@@haziqdaniel3967 deadass was somehow expected to remember grade 10/11 chem?? i was confused when i went to my first lecture yesterday and we were already getting started on some new shit with shit explanations. thank god this mans got my back
I started watching your videos to get help with pre calc last year and now I'm in college and you're helping me with chemistry. Theres literally nothing you can't do 😂
Hey, I just want to let you know that you are the ONLY reason I haven’t dropped chemistry. You are the only person who has kept me afloat for the last 8 weeks, and in the first few classes my first year of college. I wish I could give you the world truly 😂
High schools and colleges should pay these type of content creators, I’ve been doing all my learning through educational TH-cam this semester, teachers during remote learning just tell you to read a pdf if you have any question...
i was an excellent student at my university at the module of internal copution engine ... i was such an excellent student ....i worked for a period of time as a teacher to educate those who think themselves as an engineers and doctors .... i am.proud of myself and i will always be the perfect student at the module of internal computions engine and for thermofluid module .... i am proud to be a mechanical engineer ... if i forget my identity i would forget myself ... and thanks god i never ever forget that i am a mechanics and i proud to have an oil over my cloths
Professor Organic Chemistry Tutor, thank you for explaining how to find the Empirical and Molecular Formula using Combustion Analysis in AP/General Chemistry. Once again, the practice problems/examples show how to compute the Empirical and Molecule Formula using Combustion Analysis. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
BEST CHANNEL AND TUTOR E-V-E-R !!! Thank you SOOO much (I think i am going to pass my Inorganic Chem. exam in the uni because of your explanations) . if i could i would give you 1000 thumbs up !!!
I know, I know, people are saying that this guy taught more than their teachers... But seriously, my Chem 11 teacher DID NOT explain this to us (basically telling the whole class to do this ourselves). So thank you Organic Chemistry Teacher for carrying my grades throughout the years 🙏 🙏 🙏
Actually saved my life with this one I knew how to do the carbon and hydrogen but my teacher couldn’t explain how to do oxygen for the life of her, and I have a test in like 20 minutes so this is going to save my grade Thank you so much ❤
8:50, should explain why you're converted to grams (since you'll need it to calculate oxygen later on), cos as is, you get .188, which you end up getting later in the video, 11:51. same with hydrogen, 10:05, explain that we need this to calculate oxygen, since as is, you end up getting .501 12:19
A simpler way to figure out the empirical formula would be to find the composition percentage of carbon in carbon dioxide and hydrogen in dihydrogen monoxide (H2O). Then, divide the percentages by 100 so that you can then multiply that number by the amount of mass that is given for CO2 and H2O Then, you'll have the amount of grams of carbon and the amount of grams of hydrogen. To convert grams to moles just use the formula; number of moles=mass/molar mass. So, divide the amount of grams of carbon by the molar mass of carbon and divide the amount of grams of hydrogen by the molar mass of hydrogen. Now you have the quantities of both carbon and hydrogen in moles. Next, divide both mole values by the smaller one. The answer you get will be your coefficient for that element within the compound. If even one of your coefficients is not a whole number, keep on doing trial and error by multiplying that decimal number by two, then three, then four, and so on until you get a whole number. If you end up with something like 11.9 after doing trial and error, you can simply round that to 12. If u get something like 11.1, round that to 11. If u get something like 11.2, 11.3, 11.4, 11.5, 11.6, 11.7, or 11.8, you would have to multiply both the coefficients for carbon and hydrogen by a factor that would make that decimal number a whole number. For example, 11.8×5=59✔️ Then you can write out your empirical formula with your obtained coefficients. Now if you want the molecular formula from this, use the formula molecular formula factor=mass of compound (always given)/empirical molar mass. If you've done it right, you should always get a whole number for this factor. Simply multiply your coefficients by that factor and now you have your molecular formula as well. Thank you for coming to my ted talk hehe, have a great day!
bro, I'm about to buy your gear to support you. Or maybe even jump on to your patreon. I feel guilty getting so much knowledge from you for free. Good job.
at 10:16 why did you take the mass of hydrogen only 1.008 whereas, we have 2 mols of Hydrogen???But, I still want to say thank you for explaining everything so well.
