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## AP®︎/College Chemistry

### Course: AP®︎/College Chemistry>Unit 1

Lesson 3: Elemental composition of pure substances

# Worked example: Determining an empirical formula from combustion data

In combustion analysis, an organic compound containing some combination of the elements C, H, N, and S is combusted, and the masses of the combustion products are recorded. From this information, we can calculate the empirical formula of the original compound. Created by Sal Khan.

## Want to join the conversation?

• CH3 can't exist on its own though, right? Or would it just be a polyatomic ion? •  Don't mix up empirical and molecular formulas!

The empirical formula has the smallest whole number subscripts between atoms

Molecular formula is the actual formula of the molecule

The most likely molecular formula of something with an empirical formula of CH3 is ethane, C2H6
• I don't understand how having 0.128 moles of CO2 is the same as having 0.128 moles of C, and that having 0.1927 moles of H20, H2 just 0.1927 * 2?

My logic is that the moles of C02 would be larger than just O (because CO2 is comprised of 2 elements rather than just one). And the moles of H20 would be a larger number than just H2 (because H20 is comprised of two elements rather than just one).

Help! Thanks • Think of like you have a single molecule of carbon dioxide. In that one molecule you have one atom of carbon and two atoms of oxygen. So for every molecule of carbon dioxide you have one carbon, a 1:1 ratio. Likewise every molecule has two oxygens, a 1:2 ratio. Remember that the mole is a unit that just measure the amount of stuff there is, just quite a large amount. So if we take those ratios from before and scale them up to larger number, we will still have the same ratio of carbon and oxygen atoms to molecules of carbon dioxide. For example, 1,000 carbon dioxide molecules will have 1,000 carbon atoms (1:1) and 2,000 oxygen atoms (1:2).

Same logic applies to water and every other chemical.

Hope that helps.
• What determines the order of the chemical symbols in an empirical or molecular formula? e.g., why is it H2O and not OH2 • How do you determine which reactant goes on top of the ratio? • Why are there 2 moles of hydrogen for every mole of H2O instead of 2/3 moles of hydrogen for every mole of H2O? • Given the fact that the entire reaction is not balanced, How can we trust these calculations? Can I obtain the molecular formula with the same information provided in this example? • Wondering 1) why we don't have to worry about oxygen in the empiral formula, 2) why we're assuming it's O2, and 3) why we don't have to worry about the molar mass of oxygen, but are just focusing on C and H. Thanks • The question states that the compound only contains carbon and hydrogen, .

Since it's a combustion reaction we would use molecular oxygen (or oxygen gas), O2, as the other reactant. Combustion reactions always involve a fuel of some kind (here it's the carbon-hydrogen compound) and gaseous oxygen. Very rarely would we not use oxygen gas and if so the problem would explicitly state that.

Again, the compound we want to know the empirical formula of doesn't contain oxygen so knowing how many moles of oxygen was produced wouldn't help.

Hope that helps.
• can someone help me with this-
An organic compound contains element C, H and oxygen. A 4.26 mg sample of compound is burnt in oxygen.It gives 8.45 mg of carbon-dioxide and 3.46 mg of water. what is the empirical formula of the compound? • Why is it CH3 not H3C? • The way you order elements in chemical formulae is determined by the Hill system (named after Edwin A. Hill who created the system in 1900). The Hill system states that carbon atoms are listed first, hydrogen atoms next and then the number of all other elements in alphabetical order. It is the most commonly used system in chemical databases and printed indexes to sort lists of compounds. So it would be CH3 going off that system.

Of course there exceptions to the system such as ionic compounds which have the cation first then the anion.

Hope that helps. 