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How to Find Actual Yield, Theoretical Yield, and Percent Yield Examples, Practice Problems 

Conquer Chemistry
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In this video, you'll learn how to find the actual yield, theoretical yield, and percent yield by working through an example together.
It's really just a matter of knowing the equation: % yield = actual yield / theoretical yield x 100%. Actual yield is the amount of products that is actually produced in the experiment. Theoretical yield is the amount of products that can be produced if the experiment went 100% according the plan without error.
To find theoretical yield, use stoichiometry to convert the given reactants into the grams of the desired product.
By the end of this video, you'll know exactly how to find actual yield, theoretical yield, and percent yield.
If you liked my teaching style and are interested in tutoring, go to www.conquerchemistry.com/onlin...

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6 авг 2017

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Комментарии : 60   
@yaelfeder9042
@yaelfeder9042 6 лет назад
Really helping me prepare for chem finals, thanks😘
@ConquerChemistry
@ConquerChemistry 6 лет назад
Glad to help and best of luck on the final!
@ayelemoore6279
@ayelemoore6279 3 года назад
Did you do good?
@Hanin5002
@Hanin5002 2 года назад
@@ayelemoore6279 3 years later hahahahaha
@ayelemoore6279
@ayelemoore6279 2 года назад
@@Hanin5002 im curious
@-clipz-
@-clipz- Год назад
Prayers turn to God may everyone everywhere be blessed and healed and protected in Jesus name amen
@sallyb2547
@sallyb2547 4 года назад
A million Thank you! It helps me so much in my Chemistry.
@sweetpotato4977
@sweetpotato4977 4 года назад
so glad I found this video during online classes, thank you so much !
@lithzyalvarado9808
@lithzyalvarado9808 4 года назад
This helps sooo much thank you!!!! I am currently taking Organic Chem 1 and I remember we did this for General Chem but I completely forgot how to do it and we are doing this for an experiment tomorrow thank you again!!! :D
@Lia-ff3pi
@Lia-ff3pi 2 года назад
wherea are you from
@katherinebaker9250
@katherinebaker9250 2 года назад
I can't thank you enough!!! I've been searching trying to figure out how to do this and this is the FIRST TIME IT MADE SENSE
@-clipz-
@-clipz- Год назад
Prayers turn to God may everyone everywhere be blessed and healed and protected in Jesus name amen
@redirmer760
@redirmer760 3 года назад
I love you for teaching us how to solve for the actual yield
@mistytrompeter312
@mistytrompeter312 4 года назад
Very helpful! thank you
@cheerios8210
@cheerios8210 2 года назад
Thanks, was really helpful!
@hollyShmoe
@hollyShmoe 5 лет назад
THANK YOU!!!!!!!!!!!
@serinasalazar2954
@serinasalazar2954 4 года назад
My teacher isn't giving me the actual yield yet I have to find the theoretical yield
@casinovathedon8924
@casinovathedon8924 3 года назад
Its impossible to work out the actual yield without an experiment...
@toma.pudding
@toma.pudding 2 месяца назад
well you don't need the actual yield to find the theoretical yield
@ProductDesigner1989
@ProductDesigner1989 5 лет назад
Hi, why is that your theoretical yeild is 17.4g? Where did you take it?
@azca.
@azca. 3 года назад
Remove all the slashes and you get: 9.87(60.056/17*2)
@nbody5283
@nbody5283 3 года назад
@@azca. thanks, that’s the only part I didn’t understand
@TrApShotz
@TrApShotz 7 лет назад
Bro this helped me so much, i appreciate it
@ConquerChemistry
@ConquerChemistry 7 лет назад
Glad to help!
@TheWolfNations
@TheWolfNations 5 лет назад
bro im in the same boat, my teacher doesnt explain anything and she just says you should have learned this last year in pre-ap chem and shes super confusing, the rate my professor reviews online literally say the same stuff about her, thanks sooo much, u earned yourself a sub
@fzxbad6381
@fzxbad6381 6 месяцев назад
Brooooo you are THE G O A T
@fireflyschapstick
@fireflyschapstick 2 года назад
THANKU U SAVED MY LIFE
@LittleSchemers
@LittleSchemers 4 года назад
Very helpful, thanks. Kind of random, but your voice sounds exactly like my cousin.
@zahraal9208
@zahraal9208 Месяц назад
I LOVE YOU SO MUCH CONQUER VHEMISTY
@alpha4634