Hi. I hope my one-year-too-late explanation finds you well. We only take the mass of 1.008 since that's the grams per one mole of hydrogen. We did this to convert our moles to grams.
At 9:16 do you have to convert the moles back to grams or can you leave it in moles? And can you change the 3.765g of the compound to moles and solve it for moles and solve it that way?
The reason why he find grams of C and H is so that he can find the grams of O. With the grams of O, you can find the moles of O to write and empirical formula
THANK YOU SOSOSOOSSOSOSOSOSO MUCH OMG I WAS FREAKING OUT BECAUSE I FORGOT HOW TO DO EMPIRICAL AND MOLECULAR FORMULA FOR COMBUSTION ANALYSIS BUT YOU HELPED ME SO MUCH THANK YOU FOR SAVING MY GRADE IN AP CHEM
Snooyy UPnHere very late, but for the second one, you notice how the complete compound has a mass in g? Well in order to find the missing mass of oxygen you everything else you subtract also needs to be converted to grams because of the mass of the compound is also in grams. If you were to use moles instead it wouldn’t work because moles and grams are two different things and cannot be combined. Make sense?
@@teslaguy5311 but in the first question why did he also add the atoms? in his previous vids he never did convert it to atoms while finding the empirical formula :(
These was the most question I was failing to find the solution to it but thanks to the Almighty God for directing me to the good tutor ❤❤❤🙏🙏🙏I understand a lot now I will be able to answer my previous question
and in the empirical formula---- how do you do it if like theres a fixed mass of the compound-- like this---- A 170 g sample of an identified compound contains 2.7 sodium 57.34 chromium 22.33 oxygen. What is the compound's empirical formula??
hey just wanted to point out that something was miscalculated in 4:11, should've been 0.374 and not 0.7484 :) but the concept was clear and well-explained thank u, sir.
but for the first example?? should we not deduce the mass of carbon and hydrogen separately?? and then find the empirical for the two elements only?/ please help
is it possible to calculate CO level in burning process if we measuring the flue gas values like temperature, O2 and NOx? If it is what else is needed to perform it correctly (fuel type is one)
it depends on how close it is to the whole number. for example, id obviously round up 3.99 to 4, but i wouldnt round up 3.54 or round down 3.49. so multiply until you find its super duper close to the whole number. he explains it better than me in the video
In first problem we didn't × atomic mass / 1 mole of (c, h , o ) but in second one we did why ?. explain please ? So if later we get problems we can get to know either we have to do this step or not ?
Hi . so i have this problem that i really cant understand. This was suppose to be our topic this month but due to the lockdown the teacher sent us activities and do it on our own . i hope you can help .. In the process of combustion analysis of a compound containing only carbon, hydrogen and nitrogen, 12.923 g of carbon dioxide and 6.608 g of water were measured. Treatment of the nitrogen with H₂ gas resulted in 2.501 g of NH₃. What is the compounds empirical formula?
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This man is single-handedly carrying online learning. I watched for honors chem in 10th grade and now I am watching him for college chemistry
lmao same this guy is better than my lecturer
@@haziqdaniel3967 deadass was somehow expected to remember grade 10/11 chem?? i was confused when i went to my first lecture yesterday and we were already getting started on some new shit with shit explanations. thank god this mans got my back
yoooo legit tho!!. This dude literally carrys the entire online learning with his state of the art lecture.
OMG IM IN THE SAME EXACT SITUATION!!!!
Currently watching this for honors chem in 10th grade
Chemistry without him is like studying chemistry without a periodic table 🙌🙌
I started watching your videos to get help with pre calc last year and now I'm in college and you're helping me with chemistry. Theres literally nothing you can't do 😂
you alr graduated?
This is the best educational TH-cam channel, been passing all my exams ever since I subscribed, thank you, really appreciate .
