Combustion Apparatus

Now remember that combustion analysis is just an analytical process that helps us to determine the empirical formula of the compound combustion analysis is accomplished through the use of a combustion apparatus. Now this apparatus is where the sample is vaporized and goes through combustion by traveling between chambers. So here we have a basic type of combustion apparatus here it has chambers A through D. What happens first is that we have to place our sample within chamber A. And we allow oxygen gas to enter this. Remember, a key component of a combustion reaction is the presence of 0.2 as a reactant. Now, Chamber A. We've placed our sample within it and this is where the sample is vaporized. So it's changed into a gaseous state. It then travels into chamber B. And chamber B is where hydrogen is converted into water. So if this sample possesses hydrogen, it gets converted into water and the nonmetal portions of this sample gets converted into some type of gas. Now, if that gas, if that nonmetal happens to be Carden, it's normal for it to become ceo too. We talked about non hydrocarbons. So if it's a nitrogen, it would become N. 02. If it's sulfur, it becomes S. 02. If it's one of the halogen, it becomes a diatonic molecule. Now traveling out of chamber to this is where we have the splitting of water and our gas in chamber, see this is where water is being trapped and again, based on the identity of the non metal, we could make different types of gasses, C. 02 of the non metal is carbon N. 02. If it's nitrogen S. 02, if it's sulfur and then one of the di atomic halogen is if it happens to be a halogen. Now, what comes out after all of this is any excess oxygen that remains and this is a possibility. Sometimes a combustion reaction can happen in an environment that's rich in oxygen. The reaction takes place, but there's still oxygen all over the place. So just remember combustion analysis requires the use of a combustion apparatus. And now that we've seen this apparatus, let's take a look at some different types of combustion analysis questions that kind of go hand in hand with understanding this apparatus.

the compound that makes up in experimental jet contains carbon, hydrogen and oxygen. Suppose that in one experiment, the combustion of 11.5 g off the liquid gave the following results Here were given the masses of carbon dioxide and water before combustion and after combustion. Remember, when it comes to our combustion apparatus, we have our chambers, which basically collect the amount of carbon dioxide and water that's produced during our combustion reaction. Within these chambers, there had already been some carbon dioxide and some water. And what we're seeing is after combustion, how much more water and carbon dioxide was created. From this information, we need to determine the empirical formula of the unknown liquid. All right, so if we take a look at the steps, so step zero, we're going to subtract the grams after combustion by the grams before combustion to find the grams of C 02 and water. All right, so we're gonna say we have 32.815 g of co two after combustion, minus 10.815 g of CO. Two before combustion. That tells us how much co two was created when we do that We're going to get 22.0 g of CO two and then we're gonna do 23 610 g of water, minus 10.110 g of water. So that gives me 13.5 g of water. So this is how much sealed toe in water that was produced. So at this point, then continues into a regular combustion analysis type of question. We're going to do Step one or we're going to convert the grams of CO two in two moles, well, into grams actually grams off carbon. So we're gonna have 22 0 g of co two. What we want to do first is we want to cancel out. That's right. Over here. We have 22 g of co two. We want to change those grams of CO two into Moles of CO two. So one bowl of co. Two on top. At this point, we've seen carbon dioxide numerous times. So the one carbon in the two oxygen's way 44. grants grams cancel out now convert the moles of CO two into just moles of carbon. Within this formula, we have only one carbons That's one mole of C. And then finally, for every one mole of see, we're told according to the periodic table, its atomic masses 12.1 g. So when we do that, we're gonna get the massive sea, which comes out to 6.36 g of C. Next, we're going to convert grams of water into grams of hydrogen. So we have 13 point groups, 13 0.5 grams of water. We're gonna say here that one mole of water which possesses two hydrogen and one oxygen, is 18.16 g. Then we're gonna convert moles of water into just moles of H. And remember, for every one mole of water we see there's two hydrogen is within the formula. Come and then convert the one mole of H into grams of H. So when we do that, we're gonna get 1.5107 g of H Step three, if necessary, to track the grands of step one and two from the grams of the sample to determine the third element. What we're told within the question that our third element that makes up this jet fuel is oxygen. Our total amount of the liquid, which contains carbon, hydrogen and oxygen, is 11.5. So we're gonna do 11.5 g, which contains all three elements. Subtract out the grams off carbon plus the grams off hydrogen gives us 3.9857 g oxygen. Now that we have the grams of all three elements that comprise my compound, we move on to step four In step four, you convert all the masses into moles, so we're gonna bring down these masses here. So 6.36 g carbon So one mole of sea and on the bottom, the massive C. Then we have 15107 g h for every one more H. It's 1.0 g h, no, and then we have three c. So we have three 0.9857 g. Oh, and so for every one mole of oh, it's 16 g. Remember these masses air coming from the periodic table. So when we do that, we're gonna get the moles of each of these elements. This comes out to be 0.4999 moles of C. This equals 1.4987 Moles of H And this equals 0.2491 Moles of O at this point that find you divide each mole answer by the smallest mole value in order to obtain whole numbers for each element. So our smallest moles that we obtained with the moles of the oxygen So everyone gets divided by 0. So when we do that, that's gonna give us ratios of each of these elements. So that gives me two carbons, six hydrogen and one oxygen Try now, since we got whole numbers, we don't have to worry about rounding. But if we did, we look at Step six. If you get a value of 60.1 or 0.9, then you can round. But again, at this point we determine what our empirical formula is. We've determined that it's C 2860 in terms in terms of our answer