How To: Find Empirical Formula From A Combustion Reaction (Combustion Analysis)

A video made by a student, for a student. Showing how to pull an empirical formula from a combustion reaction. Kansas University. Rock Chalk Jayhawk, KU!!!!!!!!!
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Stoichiometry. Biology. Chemistry. How to work practice problem. How to convert. moles to grams
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[@zeenadthomas6818]

The number of moles of carbon does not amount to 0.073. I calculated it and it comes to 0.007 mol instead, making that the smallest value. It turns out that the empirical formula is actually CH2O2 instead. Either way, your brilliant method displayed makes up entirely for the tiny miscalculation. THANKS!

[@sciencemadeeasier7100]

+zeenad thomas That is the true test of wether this video helped you understand the concept entirely ;)

[@anli_ace]

Trust me, I'm struggling with Honors Chemistry, and never really understood the whole combustion thing, but now I've seen this here and in other problems from my teacher, so I now think I'm either confident or can be able to at least recognize this kind of problem. Thank you so much for making this video! Many likes!

[@sciencemadeeasier7100]

+Ace Forever It can look so confusing when your teacher/professor shows you I completely understand... Thats why I made these videos I felt like there was an easier way to explain these concepts. Glad I could be of assistance :)

[@anli_ace]

There really was, so thank so much! I hope I'll come to a video I'll need in the future!

[@ashoozq]

Thanks surely helped for AP Chem! 😊

[@M00971]

Now I can solve combustion reaction problems THANK YOU

[@atreyimukherji]

This video was really helpful to one of my friends! I love doing empirical formulas but never came across one of these! Thanks for the great explanation! :)

[@SahilSharma-ld4dg]

You just earned yourself a subscriber!! Thanks for helping me and keep making good videos!! 😀😀

[@pwincezzetz]

thank you so much, you really made this easier

[@42nooruleman57]

this video helped me a lot. so thank u

[@Just1MorePerson1]

so basically first find the number of grams of each element then divide each of them by there atomic masses and then by the lowest number. easy got it thank you

[@sciencemadeeasier7100]

+Just1MorePerson1 $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ baby$$$

[@delphictheman1376]

BRILLIANT VIDEO!! THANK YOU SO MUCH

[@sciencemadeeasier7100]

+Ahmed Ghazi Glad I could be of assistance :D

[@alexbarreto8774]

Thank you! Really helped to understand

[@Ajokeofwords1231]

I tried to do this format with one of my questions from my class today and I don't seem to be getting the same answers she gave to us in class :( The question was: A combustion of 4.3 g of a compound only carbon, hydrogen and oxygen. 8.59 g of co2 and 3.52 g of h2o. What is the empirical formula for the compound? We got: C2H4O

[@gossipgirlx0869]

best vid i seen, thanks!!

[@isjamarie]

Thanks for your Help!!!!

[@sciencemadeeasier7100]

+Isja Marie Glad I could help you out :)

[@aliciahelgeland1216]

I understand this until the end when you convert the grams of each element into moles, and then divide by the smallest to get your whole number integer coefficient. Why in some problems you just do this one step and round to get your coefficient, and others you have to divide by the smallest and then do an additional step of multiplying to get whole numbers? How do I know when to differentiate which to do?

[@Jerz4lunch]

+Alicia Maggio You need to divide all the molar masses by the molar mass of the element with the smallest number of moles to get one (or more) of the elements to have one mole. The reason you would sometimes multiply the number of moles of each element to get whole numbers is because there is no such thing as half an atom of one element per one atom of another. In the case of water with its two hydrogen atoms and one oxygen atom, if you didn't use whole number ratios, one could then say that we could have one hydrogen atom per 1/2 oxygen atoms. That doesn't exist. As far as how to differentiate between which to do, you have to take a look at the molar mass of each element. If the number is something like 2.499 or 2.504 or similar, multiply everything by two. If you get something like 0.334 or 0.664, multiply by three. Keep doing this until the numbers are very close to a whole number, such as 3.015. Hope this helps!

[@aliciahelgeland1216]

This helps very much, thanks!

[@lucasbarbeau954]

so my professor gave me 1g of CHO unknown, 3.14g CO2, and 1.29g H2O. doing these exact calculations shown, I end up with a very small but NEGATIVE number for oxygen... does that mean there is no oxygen?? very confused....

[@kevinoverholtzer3126]

as he or she probably told you the numbers are sometimes just pulled out of the air when it comes to grams of a sample and co2 and h20

[@mihai-alexandrugraur3057]

guys I am trying to solve this and everytime I get a diferrent answer. this is the problem: 1,621 g of compound known to contain C, H, and O, undergoes a compleate combustion to produce 1,902 g of water, 3,095g of CO2. what is the emperial formula? I found C20H6O but I am not sure :(

[@henrystjuste]

does it matter if the question gives you the mass of products in mg? should you convert to grams before multiplying by atomic and dividing by molecular weight?

[@sciencemadeeasier7100]

+Henry Saint-Juste You'll have to convert either at the beginning or end. Most answers will usually want you working with grams so I'd go ahead and do it at the beginning.

[@cationkrosh5643]

Tnx for the video, it's very helpful :)

[@meln6842]

how do you know how many decimal places to keep?

[@sciencemadeeasier7100]

+Melina Nguyen Just ask your professor what he prefers, usually two.

[@joyun20]

What if you aren't given the starting mass?

[@tejshreeborkhetaria3810]

thanks! You were kind of slow but it was really simple :))

[@sciencemadeeasier7100]

+Tejshree Borkhetaria I go slow as to lose no one

[@soominkim745]

thank you! so who is this person in the video?? :)

[@sciencemadeeasier7100]

It is me, Claire (;