@alpha4634 4 года назад
Thx bro 🙏🙏
@Con-jq9ly
@Con-jq9ly 2 года назад
This was really helpful only issue i noticed being that multiplying by 100% is the same as multiplying by one (% = /100 => 100% = 100/100 = 1) so it is time 100 not times 100% but again aside from that it is extremely helpful
@user-pc4ke5qt2l
@user-pc4ke5qt2l Год назад
How did we get to 17.4g of urea??
@enarmour
@enarmour 6 лет назад
Good basics , preparing for finals too
@enarmour
@enarmour 6 лет назад
So, thank you very much :)
@CinemaKingTheaters
@CinemaKingTheaters Год назад
thanks
@hiimaj4615
@hiimaj4615 2 года назад
Thankyouuuu 🙇🙇🙇
@geraldramos1689
@geraldramos1689 6 лет назад
why did you used 75% instead of 79% in calculating actual yield?
@Ebrech
@Ebrech 5 лет назад
It's a typo, should have been 79%
@soueivs5germscodm268
@soueivs5germscodm268 2 года назад
Why 17.4 how u calculate it
@froyvincentcorbita8651
@froyvincentcorbita8651 3 года назад
Is it possible for the actual yield to be more than the theoretical yield?
@rafshankadil1238
@rafshankadil1238 2 года назад
No
@roelwagan7153
@roelwagan7153 2 года назад
How did you get the 17.4 g of Urea ?
@babiepalanganartana5846
@babiepalanganartana5846 2 года назад
Yeah that sucks
@mirandagarcia7872
@mirandagarcia7872 Год назад
thank uuuuuuuuuuuuuu
@TeaLord.
@TeaLord. 5 лет назад
Ok, but what if the problem doesn’t give you an actual yield or the percentage yield??? I have no clue how to find the actual yield all on its own...
@euph0rya672
@euph0rya672 4 года назад
Its impossible to find actual yield if % yield isn’t given
@wizzie9231
@wizzie9231 4 года назад
I'm no scientist but that does not sound humanly possible.
@SummerDiablo
@SummerDiablo 6 месяцев назад
I have a question they didn't give me the percent nor actual yeild of the product just the moles of the reactants and I have to find percentage yeild HOWWWWWW
@noahleider5884
@noahleider5884 Год назад
the actual yield was given. I was hoping you would show us how to find the actual yield.
@srglmr
@srglmr 5 лет назад
thats not finding ACTUAL YIELD its giving in your example how to get the actual yield from the side reaction
@euph0rya672
@euph0rya672 4 года назад
I cant speak from experience the only way to find actual yield is if the % yield is given then u need to find theoretical yield then equate it and cross multiply
@khatger
@khatger 5 лет назад
i got the percentage yield one but the actual yield one was not helpful but thanks alot bro!
@SaaraEPShaundalwa-hf7ig
@SaaraEPShaundalwa-hf7ig Год назад
Well explained even though the video is not clear 😭😭..
@cassidymaclean7404
@cassidymaclean7404 4 года назад
How did you get 17.4 theoretical yeild??? I’m so confused dude
@ahmadmikhail9187
@ahmadmikhail9187 3 года назад
from the equation he made.....is ur calc giving other answers other than 7.39...?
@casinovathedon8924
@casinovathedon8924 3 года назад
Ill explain it with an easier method. . . First work out the moles of Ammonia which is n=M/Mr =9.87/17=0.581 to 3 significant figures From the balanced equation, you can see that 2 moles of Ammonia react to give 1 mole of Urea So the mole to mole ratio is 2:1 so the moles of Urea is 0.2905 Now you can work out the Maximum theoretical mass of Urea with teh formula Mass=Mr x n(moles) The relative molecular mass of Urea is 60 and the moles are 0.2905 as worked out earlier. 60 multiplied by 0.2905=17.43g % Yield = Actual mass of product (13.74)/maximum theoretical mass(17.43)X100 = 78.8% Hope this helps
@Iwanttoblowmybrainsoutrn
@Iwanttoblowmybrainsoutrn 2 года назад
Multiply everything on the top, then multiply everything on the bottom, then divide them
@FreshSongs-AND-SOON-Politics
@FreshSongs-AND-SOON-Politics 2 года назад
The semantics of this subject as it is typically taught is all over the place. It bothers me that we call this a gram to gram conversion. It makes it seem like we are measuring the same chemicals when we say conversation, or like we are switching units. The semantics of it all had me confused for years. We are dealing with gram to gram ratios. And yes, while something is being converted, it is not grams being converted. So the name, "gram to gram conversion" is unessessarily and frustratingly distracting from the overall goal.
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