Member
Posts: 1973
Mole Snacks: +222/-21
Gender:
Better living through chemistry!
Re:Theor./Percent Yield from by-product
«
Reply #3 on:
June 14, 2005, 04:28:36 PM »
You should really just weigh your product. It'll be more accurate than converting the volume via the density. Otherwise, the process seems okay.
For the dibutyl ether, what would be the balanced equation for the formation of this product?
Logged
jrpestana
Guest
Re:Theor./Percent Yield from by-product
«
Reply #4 on:
June 14, 2005, 04:52:54 PM »
It's a hyp. question in the lab manual... still completely lost.... have no clue as to the equation... thought it was the one in my post.
Logged
movies
Organic Minion
Retired Staff
Sr. Member
Posts: 1973
Mole Snacks: +222/-21
Gender:
Better living through chemistry!
Re:Theor./Percent Yield from by-product
«
Reply #5 on:
June 14, 2005, 07:12:16 PM »
Oops! I didn't see it there. Okay, you are missing one byproduct. What else is produced in the reaction along with the ether?
How many molecules of butanol do you need to make a molecule of dibutyl ether?
Logged
Print
Pages: [
1
]
Go Up
« previous
next »
Sponsored Links
Chemical Forums
Chemistry Forums for Students
Organic Chemistry Forum
Theor./Percent Yield from by-product