Combustion Analysis - Finding the Empirical Formula of a Compound
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- เผยแพร่เมื่อ 2 มิ.ย. 2011
- How to determine the empirical formula of a compound containing carbon and hydrogen using combustion analysis. The empirical formulas of compounds containing carbon, hydrogen, and oxygen can also be determined by combustion analysis.
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You could have expressed 2.4 as 12 over 5 and multiplied both elements by 5 to get C5H12
Thanks for going step by step. I was kind of confused but I was just missing something simple. Good walk through.
The organization of your work is spectacular.
I learned a lot from this video.
Thanks!
Your teaching style in amazing! I am having some trouble getting through my Chemistry class. I did fine with the first test, However now that I am getting into Stoiciometry and Empirical formulas I just bombed my second test. If you have any advice or videos that may help me I would appreciate it. I am also willing to pay your style of teaching helps me see Chemistry in a new light.
Lifesaver.. Thanku so much.
Okay Thanks!! It helps me. :) (From Malaysia)
Thanks Ben
nice video. Another way to come up with an empirical formula and avoid fractions is to realize that 2.40 can be multiplied by 5 to get a whole number. Therefore, multiply your pseudoformula by 5 and you will get the correct formula. If it ends in .33, you multiply by 3, if it ends in .20 you could multiply by 5.
I thought it was very clear!
Thanks
what happened to the oxygen ???
FINALLY A TH-cam TUTOR WHO DOESN'T HAVE A BORING VOICE!!!
great video
Yes please do!
You didnt finish :/ (Total mass) - (Grams of Hydrogen + Grams of Carbon) to find oxygen. You take that and put it in moles. You do the empirical formula with oxygen too
quick question, at the point where you have C1H2.40,..rather than dividing by all those numbers would it be okay to multiply the subscripts by 5??..so we just end up with C5H12
thank you .. can you do another contains oxygen
Cheers
@PostsThatMatter at 1:02 I said "suppose I carry out a combustion reaction with a compound that contains carbon and hydrogen". The fact that the compound does not contain oxygen is just one of the initial conditions of that particular problem. If you'd like to see one with carbon, hydrogen, AND oxygen, shoot me a message some time.
how can i shoot a message!
@@GodSahil send me an email or DM me on Instagram @benschemvideos
So taking Food Grade Hydrogen Peroxide w/ a Hydrogen Supplement could possibly increase a reaction causing the body the heat up maybe? We are trying to understand the components of the ingestion plus the supplement once a person overdose on it? Or if they did all this and smelled cigarette smoke more than normal levels or mixed it with grain alcohol. Could this maybe have a factor in raising certain gases in the body?
I dont get how you got the moles for the carbon in co2? help please?
What is the difference between analysis by mass and analysis by volume in combustion? can anyone help
I think its C2H5 is also pretty acurate
Combustion analysis of naphthalene, a hydrocarbon used in
mothballs, produced 8.80 g CO2 and 1.44 g H2O. Calculate the
empirical formula for naphthalene.
Does the question require you to know that naphthalene only contains C and H beforehand or are you supposed to figure that out? if so, how do you figure that out?
I know it has been 6 months but to answer your question when it says hydrocarbon that means it is a compound with only carbon and hydrogen.
The question tells you that your sample is indeed a hydrocarbon (meaning it ONLY contains hydrogen and carbon) ! So you know that all you're dealing with are those two elements! I hope that makes sense!
Funny that you use a sock to erase :) great videos!
the oxygen doesn't matter, unless you are looking for the amount of oxygen used. The compound only had carbon and hydrogen, so there wasn't any oxygen in the initial compound.
Please do!
There is one carbon atom in a CO2 molecule. So, in 10 CO2 molecules, there would be 10 carbon atoms. In one mole of CO2 molecules (6.022*10^23 molecules), there is one mole of carbon atoms.
Couldnt you have just multiplied 1 and 2.4 by 5 to get both into whole numbers?
Is there a video where you have a problem that you have to find the total mass of oxygen? Someone please send a link my way! :)
your method at the end is a little bit more involde. just divide by the smallest number. Then multiply by some number till you get a whole number
this is how it's done out of a textbook in the U.S., but it's NOT the best way to explain the process.
Todd Crawford
Why do you have to bring in the US??
todd fuck me
Don't do it Todd he has chlamydia
But how do you get oxygen!
you sound like mike ross from suits
How do i find mol of O when it doesnt give me a starting mass! thats the only thing pissing me off i get how to do these but what about O!!! I can find O if they give starting mass but when they dont give strting mass i dont get it
Zaryab B In Combustion reactions, you are always using oxygen, which is O2. So you don't need to know the starting mass to find the empirical formula. You find the empirical formula for C and H and add O2 to it, then balance with the CO2 and H2O products. Hope that helps!
Pleaaaase :c
I wish I could marry you!
Lol uhh you missed a step there.. you were supposed to convert the moles to grams of the single atom in the compound.. and then divide by smallest value to get the empirical ratio thing
if you help me get a 4 in ap chem, ill marry you (and im a straight guy)
not the combustion formula u didnt find o completely useless
Thanks for your feedback!
I would suggest working on your English a bit more, and then watching this video again, or my other videos on combustion analysis.
The links to the other videos are below.
Part I: th-cam.com/video/BFalll3tSF4/w-d-xo.html
Part II: th-cam.com/video/TvgUbv0c5gE/w-d-xo.html
Thanks again!
-Ben
It's not useless, he explained that O doesn't always have to be part of the original compound.