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?(10 votes)
- 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(30 votes)
- What determines the order of the chemical symbols in an empirical or molecular formula? e.g., why is it H2O and not OH2(6 votes)
- The positively charged ions or the metals are placed first and the negatively charged ions or anions are placed later for inorganic compounds.
For organic compounds, carbon is placed first and all the other attached groups are placed later.(13 votes)
- 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(7 votes)
- 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.(11 votes)
- How do you determine which reactant goes on top of the ratio?(7 votes)
- What you are converting TO always goes on TOp. Eg 5.65 g CO2 x (1 mol CO2 / 44.01 g CO2), this is converting from grams to moles so moles go on top.(4 votes)
- Why are there 2 moles of hydrogen for every mole of H2O instead of 2/3 moles of hydrogen for every mole of H2O?(3 votes)
- Because that’s not how moles work.
If you broke H2O up into the atoms that make it up you would have H2O -> H + H + O
From 1 molecule of H2O you get 2 atoms of H
So from 1 mole of H2O molecules you get 2 moles of H atoms(10 votes)
- 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?(4 votes)
- Yes you don't know need the equation to be balanced to solve for the empirical formula given the information about the products provided. For the molecular formula though you would need more information most likely using mass spectrometry to build on the empirical formula.
Hope that helps.(3 votes)
- 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?(2 votes)
- using the masses of CO2 and H20, find the respective amounts of carbon, hydrogen, and oxygen through dimensional analysis. Use the method to find empiricle formulas (divide masses of elements by respective atomic masses, then divide them all by the smallest quotient). thats how many of each element is in the empirical formula(3 votes)
- Why is it CH3 not H3C?(2 votes)
- 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.(3 votes)
- 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(1 vote)
- The question states that the compound only contains carbon and hydrogen,0:00.
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.(3 votes)
So how does one perform combustion analysis on an unknown substance that contains carbon, hydrogen, and oxygen?(2 votes)
- [Instructor] We are told that a sample of a compound containing only carbon and hydrogen atoms is completely combusted, producing 5.65 grams of carbon dioxide and 3.47 grams of H2O or water. What is the empirical formula of the compound? So pause this video and see if you can work through that. All right, now let's just try to make sure we understand what's going on. They say that I have some mystery compound. It only contains carbon and hydrogen, so it's going to have some number of carbon, I'll call that x. Some number of hydrogens, I'll call that y. We're going to put it in the presence of molecular oxygen and it's going to combust and after it's combusted, I'm going to end up with some carbon dioxide and some water. And what I just drew here, this is a chemical reaction that I'm describing. I haven't balanced it. I could try to, even with the x's and y's, but that's not the point of this video. The point of this video is they tell us how many grams of the carbon dioxide we have and how many grams of the water we have, they tells us that right over there. And so what we need to do is say, all right, from that, we can figure out how many moles of carbon dioxide we have, how many moles of water we have, and from that, we can figure out how many moles of carbon did we start with and how many moles of hydrogen did we start with? And if we look at those ratios, then we can come up with the empirical formula of the compound. So just to start, because I'm going to be thinking about molar moles and molar masses and the mass of a mole of a molecule or an atom, let's just get the average atomic mass for carbon, hydrogen, and oxygen for us to work with. So I'll get out our handy periodic table. We can see hydrogen has an average atomic mass of 1.008. Let me write that down. So we have hydrogen is at 1.008, and then we have carbon, and carbon is at 12.01. So carbon is at 12.01, and we could also think about them in terms of molar masses. We could say this is gram per mole, grams per mole, and then last, but not least, we have oxygen and then oxygen is at 16.00 grams per mole. It's the average atomic mass, but whether we can think of that as molar mass, that number is molar mass. So oxygen is at 16.00 grams per mole. And so now we can try to figure out how many moles of C in the product do we have? So we can see that all of the carbon in the product is in the carbon dioxide