How To Calculate Theoretical Yield and Percent Yield 

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Good luck on your test tomorrow

you've taught me more in a 6 minute video then my chem teacher has this entire year

Thank you so much. I'm currently failing chemistry and have studied for 2-3 hours every night. This subject is super hard for me. I spent 5 hours studying this tonight, watching videos and reading guides and doing problems. Whatever you did differently, it helped. Something in my head clicked and I understand this now! Again, thank you SO much!

I’m gonna fail this final exam dawg 😭

How is it that the education system could be so fucked that the majority of a class won't be able to properly grasp a concept through hours of "teaching" yet can develop a perfect understanding from this 6 and a half minute RU-vid video. Teacher incompetence should really be addressed in some schools...

my method: 1.) balance the reaction 2.) find the ratio (how many moles of this make how many moles of that) 3.) find the mr(s) of the substances you want to work with 4.) just put the mass given into the ratio after converting mass into moles brief version: 57 kg of aluminum oxide gave 17 kg of aluminum via electrolysis: 1.) 2Al²O³ ---> 4Al + 3O2 (balanced equation 2.)2 moles of Al²O³ give 4moles of Al so the ratio is 2:4 or 1:2 3.)Al²O³ Mr = 102 Al Mr = 27 4.) 57kg = 57000g 57000g ÷ 102(Al²O³ Mr) = 500 so 500 moles of Al²O³ were used ratio = 1:2 (1mole of Al²O³ gives 2 moles of Al) so 500moles of Al²O³ gives 1000 moles of Al this mass of Al= 1000 moles × 27(Mr of Al) mass = 27kg actual mass / theoretical mass so: 17kg/27kg ×100% = 62.9 % long explanation: (sorry the numbers appear as powers) 57kg of aluminum oxide gave 17kg of aluminum via electrolysis step1.) balance the reaction 2Al²O³ ---> 4Al + 3O² step2.) find the ratio of the elements we want to work with 2 moles of aluminum oxide make 4 moles of aluminum so the ratio is 2:4 or 1:2 (look before the element/comound in the balanced reaction and use the numbers to figure this out) step3.) find the Mr(s) Al²O³ has an Mr of 102 Al has an Mr of 27 (technically this is aluminums Ar since it is not a compound) step 4.) it says 57 kg of Al²O³ was used in electrolysis so we wanna convert this into moles: 57kg is the same as 57000g so 57000/ 102 (the Mr of Al²O³) gives us 500moles. so we now know that 500 moles of aluminum was used. lets put this into our ratio that we found out in step 2. ratio is 1:2 (for moles of aluminum oxide vs moles of the aluminum produced) now just use this ratio to find how many moles of just aluminum were produced. 1:2 -----------> 500:1000 since the ratio states that 1 mole of aluminum oxide gives 2 moles of aluminum. so 500 moles of aluminum oxide should theoretically give 1000 moles of alumimun. now we now that 57kg of aluminum oxide was used and that this equates to 500 moles of it so now lets just find out how much aluminim was produced in mass convert moles into mass ----> 1000×27(mr of just aluminum) gives 27kg (theoretical mass) we were told that 17kg of aluminum was produced so niw what we do to find %yield is divide 17kg(actual mass given) by 27kg and we turn that into a percent to get the percebt yield which should be 62.9 percent.

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I know I'm like 4 years overdue but THANK YOU! Isn't it weird that RU-vid videos are MUCH better than our actual teachers in school? Like all of the lessons I learned was from RU-vid😅

How come professors cannot explain these concepts this simply?

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I've watched so many videos on this and yours is the only one that makes sense. Thank you!!

Kinda complicated method... simply put; [[Mol = Mass(g) ÷ Molar Mass]] ∴ 30 / 44 = (30/44) [= Moles of C3H8] 3(30/44) = 90/44 [= Moles of CO2 since there is a 1:3 ratio] 90/44 * 44 = 90g [Theoretical yield since molar mass of CO2 = 44] It was just coincidence that the molar mass of Both C3H8 and CO2 are the same

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This dude works wonders. My prof teaches for 3 hours every lecture, and somehow makes the topic more confusing then it needs to be. Thank you for making things clear and organized.

Absolute life saver, my chem lab professor works multiple chapters ahead of lecture and did not want to read more than necessary for my upcoming lab. Thank you! I should ace this quiz now.

"you know". Nah but out of all honesty you're teaching me better then my chem teacher can/has since the start of the semester

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My professor with a PhD in chemistry cannot explain any mathematical processes in laymen's terms. He makes everything SO difficult and hard to understand. Thank you, stranger on the internet, for doing this for all struggling chemistry students.

Beautifully explained as always! You're the only RU-vid channel I can understand so clearly.

Thank you so much, as a maths lover it irritated me that I could not work out complex percentage yield and now, a day before my GCSEs, your video has saved me. Much appreciated.

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Very good video, paced just right for me to understand what's going on and not lose focus, thank you!

