Combustion analysis help?

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McKenna disc 1C
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Joined: Fri Sep 29, 2017 7:04 am

Combustion analysis help?

Postby McKenna disc 1C » Thu Oct 05, 2017 12:59 pm

Hi all,
I am confused about the process of using combustion analysis to determine molecular formula. It's been awhile since I've taken high school chem! Could somebody please explain how combustion analysis works (even a broad explanation is fine-- I'd just like to hear it in someone else's words rather than the textbook!)

Yashaswi Dis 1K
Posts: 56
Joined: Fri Sep 29, 2017 7:04 am

Re: Combustion analysis help?

Postby Yashaswi Dis 1K » Thu Oct 05, 2017 1:42 pm

Sure. So Combustion Reactions generally involve organic compounds or hydrocarbons (they can also involved other molecules too) that burn in the presence of oxygen. So whenever you hear the words burning, combustion, etc. always think of the compound given reacting with oxygen gas (O2). Now, the products that are always produced are CO2 (carbon dioxide), H2O (water), and if Nitrogen is in the original compound, then N2 (nitrogen gas) is also produced.

So to determine the molecular formula, you need to first find the empirical formula and that can be done by converting given masses of C, H, N (if present), and O (sometimes O won't be given) to moles and then to grams. If O's mass isn't given, you may be given the total mass of the compound and subtract the grams found of each element from the total mass to find Oxygen.

Then, given all of the moles of each element in the compound, you can try to find the smallest whole-number ratio by dividing each element in the compound by the smallest moles that's present. If the numbers aren't exact whole-numbers (which they usually aren't) either: round like 2.1 ~ 2 and 5.9 ~ 6 or multiply by a number if you see numbers like 1.33 (multiply by 3 to get 4) to get a whole number. After that you should get your empirical formula.

If a molar mass is given of the unknown molecular formula, then divide the molecular formula's molar mass by the empirical formula's molar mass to get a whole-number ratio.

Finally, to get the actual molecular formula, multiply that whole-number ratio to the element's subscripts (like if it's 2 for the ratio of the molar masses and the empirical formula is CH3O, the molecular formula is C2H6O).

Hopefully that helped in finding empirical and molecular formula from combustion reactions. If you need more help, I suggest trying the problems in the textbook chapter that have detailed explanations or looking online for more videos for practice. The more you practice, the more I think you will get a hang of it.

Hope this helps!

Aleisa Quach 3C
Posts: 3
Joined: Sat Jul 22, 2017 3:00 am

Re: Combustion analysis help?

Postby Aleisa Quach 3C » Thu Oct 05, 2017 2:53 pm

How would you use combustion analysis if there are other elements besides C, H, and O present?

Isabella Sanzi 2E
Posts: 54
Joined: Thu Jul 13, 2017 3:00 am

Re: Combustion analysis help?

Postby Isabella Sanzi 2E » Thu Oct 05, 2017 2:58 pm

I believe that the answer is that it depends on the element that is being added in addition to C, H, or O. For example, with an element like Nitrogen in a combustion analysis, the product would be N2(g). For other elements, however, it may be different.

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