General Chemistry 1 Review: bit.ly/3DNmZqb PDF Worksheet - 160 Questions: bit.ly/37SVLn6 Final Exams and Video Playlists: www.video-tutor.net/

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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 KZfaq TV with the Organic Chemistry Tutor.

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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

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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!

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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.

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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.

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In problem 2 you converted to grams in the first step for no reason, then converted back to moles lol

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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

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i am confused on when to round the empirical formula vs. when to multiply it by a number

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one addition . if there is 0.666 multiply by 3. great lesson

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What happends when i get 1.15 as a result for one of the elements?

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My teacher gave me a combustion analysis with a compound containing CH and Fe. Yet he only taught us to solve problems with CH and O.

What if we get a decimal instead of a while number after deciding the Molecular Formula Mass by the Empiracle formula mass?

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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?

In your problems you didnt say that it reacts with oxygen gas

Very helpful!!

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Our savior

Super helpful!