hi everyone in this last video I just wanted to do one more quick example of combustion analysis in this example we have a compound that contains carbon hydrogen and also oxygen so when 4.18 grams of this compound is burned we're able to collect 2.5 grams of water and 6.13 grams of carbon dioxide we're told the molar mass of the substance we want to figure out what the molecular formula is of the compound so just as we did in the previous video we want to use the mass of the water and carbon dioxide collected to figure out how many moles of hydrogen and carbon were in our original compound so that's where we're going to start to find the moles of carbon we use the mass of carbon dioxide that was collected in the combustion analysis which is 6.13 grams as we did previously we'll want to use the molar mass of carbon dioxide in order to convert this to moles the molar mass is 44.01 grams for every one mole of carbon dioxide so grams of carbon dioxide cancel and we have moles and then we write the stoichiometric factor based upon the molecular formula of CO2 to figure out how many moles of carbon this is we know that for every one mole of CO2 we have only one mole of carbon so this will give us the moles of carbon that were collected in the carbon dioxide this is 0.139 moles we can do the same thing with the water to figure out moles of hydrogen so starting with the mass of water which is 2.50 grams we're able to figure out how many moles of Water by dividing this by the molar mass of water which is 18.02 grams for every one mole of H2O grams of water cancel and then we can use the stoichiometric factor based on the molecular formula of water to know that this for every one mole of water we have two moles of hydrogen doing this calculation we find that we have 0.277 moles of hydrogen now when we got to this point in the last problem we were able to take the molar ratio to determine the empirical formula right the numbers of carbon and hydrogen that should be in our empirical formula however if we know that we have a certain mass of our sample that is made up of carbon hydrogen and oxygen we can subtract the mass of the carbon and the mass of the hydrogen and what should be left would be the mass of the oxygen so let's use these numbers of moles to figure out the mass of hydrogen and the mass of carbon the mass of carbon we can find by taking this number of moles and using the molar mass of carbon to figure out the mass so here we would say for every one mole of carbon right the mass would be 12.011 grams so this gives me a mass of 1.67 grams of carbon we can do the same thing for the moles of hydrogen from 0.277 moles of hydrogen I can use the molar mass of hydrogen where every one mole of hydrogen is 1.0079 grams so this tells me that the mass of hydrogen is 0.279 grams all right so now if we want to figure out the mass of oxygen we take our sample Mass which is 4.18 we subtract the mass of carbon which we found is 1.67 grams and we subtract the mass of hydrogen which is 0.279 grams so the mass of oxygen should be 2.231 grams and just as we did before for carbon and hydrogen we'll need to know how many moles this is so the moles of oxygen would be 2.231 grams and we use the molar mass of oxygen where one mole of oxygen is 16.00 grams grams cancel and this tells me that the number of moles of oxygen is 0.139 and as we learned previously if we want to figure out what our empirical formula is we can write our empirical formula with the number of moles that we've calculated and then we can divide all of these numbers by the smallest one so the smallest number is 0.139 and doing these calculations we find that the empirical formula should be c h two o now our problem told us that the molar mass of the substance is 180.156 and they want to know the molecular formula we just found the empirical formula in order to determine the molecular formula we would want to take the molar mass which is 180 and divide it by the mass of our empirical formula so this is 30. 03 grams per mole so this is the mass of our empirical formula ch2o when I do this calculation it's going to tell me the number by which I need to multiply all of my subscripts and so this calculation comes out to 5 which means that to arrive at my molecular formula for the compound I would need to multiply all my subscripts by 5. so this becomes C5 h10 o5