Underrated channel
Hey, I just want to let you know that you are the ONLY reason I haven’t dropped chemistry. You are the only person who has kept me afloat for the last 8 weeks, and in the first few classes my first year of college.
I wish I could give you the world truly 😂
I’m in 9th grade your so lucky
Agree
@@nuraz5202 you're*
@@dystint8817 right.
@@nuraz5202 indeed
Who’s here for AP Chem? ... Beautiful explanation
well i'm here for college chem but goodluck with ap chem ^^
9th grade chemistry 😭😭😭😭
IB anyone?
A level anyone?
@@JoaquinRevello Eyyy it me.
High schools and colleges should pay these type of content creators, I’ve been doing all my learning through educational TH-cam this semester, teachers during remote learning just tell you to read a pdf if you have any question...
Thank you this made so much sense out of everything. Now i can pass chem.
You saved a life! Thank you! Best video I’ve seen
thank you
I swear man this guy is really helpful with all this online classes happening
i was an excellent student at my university at the module of internal copution engine ...
i was such an excellent student ....i worked for a period of time as a teacher to educate those who think themselves as an engineers and doctors ....
i am.proud of myself and i will always be the perfect student at the module of internal computions engine and for thermofluid module ....
i am proud to be a mechanical engineer ...
if i forget my identity i would forget myself ...
and thanks god i never ever forget that i am a mechanics and i proud to have an oil over my cloths
nga tf is you on?
damn u really out here saving my grade in ap chem
Professor Organic Chemistry Tutor, thank you for explaining how to find the Empirical and Molecular Formula using Combustion Analysis in AP/General Chemistry. Once again, the practice problems/examples show how to compute the Empirical and Molecule Formula using Combustion Analysis. This is an error free video/lecture on TH-cam TV with the Organic Chemistry Tutor.
BEST CHANNEL AND TUTOR E-V-E-R !!! Thank you SOOO much (I think i am going to pass my Inorganic Chem. exam in the uni because of your explanations) . if i could i would give you 1000 thumbs up !!!
This helped me so much, thank you for the simple yet extremely helpful video!
I know, I know, people are saying that this guy taught more than their teachers...
But seriously, my Chem 11 teacher DID NOT explain this to us (basically telling the whole class to do this ourselves).
So thank you Organic Chemistry Teacher for carrying my grades throughout the years 🙏 🙏 🙏
Saving us time and time again!!! Thank you Organic chem tutor!
Actually saved my life with this one
I knew how to do the carbon and hydrogen but my teacher couldn’t explain how to do oxygen for the life of her, and I have a test in like 20 minutes so this is going to save my grade
Thank you so much ❤
saved my life. Always struggled with finding the O2 thats in both compounds.
I still struggle but not as much
thanks bro 🙏🧸
i was about to cry but then i remembered this channel existed
Bro u made things so much easier when things seemed impossible to figure out, I love u so much
8:50, should explain why you're converted to grams (since you'll need it to calculate oxygen later on), cos as is, you get .188, which you end up getting later in the video, 11:51. same with hydrogen, 10:05, explain that we need this to calculate oxygen, since as is, you end up getting .501 12:19
this legend carrying chem, beast helped me understand DP chem
you saved my exam man !!!!!!!
Years later and here I am using your videos to get an A in chem. Thanks for the help!
this man is doing well ,may god bless you
I don't have much to say all i can say is thank you... 'you have saved a life'
This video helped me a lot, thank you for your great effort🤗شكرا لك على مجهودك العظيم
A simpler way to figure out the empirical formula would be to find the composition percentage of carbon in carbon dioxide and hydrogen in dihydrogen monoxide (H2O).
Then, divide the percentages by 100 so that you can then multiply that number by the amount of mass that is given for CO2 and H2O
Then, you'll have the amount of grams of carbon and the amount of grams of hydrogen.
To convert grams to moles just use the formula; number of moles=mass/molar mass. So, divide the amount of grams of carbon by the molar mass of carbon and divide the amount of grams of hydrogen by the molar mass of hydrogen.
