Writing Empirical Formulas From Percent Composition - Combustion Analysis Practice Problems

This chemistry video tutorial shows you how to determine the empirical formula from percent composition by mass in grams. This video also shows you how to determine the molecular formula from the empirical formula using the combustion analysis technique of a compound. This video contains plenty of examples and practice problems that can help you on your next upcoming worksheet assignment / quiz.
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...
_________________________________
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!
_________________________________
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...
_______________________________________
Final Exams and Video Playlists:
www.video-tuto...
Full-Length Videos and Worksheets:
/ collections
Stoichiometry Formula Sheet:
bit.ly/3Z7uMw8

[@TheOrganicChemistryTutor]

Stoichiometry Formula Sheet: bit.ly/3Z7uMw8
Final Exams and Video Playlists: www.video-tutor.net/

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25:00, wouldn’t the moles of H2O be the same as the moles of Oxygen because for every mole of H2O, isn’t there only 1 oxygen atom?

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Good evening Sir, can l ask which special integers to take note of when writing the Emperical formula?

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At the very last we find out the molar mass of oxygen(Present in compound) to be something by subtraction the molar masses of C and H from reactants. Where does the other oxygen in air reacting with the compound go?

[@ainjeffery]

+scienceguy
I think I solved this problem by watching the video again carefully. If there's a compound of some amount, say, 200 grams and there's C,H and O present then we can break down the 200 into each of those masses. C =x H=y O=z such that x+y+z=200.
First we find the masses of C and H from products which turn out to be, say, 150, so that means the sum of x and y has to be 150. Now we just do the arithmetic : x+y+z=200 150+z=200
z=200-150 = 50.
So we that means that the 50 mass of oxygen is from the compound and not from the products.
*I hope it helped*

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why do we round up for "H" (minute 4:01) it when from 1.99 to 2. However, for 9:51 "H" why doesn't it round up to just 3?

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for the last one what if were not given the mass of the compund 4.5g

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shouldnt you have reported 2 sf for the molar mass of Carbon? at 2.59

[@davidvalle3068]

i used 12.01 for the conversion factor

[@luluo8056]

This was a bit long. Since we know the total mass of the molecular compound we can multiply the percent composition of each element by 88grams giving us the total amount of grams for each element in the compound. We can then divide that by the mass per mole of each element and figure out how many atoms of each element there are which gives us the complete molecular compound. Then we simplify rations to derive the empirical formula.

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how do you know there is 2 carbons?

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Where is the 216coming from

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0.75 g of hydrocarbon compound contains 0.60 g of carbon.
(Ar: c=12:0; H=1.0)
Which one of the following could be the molecular formula of the hydrocarbon compound?
A. C3 H8
B. C2 H4
C. C2 H6
D . C H4
E . C2 H3
Please can you help me solve this question?

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3:13 it's actually 16g O not 16g H.

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Im still confused with how and why you multiplied 3 with the empirical formula with decimal.

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me too, I thought you would just round 2.67 to 3?

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