Introduction to Combustion Analysis, Empirical Formula & Molecular Formula Problems
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- Опубликовано: 27 янв 2025
- This chemistry video tutorial explains how to find the empirical formula and molecular formula using combustion analysis. It explains how to calculate the number of moles of each element given the mass in grams of CO2 and H2O. Examples include compounds containing Carbon, Hydrogen, and Oxygen. This video contains plenty of practice problems.
Stoichiometry - Free Formula Sheet:
bit.ly/3Z7uMw8
Chemistry 1 Final Exam Review:
• General Chemistry 1 Re...
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Introduction to Moles:
• Introduction to Moles
How To Calculate the Molar Mass:
• How To Calculate The M...
How To Convert Grams to Moles:
• How To Convert Grams T...
How To Convert Moles to Grams:
• How To Convert Moles t...
Moles to Atoms Conversion:
• Moles To Atoms Convers...
Grams to Molecules Conversion:
• Grams to Molecules and...
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Grams to Atoms:
• How To Convert Grams t...
Moles, Atoms, & Grams Conversions:
• How To Convert Between...
How To Balance Chemical Equations:
• How To Balance Chemica...
Stoichiometry - Basic Introduction:
• Stoichiometry Basic In...
Avogadro's Number:
• Avogadro's Number, The...
_________________________________
Limiting Reactant Problems:
• Limiting Reactant Prac...
Excess Reactant Problems:
• How To Find The Amount...
Theoretical & Percent Yield:
• How To Calculate Theor...
Percent Yield - More Examples:
• How To Calculate The P...
Percent Error:
• Percent Error Made Easy!
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Percent Composition by Mass:
• Percent Composition By...
Empirical Formula Problems:
• Empirical Formula & Mo...
Empirical Formula - Hydrated Compounds:
• How To Find The Empiri...
Combustion Analysis:
• Introduction to Combus...
Stoichiometry Practice Test:
• How To Solve Stoichiom...
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Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections
Stoichiometry - Free Formula Sheet: bit.ly/3Z7uMw8
General Chemistry 1 Review: bit.ly/3DNmZqb
Print The Exam - 160 Questions: bit.ly/3VTpcbj
Chemistry 1 Final Exam Review: ruclips.net/video/5yw1YH7YA7c/видео.html
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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 RUclips 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
Hi. Can you explain why he converted to grams instead of leaving them as moles of carbon or hydrogen?
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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.
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
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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.
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16:11
Loved the way you said 40
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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
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In problem 2 you converted to grams in the first step for no reason, then converted back to moles lol
good stuff straight to the point I like it.
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at 4:30 why do you divide by the smallest number?
thats just how you do it. also your pfp is garbage
one addition . if there is 0.666 multiply by 3. great lesson
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?
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No problem
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funny how most people have to learn this in college, while i must learn this for my g10 cycle tomorrow
because people have majors in science which is required if you were doing the same thing, you'll be taking loads of chem labs and classes
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 :(
i am confused on when to round the empirical formula vs. when to multiply it by a number
Thank you sooooooo much 🤧
thank you very much sir
What happends when i get 1.15 as a result for one of the elements?
Super helpful!