Now you have the quantities of both carbon and hydrogen in moles.
Next, divide both mole values by the smaller one. The answer you get will be your coefficient for that element within the compound.
If even one of your coefficients is not a whole number, keep on doing trial and error by multiplying that decimal number by two, then three, then four, and so on until you get a whole number.
If you end up with something like 11.9 after doing trial and error, you can simply round that to 12. If u get something like 11.1, round that to 11. If u get something like 11.2, 11.3, 11.4, 11.5, 11.6, 11.7, or 11.8, you would have to multiply both the coefficients for carbon and hydrogen by a factor that would make that decimal number a whole number. For example, 11.8×5=59✔️
Then you can write out your empirical formula with your obtained coefficients.
Now if you want the molecular formula from this, use the formula molecular formula factor=mass of compound (always given)/empirical molar mass. If you've done it right, you should always get a whole number for this factor.
Simply multiply your coefficients by that factor and now you have your molecular formula as well.
Thank you for coming to my ted talk hehe, have a great day!
bro, I'm about to buy your gear to support you. Or maybe even jump on to your patreon. I feel guilty getting so much knowledge from you for free. Good job.
Crushed it. Clear as can be. Thank you
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Homeschooling 15 year old daughter in 10th grade. Your chemistry and math vids are a godsend. Thank you!
You have saved my life many times sir 🥺🥺
Thank yoouuu
This seems easy but Imma have to revisit quite a lot of times!
7:00 really helped
Thanks for existing. God highly bless you
Thanks for the video!!! I'm gna ace my next chem test all because of you
This was really helpful! - thank you.
Dude thx have an exam in 2 days helped me prepare with different types of problems
Much appreciate sir , you really help us to do well in our studies
THANK U FROM THE BOTTOM OF MY HEART AND SOUL
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Loads of Love @Organic Chemistry Tutor
thank you so much for doing this. really thank you
Broooooo. super clutch chem action on your part!
at 10:16 why did you take the mass of hydrogen only 1.008 whereas, we have 2 mols of Hydrogen???But, I still want to say thank you for explaining everything so well.
Hi. I hope my one-year-too-late explanation finds you well. We only take the mass of 1.008 since that's the grams per one mole of hydrogen. We did this to convert our moles to grams.
1 mol of hydrogen is 1.008 and we want to cancel the moles of H to get grams of H
Lifesaver thnx. Even with a few views your vid makes an impact. Thanku
Saved my final
THANK YOU
U such a life savior God bless u so so much
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I've been studying this kind of problems in Morrison & Boyd Organic Chemistry for IChO. You made this so easy for me. Thank You a ton.
Do we Morrison and Boyd Organic Chemistry in PDF?
save my life again! Thanks a lot!!!
THANK YOU! This was so incredibly helpful! I appreciate your work so much.
How are you doing
Hello
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Best tutor ever 👏👐
What happends when i get 1.15 as a result for one of the elements?
When you multiplied the first problem's empirical formula by 5 to get the whole number of H atoms, did you also need to multiply the other side by 5?
At 9:16 do you have to convert the moles back to grams or can you leave it in moles? And can you change the 3.765g of the compound to moles and solve it for moles and solve it that way?
The reason why he find grams of C and H is so that he can find the grams of O. With the grams of O, you can find the moles of O to write and empirical formula
i am confused on when to round the empirical formula vs. when to multiply it by a number
taught this so well. thank you.
16:11
Loved the way you said 40
THANK YOU SOSOSOOSSOSOSOSOSO MUCH OMG I WAS FREAKING OUT BECAUSE I FORGOT HOW TO DO EMPIRICAL AND MOLECULAR FORMULA FOR COMBUSTION ANALYSIS BUT YOU HELPED ME SO MUCH THANK YOU FOR SAVING MY GRADE IN AP CHEM
Thx man
Thanks dude you helped me alot, cause im kinda bad at chemist 😅
Thank you so much! This really helped!
the carbon and hydrogen in the first question was finalized into moles. BUT the second question you did an extra step to make them into grams. Done
Snooyy UPnHere very late, but for the second one, you notice how the complete compound has a mass in g? Well in order to find the missing mass of oxygen you everything else you subtract also needs to be converted to grams because of the mass of the compound is also in grams. If you were to use moles instead it wouldn’t work because moles and grams are two different things and cannot be combined. Make sense?
