Rob Lederer: To find the molecular formula, what you need to realize is that the empirical formulae is the lowest whole number ratio for that formula that's possible. CH2Cl is lowest whole number ratio, but actually might not be that formulae. We were told in the question that you through mass spectroscopy perhaps, which is just take a gaseous form of this compound and inject into a big machine and it breaks it down and calculates the molar mass. It gives you molar mass of 98.96 grams per mole. Well, that's not this number here which is actually the molar mass for this compound. But it's pretty obvious, isn't it? You take the empirical molar mass and divide it into the molar mass that you get for the molecular compound to get the real molecular formula, this way. obviously this divided into this gives you the number two which means that the molar mass is twice as big as this number which means the formula is twice as big which means you multiply every element by two. So the real formula for the compound is C2H4Cl2 and in the organic unit, that is actually a substance that is alkane, but it is disubstituted with chlorine. So would actually be di chloro be one, 1-dicholoro or 1, 2 dicholoro ethane. Where do I run and hide? Don't worry. If this question looks really tough, it can be really broken down into simple parts to be able to answer. But right away, it can be quite intimidating because here you got a compound, you don't know how many carbons, hydrogens and oxygens are in this formula. You know the mass of it, and you are told that it undergoes combustion to form this much Co2 and this much H2O. Now go back and find the formula for that compound. Okay, you have to be told and you were in this question that, all of the carbon in the combustion is then found in the Co2 and all of the hydrogen present in the combustion is found in the H2O. If you told that, the complete combustion occurs, okay. Now, you can set it up this way. You know that this is the mass of CO2, but do you know the mass percent of carbon in Co2. Because, if you do, you can actually figure out how much carbon is present in that sample of 1.53 grams. This is how you do it? 1.53 grams of Co2. Here is the mass percent really, okay of C in Co2. Its 12.01 grams of C divided by the entire molar mass that is the mass percent, isn't it? Yeah, we got over that. Okay, so when you do that for the Co2 and you do this for H2O, its 939 grams times, here is the mass percent of hydrogen in H2O. Look at, there is H2, 2.02 divided by 18.02 right? You get two numbers that give you the mass of carbon and hydrogen that are present in this and this. What we were told, all of the carbon here is in here. All of the hydrogen here is in here. So therefore, those are the masses of the carbon and the hydrogen in that compound. All we have to do is find the oxygen and we have got the mass of each of those elements in the compound. Then it is going to be just like an empirical type of formula question. Oh, we can do that. So, how do you find out how much oxygen was in here. Well, quite simply 0.800 grams of total minus those two masses here added together on the carbon and hydrogen give you the oxygen. So, here is the oxygen at 0.278 grams. Now to finish.