that's in the product, and so we have 5.65 grams of CO2. We can think about how many moles of CO2 that is, so times one mole of CO2 for every how many grams of CO2. Well, we just have to think about, actually, let me just put it right over here. CO2, you're going to have, let's see, you have one carbon and two oxygens. So it is going to be 12.01 plus two times 16. Two times 16.00 grams per mole, and so let's see. This would get us to, this is 32 plus 12.01, so that is 44.01. And that's grams per mole, but now we're thinking about moles per gram, so it's going to be one over 44.01 and so if we did just this, the grams of CO2 would cancel the grams of CO2 and this would give us moles of CO2, but I care about moles of carbon in the product. So how many moles of carbon are there for every mole of CO2? Well we know that we have one mole of carbon for every one mole of CO2. Every carbon dioxide molecule has one carbon in it, and so what is this going to get us? So we have 5.65 divided by, divided by 44.01, and then times one, so I don't have to do anything there. That is equal to, I'll round it to three digits here, so 0.128, so this is 0.128 and my units here are, let's see. The grams of carbon dioxide cancel the grams of carbon dioxide. The moles of carbon dioxide cancel the moles of carbon dioxide, so I am exactly where I want to be. This is how many moles of carbon that I have. And you can do the dimensional analysis, but it also makes intuitive sense, hopefully. If this is how many grams of carbon dioxide we have and a mole of carbon dioxide is going to have a mass of 44.01 grams, well then 5.65 over this is going to tell us how, what fraction of a mole we have of carbon dioxide and then whatever that number of moles we have of carbon dioxide is gonna be the same as the moles of carbon, 'cause we have one atom of carbon for every carbon dioxide molecule, so that all makes sense. And now let's do the same for hydrogen. So let's think about moles of hydrogen in the product, and it's going to be the same exercise. And if you're so inspired and you didn't calculate it in the beginning, I encourage you to try to do this part on your own. All right, so all of the hydrogen is in the water, so and we know that we have, in our product, 3.47 grams of water. 3.47 grams of H2O, and now let's think about how many moles of H2O that is. So that's going to be, let's see, every one mole of H2O is going to have a mass of how many grams of H2O? And we could do that up here. H2O, it's going to be, we have two hydrogens, so it's going to be two times 1.008 plus the mass, the average atomic mass of the oxygen is going to be plus 16, but we can also view that as what would be the mass in grams, if you had a mole of it? And so this is going to be in grams per mole, and so this is going to be, let's see, two times 1.008. This part over here is 2.016, and then you add 16 to it. It's going to be 18.016. 18.016, and then if I just calculated this, this would give me how many moles of water I have in my product, but I care about moles of hydrogen. And so how many moles of hydrogen do I have for every mole of water? So for how many moles of hydrogen for every mole of water? Well I'm going to have two moles of hydrogen for every mole of water because in each water molecule, I have two hydrogens. And so that's going to cancel out with that and we're just going to be left with moles of hydrogen. And so this is going to be equal to, I'll take my 3.47 grams of water, divide it by the, how much, how many grams a mole of water, what its mass would be, so divided by 18.016. This is how many moles of water I have. Now for every, for every molecule of water, I have two hydrogens, so then I will multiply by two. Times two is equal to zero point, I'll just round three digits right over here. 0.385. 0.385 moles of hydrogen. So now we know the number of hydrogen atoms. We know the number of carbon atoms, and to figure out the empirical formula of the compound, we can think about the ratio between the two, so I'm gonna find the ratio of hydrogens to carbons. And that is going to be equal to, I have 0.385 moles of hydrogen over, over 0.128 moles of carbon. 0.128 moles of carbon, and what is this equal to? And it looks like it's going to be roughly three, but let me verify that. In my head, that seems so, I already have the hydrogen there, and so if I divide it by 0.128, I get, yep, pretty close to three. Now actually, I think if I, yep, pretty close to three, so there you go. This is approximately three, and so I can, with pretty good confidence, this was very close to three. I could say for every carbon, I have three hydrogens in my original compound, in this thing right over here. So the empirical formula of our original compound for every one carbon, I have three hydrogens. So CH3, and we are done.