@@teslaguy5311 but in the first question why did he also add the atoms? in his previous vids he never did convert it to atoms while finding the empirical formula :(
What if we get a decimal instead of a while number after deciding the Molecular Formula Mass by the Empiracle formula mass?
These are so helpful
Got an AP Chem Test tomorrow on this, If I get an A on the Test I'll like the vid, lets see
How'd your test go?
@@ApplePi1 idk still awaiting results
@@ReaperResidence Hope you did well!
you are better than my chemistry teacher!
btw, what app do you using in this video?
carrying me through college chem, thank you ochem daddy
These was the most question I was failing to find the solution to it but thanks to the Almighty God for directing me to the good tutor ❤❤❤🙏🙏🙏I understand a lot now I will be able to answer my previous question
beats online Aleks hands down.....glad i youtube'd combustion analysis.
This video was explained so well
at 4:30 why do you divide by the smallest number?
thats just how you do it. also your pfp is garbage
I appreciate you so much I am actually passing chem 2045 in college because of you. Thank you!!
and in the empirical formula---- how do you do it if like theres a fixed mass of the compound-- like this---- A 170 g sample of an identified compound contains 2.7 sodium 57.34 chromium 22.33 oxygen. What is the compound's empirical formula??
Convert each of those elements to moles and you have a ratio of Na to Cr to O. Then simplify it to get the empirical formula
hey just wanted to point out that something was miscalculated in 4:11, should've been 0.374 and not 0.7484 :) but the concept was clear and well-explained thank u, sir.
You're the best!!!❤️❤️🙏🏽🙏🏽
if the mass is in mg do we have to convert?
one addition . if there is 0.666 multiply by 3. great lesson
but for the first example?? should we not deduce the mass of carbon and hydrogen separately?? and then find the empirical for the two elements only?/ please help
You are the BEST!
is it possible to calculate CO level in burning process if we measuring the flue gas values like temperature, O2 and NOx? If it is what else is needed to perform it correctly (fuel type is one)
How do you know whether it's to round off or multiply by number
it depends on how close it is to the whole number. for example, id obviously round up 3.99 to 4, but i wouldnt round up 3.54 or round down 3.49. so multiply until you find its super duper close to the whole number. he explains it better than me in the video
In first problem we didn't × atomic mass / 1 mole of (c, h , o ) but in second one we did why ?. explain please ?
So if later we get problems we can get to know either we have to do this step or not ?
thank you so much this helped a lot..
This Video Presentation was Great ,Enjoyed And Understood The Topic To The Fullest ..150%,,,,,,Zikomo Kwambili
Hi . so i have this problem that i really cant understand. This was suppose to be our topic this month but due to the lockdown the teacher sent us activities and do it on our own . i hope you can help
.. In the process of combustion analysis of a compound containing only carbon, hydrogen and nitrogen, 12.923 g of carbon dioxide and 6.608 g of water were measured. Treatment of the nitrogen with H₂ gas resulted in 2.501 g of NH₃. What is the compounds empirical formula?
thank you sir.... very well explained
why do you multiply the moles of hydrogen by 1 when converting it to mass? why didn't you multiply it by 2 since we started it out by h2?
Thanks u saved a life
Well i got C3.8H11.4O1 plz tell me if i multiply it with 2 then i get C7.6H22.8O2 so can i write it as C8H23O2
Is this correct
why can't i do 8.272 +4.515 -3.765 for the second problem to find the mass of oxygen?
hey,,, i have a smilar problem but when i get a number for the hydrogen its a huge number like 40
Been stressing cause we have a quiz tomorrow and our modules are useless. Thank you very much for this 😭😭😭.