Combustion Apparatus - Video Tutorials & Practice Problems
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A Combustion Apparatus vaporizes a sample into gases to later determine its formula.
Combustion Apparatus
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concept
Combustion Apparatus
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Now remember that combustion analysis is just an analytical process that helps us to determine the empirical formula of a 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 o two 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 carbon, it's normal for it to become c02. We talked about non hydrocarbons. So if it's nitrogen, it would become n o 2. If it's sulfur, it becomes s o 2. If it's one of the halogens, it becomes a diatomic molecule. Now, traveling out of chamber 2, this is where we have the splitting of water and our gas. In chamber c, this is where water is being trapped. And again, based on the identity of the non metal, we could make different types of gases. C o 2 if the non metal was carbon, n o 2 if it's nitrogen, s o 2 if it's sulfur, and then one of the diatomic halogens if it happens to be a halogen. Now what comes out after all of this is any excess oxygen that remains, and this is 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.
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example
Combustion Apparatus Example 1
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The compound that makes up an experimental jet contains carbon, hydrogen, and oxygen. Suppose that in one experiment, the combustion of 11.5 grams of the liquid gave the follow-up results. Here, we're 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 a 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. Alright. So if we take a look at the steps. So step 0, we're going to subtract the grams after combustion by the grams before combustion to find the grams of c o two and water. Alright. So we're gonna say we have 32.815 grams of c o 2 after combustion minus 10.815 grams of c o 2, before combustion. That tells us how much c0 2 was created. When we do that, we're going to get 22.0 grams of c 0 2, and then we're gonna do 23.610 grams of water minus 10.110 grams of water. So that gives me 13.5 grams of water. So this is how much c o two and water that was produced. So at this point, it then continues into a regular combustion analysis type of question. We're going to do step 1 where we're going to convert the grams of CO 2 in 2 moles or into grams, actually, grams of carbon. So we're gonna have 22.0 grams of CO 2. What we wanna do first is we wanna cancel out that's right over here. We have 22 grams of c02. We wanna change those grams of c o 2 into moles of c o 2, so one mole of c o 2 on top. At this point, we've seen carbon dioxide numerous times. So the one carbon and the 2 oxygens weigh 44.01 grams. Grams cancel out. Now convert the moles of c o 2 into just moles of carbon. Within this formula, we have only 1 carbon, so that's 1 mole of c. And then finally, for every 1 mole of c, we're told according to the periodic table, its atomic mass is 12.01 grams. So when we do that, we're gonna get the mass of of c which comes out to 6.0036 grams of c. Next, we're gonna convert grams of water into grams of hydrogen. So we have 13 point oops. 13 0.5 grams of water. We're gonna say here that 1 mole of water, which possesses 2 hydrogen's and one oxygen, is 18.016 grams. Then we're gonna convert moles of water into just moles of h, and remember for every one mole of water, we see that there's 2 hydrogen's within the formula. And then convert the 1 mole of h into grams of h. So when we do that, we're gonna get 1.5107 grams of h. Step 3, if necessary, subtract the grams of step 12 from the grams of the sample to determine the 3rd element. Well, 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 grams, which contains all three elements, subtract out the grams of carbon plus the grams of hydrogen gives us 3.9857 grams oxygen. Now that we have the grams of all three elements that comprise my compound, we move on to step 4. In step 4, you convert all the masses into moles. So we're gonna bring down these masses here. So 6.0036 grams carbon, so 1 mole of c and on the bottom, the mass of c. Then we have 1.5107 grams h. For every 1 mole h, it's 1.008 grams h. And then we have 3 let's see. So we have 3.9857 grams o. And so for every one mole of o, it's 16 grams. Remember, these masses are 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 0.499 9 moles of c. This equals 1.4987 moles of h, and this equals 0.249 1 moles of o. At this point step 5, 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 were the moles of the oxygen, So everyone gets divided by 0.2491. So when we do that, that's gonna give us ratios of each of these elements. So that gives me 2 carbons, 6 hydrogens, and 1 oxygen. Now, since we got whole numbers, we don't have to worry about rounding. But if we did, we'd look at step 6. If you get a value of 0.1 or 0.9, then you can round. But again, at this point, we've determined what our empirical formula is. We've determined that it's c286o internal in terms of our answer.
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Problem
Problem
The following results are obtained from burning 5 g of an organic molecule (MW = 233.07 g/mol), which contains C, H, and O, in a combustion chamber:
Determine the molecular formula of linolenic acid.