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Literally 2 mins into the video and I get it already...Hope you hit 1 million .

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yeah i’m done w chemistry i still don’t know what i’m doing💀🙏

You explain it so well and your voice is very pleasent unlike my proffessor :)

Great video! My lab instructor is a new TA and he just gave us one example and turned us loose on solving a ton of unknowns. No one understood. Unfortunately, having a Dr. teach us in the lecture and his TA do our lab causes a huge disconnect. The TA does things differently than the Dr. who breaks things down in a great way. It is killing my grade! I am struggling to keep an A and that is not me! I have never made below 97% in a college course. Until now, I have pulled the grade up to 93% and that took more work than I am used to. Thanks for the great video, the theoretical yield is pretty simple, only he could make it hard! He NEVER mention the limiting reactant once! Pretty important information!

Thank you!! Sometimes it just helps hearing it explained just a slightly different way to get the concept. This'll really help with my exam tomorrow!

Thanks so much, this finally makes sense to me!

This was very helpful! So glad RU-vid exists, and people like this to makes these vids. Hope everyone understood!

I’ve never been able to solve a percentage yield question until now. Thank you!

That is amazing. Understood a concept I’ve been struggling for weeks on in six minutes. Thank you!!!

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It's a great explanation but you could get the theoretical yield by just using the molar ratio in grams. 30g of C3H8 and the molar ratio is 1, x grams of CO2 and the molar ratio is 3. Cross multiply and you'll get 90. it's way more simple

Thank you so much, I finally understand how to do this and I have my exam on Wednesday...I can’t believe how simple this actually, why can’t my teacher can’t explain it like this 😪

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You've been my teacher for my whole chem course lol, my chem teacher can't teach.

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Thank you! btw you sound soo much like Mark Wahlberg! lol

Thank you for the step by step. it was so very helpful.

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this video saved me lmao. my first AP chem test is tm and this finally made sense to me

to all of those wondering. here's a quicker way to do this. Actual Mass of C3H8 / molar mass (relative formula mass) of C3H8 using your answer times that by the Molar ratio difference (Example. H20 and 2H20 = 1:2) then time by that by the molar mass (relative formula mass) by the one you are comparing your molar mass to. as if it had a molar mass of 1. Here's how you do it. 1C3H8 = 30g compared to 3CO2 30 (g)/ 44 (relative mass of C3H8 12x3=36 + 1x8=44)= 0.681 (continuous) molar ratio = 1:3 (1C3H8 3CO2) 0.681 (continuous) x 3 = 2.045 (continuous) time that by the relative mass of the product (in this case 3C02 if it was 1CO2 which coincidently has the same relative mass as 1C3H8) CO2 = 44 (12 + 16x2= 44) 2.045 x 44 =90g 90% theoretical yield. the actual that we got is 70g/70% so its 70/90 (the smaller number is first.) x100% = 77.7% percentage yeild. there you go. hope this helps. (also I am not a teacher, I am a college student who just grasped this.)

Needed some clarification on this topic for my unit test tomorrow. Thanks once again OCT!

oh my god thank u so much I have a lab report due on Monday and I literally don't understand anything especially percentage yield but now I understand a lot better!!!

why is it that you did not get the g of O2?? like you did in C3H8??

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Thank you so much for this video. The Way you explained really helped me understand. Most teachers just tell you what to do with out telling you why.

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Don't mind me, just taking notes because I slept in for my Ap Chem class and wasn't able to know what everyone was doing

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Thank you! Chemistry is super hard when there’s all this maths.

My test us actually tomorrow

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another simpler method guys C3H8:3CO2 44 : 132 30 : x 132×30/44= x 90=x

Thank you so much! I had no idea how to do this, and you made it so simple!

Thank you Julio, I owe you so much, 3rd year chemical engineering student here.

I am still so confused :'( Like... I understand percentage yield no problem. It's the calculating of TY I can't grasp. The whole... 30g over 1 and 1 over 44, and 3 over 1.... x g / mol... I'm so so lost. Help! Someone, anyone!

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Didn’t get a single damn thing and I’m running out of time

I swear i was looking for the video about percentage yeild and percentage purity i have test tomorrow of chemistry about this very good explanation i understand easier you are better than my teacher

Watched this video while doing my exam because I didn’t study at all and was gaming in classes, passed with 97/100 appreciate it i picked it up easily thanks to the video

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He didn't put the correct molar mass of CO2...... He filled it in with the molar mass of C3H8

Thank you so much you helped me understand this concept.

Appreciate the quick video as review for my AP Chem clsss

Thank you. To be funny: 0:40 8 hydrogen atoms, eight has five letters, 8-5= 3, 3 sides to a triangle. Illuminati confirmed!

This was super helpful, thanks!!

Thanks This video really helped me understand the concept But I have a qns. Is there a formula to calculate theoretical yield?

I already watched this video back in highschool on the subject General Chemistry, and I'm back to rewatch it again on our Chemistry for Engineers now in college.

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This helped so much. Thank you!