Theoretical, Actual and Percent Yield Problems - Chemistry Tutorial

[@sabrinar23]

lol I still don't get it - just going to accept reality and fail chem tomorrow 😩

[@Nickelobel]

It's like having Luna Lovegood teach you chemistry.

[@TheChemistrySolution]

nope, I have a phd in biochemistry and already have a job. these videos go along with my website- I actually just like helping students understand chemistry.

[@TonyAKA30]

What a beautiful voice. I'm no longer stressed for my chemistry midterm!

[@TheChemistrySolution]

Please help us continue making FREE chemistry videos by clicking on the link below and voting for Big Head Business Solutions. It will take less than 10 seconds, I promise! Thank you! Thank you! ~Missy (The Chemistry Solution) www.missionmainstreetgrants.com/business/detail/33553

[@lemonfruiit]

This was extremely helpful! Thank you so much! My teacher goes way too fast.. and I missed a few days for a family trip so this really helped me catch up! Great job :D

[@joecoburn5510]

thank you!! it makes sense now! Idk why this took me so long to grasp but I couldnt find a good explaination ANYWHERE...you're a great teacher!

[@damolux1]

that's a very complicated method of doing this problem. this is how i teach it: (n=no. of moles, m = mass, gfm = gram formula mass) 1) n(CaCO3) = m(CaCO3) / gfm(CaCO3). 2) mole ratio is 1 CaCO3 :1 CaO therefore n (CaCO3) = n(CaO). 3) m(CaO) = n(CaO) X gfm(CaO)

[@jharding578]

100.1g CaCO3 is from the molecular mass/weight of the entire compound- Ca= 40.1 C= 12.01 + O3= 48 ------------------- CaCO3= 100.11

[@hoangtran95]

Ignore my past comment Doc. I figured it out after going over and over your video. Thank you for such an amazing job on helping us understand chemistry.

[@mrbrethmint]

Luna Lovegood?

[@st81982001]

This video is so much clear than in class lol thx I finally get it

[@r_wael7094]

Thanks for this helpful video. I finally understand theoretical yield.

[@lije08]

what do you calculate to get 8.97g CaO

[@SimraKhadam]

Omg thank you so much I actually worked it out before I looked at your answer and it was right!!!!

[@Firestorm1820]

Thank you! Very informative and easy to understand.

[@rinasay98]

Thank you thank you thank you! I would never understand my chem lab without you ^^

[@TheChemistrySolution]

True, your actual yield can never be higher than your theoretical yield, as long as you have a completely pure product. But if in the lab you measure your yield and there is a contaminating substance present with your product (because you are not able to completely isolate your product from the reaction mixture) then you may at times measure a value higher than your theoretical yield.

[@brentpolhemus2774]

You should seriously be my chemistry teacher...

[@SoraShadowdancer]

I am confused. How did you come up with 56.1 grams of CaO? That number wasn't given in the original equation. Is that the number you get when you add the amu from Ca and O together?

[@wheredah0odat]

your voice is very pleasing....for some reason I understand all your videos much better than others.... LOL

[@PotatoDweller]

How do you find the formula weight??

[@Aliaisawesomeable]

Thank you so much! Just saved me a few questions on my final! :DD

[@Sarranwrap_dc4]

When they say an excess of an element what do they mean?

[@rufusbrooks8460]

She sounds like Quinn from Glee :D

[@adamzidine1234]

Hamdullah I understand now

[@aikhan690]

Your voice is so sweet!!

[@JimmyTMach]

You should definitely update your website, I would become a member for more tutorials!

[@group5and6JADE]

THANK YOU! Your videos help me a lot!

[@hoangtran95]

I have a question. In step 3 where you convert gram to mole, will you put another numbers instead of 1/1 if the equations have different coefficients? Let's say we have another equation that we have to balance. Thank you. You have made chemistry easier.

[@fuzzwahmrp]

Most professors teach that the actual can never be higher than the theoretical, but I've heard one point out that in the lab it can be higher due to contamination.

[@colossulcrush]

is this method used to find the theoretical yield?

[@wheredah0odat]

I would like to ask you a question as well mrs. Are you doing these videos to build a good resume for a future job? Are you a science major as well? I would appreciate a response thank you :)

[@Nami328]

I have a question, if I have to calculate the percentage yield for the decomposition of KClO3 but I heated it and did this twice, I measured the mass of KClO3 before heating and after heating both times. So now that I have two data, how can I calculate the percentage yield using this? In the first trial I used 2.235g of KClO3 and after heating produced 1.182g of KClO3, in the second trial I used 2.279g and produced 1.876g of KClO3. Can you please help me with this?

[@courtwiz]

This was very helpful! Thanks

[@123reletive123]

Anyone know how she got 8.97g of Cao

[@knightgaminghd8749]

she did the mass of CaCO3 divided by its mr to get the moles, then as the molar ratio is 1:1, CaO has the same number of moles. Then you convert this to g by multiplying moles by the mr of CaO =8.97

[@celiaellsworth418]

How did you get 100.1 g?

[@VisakanThayalan]

Keep up the great work

[@haquet1]

AHHHHH why didnt i see this sooner... i had a test yesterday and i totally left this type of question blank :/

[@sagala40]

very good explanation!!

[@Chiwiwee]

how did she get 101.1 g?

[@dr.meathany6677]

Thank you ❤️❤️❤️❤️❤️

[@pk4lyfe1]

thanks made it clear for me

[@mandradep97]

Thank you a looot. Very helpful :)

[@CJVarner98]

This helped a lot!!!

[@group5and6JADE]

Thank you so much!!!!

[@qwerpoiuROX]

Oh thank you I finally understand

[@Yellowexterior]

i like your voice

[@nicoletara353]

thanks!

[@nicholaskampouris6343]

THANK YOU!

[@SolutionsInChemistry]

Good stuff

[@Tylanjamison96]

your brilliant

[@wadehamilton5084]

Great

[@brightomens]

I can't understand half of what she's saying because of the way she pronounces her s's.

[@ahmedoman8593]

good

[@MegaBigbaby1]

fourth video still lost

[@lenafaith99]

yup im Einstein now

[@MsBrieBrutal]

why do you try not make any mistakes while you talk? it really looses my concentration

[@iamanhar]

shittt your voice man

[@Beth096]

omg so confused

[@0318kent]

She sounds hot

[@joshuajeyaranjan9167]

This girl sounds hot

[@maggieboolovesyou518]

her voice kinda gets on my nerves ,

[@juliannabhullar8701]

this helped so much!!