How To Calculate Theoretical Yield and Percent Yield

[@TheOrganicChemistryTutor]

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Because schools don't check if the teachers are able to competently teach their students

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[@spaidory874]

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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[@bailey125]

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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[@khansaboo2030]

what if the ratio was 4:3 then??

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[@ayaanlasheen2903]

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

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[@blackcat738]

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.)

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Thank youuuu Organic Chemistry Tutor!!! I have an exam about this topic in Chemistry and I easily understood everything you're saying 😭 what a lifesaver

[@m.bilawalkhan7789]

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

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This is wrong and definitely isn't a simpler method

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Thanks bro

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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!

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

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the way u explain it is SO SIMPLE and easy to understand u should be a hs teacher

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Good luck on your tests everyone!!!!❤

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bro saved my chemistry exam

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This explanation was so helpful! I worked the problem along with the video, understood the concept and got the right answer. The issue for me is how easy he makes it look….the heavy lifting is done as he describes exactly what’s happening. Then if I try it myself, it’s like ok no…chemistry is not easy when JG is not here to break it down for me 😝 But at least I was able to follow his example!!!

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who come here looking for yield calculation in finance? its like you entering a wrong class

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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.

[@alysfave7316]

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

[@TheOrganicChemistryTutor]

it's the excess reagent. We don't need it to calculate the percent yield. There's plenty of oxygen in the air. I hope that answers your question :)

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Mark Wahlberg knows his chemistry.

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Anyone who is here 2024

[@sohret33]

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

[@darcybehati2083]

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!

[@IdrhysZK]

Simplest way to do it is (Moles = Mass x Mr- then use the stoichiometric equation to find the ratio. Then using the moles from the ratio, and the equation Mass = Mr x Moles, you can find your theoretical yield.

[@nitamac1312]

you need to learn dimensional analysis

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

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My test us actually tomorrow

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Thanks dude this was helpful, you teach better than my chem teacher

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I am sooo uncooked

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

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Don’t expect to get help with that attitude.

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@@AKCman35 2 months ago

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@@josh-qj2zu im in the same position rn

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this video helped me lol now i get the concept

[@JS-kf9ee]

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

[@gretchensich7144]

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

[@findo9615]

Alright kiddy. Before you go calling out our professor here,,,,,, c = 12 o = 16 h = 1, if we're rounding to whole numbers. (12+32) = (36 + 8)

[@Bilbus7]

32+12=44 numb nuts

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lol

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lmao!

[@Niggapicen]

Gretchen Sich if you didnt notice, the molar mass of CO2 and C3H8 happen to be the same

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You just saved me form failing my homework. Thanks!

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GOD BLESS YOUUU I HAVE AN EXAM TOMORROW

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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!!!

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

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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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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!

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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.

[@TheOrganicChemistryTutor]

You're welcome

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Thank You so much! I have a test in 5 minutes and I’m study right now 😁

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God bless you, you help me through my suffering in chemistry

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