How To Calculate Theoretical Yield and Percent Yield

This video shows you how to calculate the theoretical and percent yield in chemistry. The theoretical yield is the maximum amount of product that can be produced in a reaction. The percent yield is equal to the actual yield divided by the theoretical yield times 100%.

Full-Length Math & Science Videos:

Stoichiometry Formula Sheet:
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Introduction to Moles:

How To Calculate The Molar Mass:

How To Convert Grams to Moles:

How To Convert Moles to Grams:

Moles to Atoms Conversion:

Grams to Molecules Conversion:

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Grams to Atoms:

Moles, Atoms, & Grams Conversions:

How To Balance Chemical Equations:

Stoichiometry - Basic Introduction:

Avogadro's Number:

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Limiting Reactant Problems:

Excess Reactant Problems:

Theoretical & Percent Yield:

Percent Yield - More Examples:

Percent Error:

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Percent Composition By Mass:

Empirical Formula Problems:

Empirical Formula - Hydrated Compounds:

Combustion Analysis:

Stoichiometry Practice Test:

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Final Exams and Video Playlists:

[sairyortiz]

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

[screamingbullfrogs]

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!

[larsenswann]

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 YouTube video. Teacher incompetence should really be addressed in some schools...

[Techbuster]

i love it when mark wahlberg teaches chemistry

[spaidory]

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.

[samiam]

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

[nouthao]

You got a nice ass voice. It's comforting before a chem exam.

[pabloguerrero]

Dude, bless your soul. You just saved me from failing an assignment big time

[lauren]

How come professors cannot explain these concepts this simply?

[calebgeer]

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.

[rapidfireplayz]

"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

[bailey]

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

[kobynmalone]

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.

[kittymowmow]

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.

[jabary]

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

[sukhmanpreet]

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.

[piczohun]

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

[ayaanlasheen]

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

[blackcat]

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.)

[Marker-]

I'm in seventh grade and you don't know how much this has helped me not fail my finals <3