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Worked example: Determining an empirical formula from combustion data

AP.Chem:
SPQ‑2 (EU)
,
SPQ‑2.A (LO)
,
SPQ‑2.A.3 (EK)

Video transcript

we are told that a sample of a compound containing only carbon and hydrogen atoms is completely combusted producing five point six five grams of carbon dioxide and three point four seven 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 hydrogen's 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 its 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 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 tell us that right over there and so what we need to do is say alright 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 and 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 that one point zero zero eight and then we have carbon and carbon is at 12.01 so carbon is that 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 point zero zero grams per mole this is average atomic mass but then we can think of that as molar mass that that number is molar mass so oxygen is that 16 point zero zero 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 the carbon in the product is in the carbon dioxide that's in the product and so we have five point six five 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 say you have one carbon and two oxygens so it is going to be twelve point zero 1 plus 2 times 16 2 x 1600 grams per mole and so let's see this would get us to this is 32 plus 12 point oh 1 so that is 44 point oh one and that's grams per mole but now we're thinking about moles per gram so it's going to be one over forty four point zero one and so when we did just this the grams of co2 would cancel with 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 five point six five divided by divided by forty four point zero one 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 zero point one two eight so this is zero point one two eight and my units here are this e the grams of carbon dioxide cancel with the grams of carbon dioxide the moles of carbon dioxide cancel with 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 0.01 grams well then five point six five 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's going to be the same as the moles of carbon because 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 it's going to be the same exercise and if you're so inspired and if you didn't calculate it in the beginning I encourage you to try to do this part on your own alright so all of the hydrogen is in the water so and we know that we have in our product three point four seven grams of water three point four seven 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 hydrogen's so it's going to be two times 1.008 plus the mass the average atomic mass of the oxygen it's going to be plus sixteen 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 one point zero zero right this part over here is two point zero one six and then you add 16 to it it's going to be eighteen point zero one six eighteen point zero one six 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 mole moles of hydrogen and so how many moles of hydrogen do I ever 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 hydrogen's 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 three point four seven grams of water divide it by the how much how many grams a mole of water about what its mass would be so divided by eighteen point zero one six this is how many moles of water I have and now for every for every molecule of water I have two hydrogen's so then I will multiply by two times two is equal to zero point I'll just around three digits right over here zero point three eight five zero point three eight five moles of hydrogen so now we 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 hydrogen's to carbons and that is going to be equal to I have 0.38 five 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 zero point one two eight 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 conference this this was very close to three I can say for every carbon I have three hydrogen's in my original compound and this thing right over here so the empirical formula of our original compound for every one carbon I have three hydrogen's so ch3 and we are done