the next analysis that will give us data that we can use to find empirical and molecular formulas is called combustion analysis this is also a type of analysis that chemists do regularly and it is very similar in strategy to finding the empirical and molecular formula like we did from elemental analysis but because the data comes from a different method there are some tweaks to the beginning and what's different here is that combustion is a reaction with oxygen gas so o2 and the products are co2 and h2o assuming that you only have carbon hydrogen and oxygen in your sample we'll see some that are a little more complicated later on so this reaction which is right this is just we take the sample in we burn it we collect the carbon dioxide and water and the amount that we get depends on the amount that was in the sample that we combusted to begin with so our strategy step one the mass of carbon we will be able to get from the mass of co2 that's produced the mass of hydrogen will come from the mass of h2o that's produced the mass of oxygen if you have any you cannot use co2 or h2o to find the mass of oxygen in your original sample because you're adding oxygen when you do the reaction so step two if you need to find a mass of oxygen you'll need the mass of the sample and so this is where we will use conservation of mass if you saw that and remembered it from way long ago if you did any of the early readings that are associated with the video lectures the mass of the sample it's completely made up from the mass of the elements that are in there so if we know how much of it is carbon and how much of it is hydrogen the rest of it has to be oxygen so we'll set up that calculation then steps three through five are the exact same as elemental analysis we will have the mass of each element so we will calculate the moles of each element use the number of moles to write a preliminary formula which is kind of like a step 3.5 that's not written then find the empirical formula and if we have enough information find the molecular formula let's do this with vitamin c which has a molar mass of 176.12 grams per mole now you might have a question at this point how do you know the molar mass if you're doing an analysis to find out what the molecular formula is and look that's a really really good question because it seems arbitrary and fake that we would give you a molar mass and then ask you for the formula of something where so far you have needed the formula to find the molar mass it's not fake there are methods and experiments that we won't get to until second semester that you could take an unknown compound and do this experiment and it would tell you what the molar mass is so i promise that it's not just made up stuff for making the beginning of chemistry terrible there are experiments you can do that will give you a molar mass without knowing the molecular formula okay anyway sidebar over back to our problem step one was taking the grams of carbon dioxide and using it to find the grams of carbon we're going to apply a couple of things here that are from previous videos one is that we're going to need the molar mass of carbon dioxide so we know how much how many grams of carbon dioxide were produced because i didn't finish reading it in the problem but it was 1.50 grams of carbon dioxide that would let me convert to moles of carbon dioxide and then once i know how many moles of carbon dioxide i have because i know the relationship between the number of moles of carbon per mole of carbon dioxide i can get to moles of carbon and then from moles of carbon i could get to grams of carbon now you can see from how i've set up my calculation right of 1.5 grams of carbon dioxide times one mole of carbon dioxide over 44.01 grams of carbon dioxide times one mole of carbon over one mole of carbon dioxide i've stopped and just written the number of moles of carbon because in step three when we make a preliminary formula we will need moles of carbon so i'm not going to recalculate that later i'm going to write it down here and save it however in order to find the mass of oxygen i will need to know the mass of carbon so i do need both pieces of information i'm just going to make sure that i don't have to calculate the number of moles twice so my second line of the calculation is taking moles of carbon and converting it to grams of carbon for hydrogen we're getting this from the amount of water that was produced it was 0.41 grams of water and the method is the same i can go from grams of water to moles of water and then i can go from moles of water to moles of hydrogen but as you can see because i have a giant red box giant red stars at every corner this conversion factor is the number one mistake in calculations for combustion analysis there are two hydrogens for every one mole of h2o you cannot forget the two you will absolutely go down some crazy paths and end up with weird answers and potentially end up with an unsolvable problem if you do not have the two in here so two moles of hydrogen for every one mole of water this would be 0.046 moles of hydrogen again saving that mole value because i'm going to need it and then converting the moles to grams because i will also need the mass value the mass of oxygen again you can't use the mole ratios from carbon dioxide or water because the oxygens in carbon dioxide and water some of them came from vitamin c but some of them came from the o2 that we put into the reaction and we are just using massive amounts of o2 we're not measuring it so there's no way to tell which oxygens came from vitamin c and how many of them came from vitamin c from the carbon dioxide and water output we do however know that our entire sample weighed one gram and out of that one gram we know that 0.409 grams was carbon and 0.046 grams were hydrogen so anything that's left has to be the mass of oxygen this means that our mass of oxygen is .545 grams step three is to find the moles of each element now carbon and hydrogen we already have because we saved those values in step one for oxygen we just found the mass of oxygen in that sample but if we divide by 15.999 grams per mole we will end up with the number of moles of oxygen step four is to get to the empirical formula so first i'm going to set up my preliminary formula which is carbon with a .0341 mole subscript hydrogen with 0.046 moles oxygen with 0.0341 moles then the next step is to divide by the smallest number which is 0.0341 and what this gives me in my preliminary formula is carbon with one as its subscript hydrogen with 1.34 so not a whole number we're gonna have to do something there and oxygen with one so this is a case i told you we would get to one where you have to also apply the multiply to get a whole number step you can do a couple of things there are some tables that will show you if you have this as a subscript this is what you need to multiply you might be able to do it quote unquote by inspection so if you had let's say that it was 1.5 it might be obvious to you that you need to multiply by 2 to get to the next whole number the other thing that you can do is just try values and see what you get so if i were to multiply through by 2 well carbon and oxygen are fine but multiplying 1.34 by 2 gives me 2.68 so i'm still not close enough to a whole number so it's 2 doesn't work so then i would try 3 and it turns out if you multiply by 3 you do end up near a whole number so c3h4o3 is my empirical formula if three didn't work then i would try four if four didn't work i would try five hopefully you don't have to go up too far to get the numbers to turn into whole numbers in fact you should not have to go too far right obviously like 10 is probably not going to work depending on what you have for an answer but if you hit 10 and it's still not working go double check your other work and make sure that you've got all the pieces there so i have my empirical formula let's look for the molecular formula first by checking the molar mass of my empirical formula so three carbons four hydrogens and three oxygens would have a molar mass of 88.032 which does not match my actual molar mass so if i compare them by dividing so the actual molar mass divided by the empirical i get a factor of two so i need two of my empirical formulas to make the molecular formula so going through and multiplying all of those subscripts by two gives me a molecular formula of c6h8o6