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General Chemistry

Learn the toughest concepts covered in Chemistry with step-by-step video tutorials and practice problems by world-class tutors

3. Chemical Reactions

Combustion Analysis

Under Combustion Analysis, the empirical formula of a compound is determined through a combustion reaction. 

Combustion Analysis
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Combustion Analysis

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combustion analysis represents yet another way to determine the empirical formula of a compound. It represents an analytical process that determines the empirical formula of a compound. It does so by utilizing what we call combustion reactions. Now. Normally, combustion reactions involve a compound composed of just carbon and hydrogen. These we call hydrocarbons because it's hydrogen and carbon together, or a compound composed of carbon, hydrogen and oxygen together reacting with gas. Now here, when we have ah, hydrocarbon or compound composed of carbon, hydrogen and oxygen. When it reacts bio to buy combustion reaction, the products form will be CO two and water. Now if we take a look here at some common examples of combustion reactions, Remember, we talked about octane when we discussed the molecular formula. Octane, it was C eight h 18. So this is Ah, hydrocarbon. It has only carbons, and hydrogen is with it. When it reacts with 02 it creates CO two and water. Another compound that we talked about previously is glucose. Remember the molecular formula glucose is C six, H 12. 06 This compound is comprised of carbon, hydrogen and oxygen. So when it reacts with 02 it produces also, see co two and water here. We're just concerned with the fact that we have a hydrocarbon and then a compound of carbon, hydrogen and oxygen when they react with we produce as a result, CO two and water. So that's the basic idea behind Ah, typical combustion reaction. We won't worry about balancing these chemical reactions were just concerned with the products that are being formed. So just remember, we're gonna talk about combustion analysis a little bit more, but it just represents yet another way to find empirical formula, and it does sold to the use of combustion reaction.
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Combustion Analysis Example 1

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will now take a look at the combustion of carbon compounds. Here in this example, problem in states, a 0.2500 g sample contains carbon, hydrogen and oxygen and undergoes complete combustion to produce 0.3664 g of carbon dioxide and 0.1500 g of water. What is the empirical formula of the compound? All right, so this is typical of any combustion analysis type of question, and usually we'll have the production of sealed to in water. Now, these are the steps we're gonna have to undergo in order to determine the empirical formula off our compound. Now, Step One says that we have to convert the grams of carbon dioxide, which is co two into just grams of carbon now within C 02 We have a C in there, so we're just trying to figure out how many grams of C R within it. We're going to start out with 0.3664 g of co two. The first step is to get rid of these grams of co two. So grant of seal to go on top and on the bottom and on top, we have one mole of CO. Two now. At this point, we need to determine the molar mass of carbon dioxide. Carbon dioxide is composed of one carbon and two oxygen. Their atomic masses from the periodic table are 12.1 g and 16 g, respectively. Multiply everything out and then add those numbers together and we'll have the mass of carbon dioxide as being 44 g. So we're gonna be dealing a lot with carbon dioxide when it comes to combustion analysis type of questions. So it's pretty important to remember the Mass of CO two comes from one carbon into oxygen's. At this point, the grams cancel out and we have moles of Co. Two. We're going to stay here for every one mole of co two. We have exactly one mole within its formula. If you look at co two, there's one carbon for two oxygen's. So for every one mole of the whole thing, we see that there's only one card, one mole of carbon. At this point, moles of carbon dioxide cancel out and now we have moles of C. Remember, we don't want to just stop it. Moles of See, we want to stop at grams of C. We're going to stay here for every one mole of carbon. It's mass again. According to the periodic table is 12. with moles of carbon. Cancel out. We'll have now the grams of carbon, which comes out to be 0. g of C first step to we're gonna do the same thing with water, but here, we're gonna find the grams off hydrogen. Now, the reason that we're finding grams of carbon and hydrogen first is because they're located in only one place. See is only located in CO two and H is only located in water. It be hard for us to determine the amount of grams of oxygen because oxygen is present within both compounds. Okay, so again we find carbon and H first because they're located in only one of each of those compounds we have here 1500 g of water. We're gonna convert grams of water into moles of water, so grams of water go on the bottom. Now we need to determine the Moller Mass of water. Water has in it to hydrogen and one oxygen their masses, according to the periodic table are 1.8 g and 16 g. So that is 2016 g and 16 g. So that comes out to 18 06 g. So here grams of water, cancel out. Now I have moles of water. At this point, we're gonna say for every one mole of water, we can see that the formula contains two hydrogen. So for every one mole of water, we have two moles of H. So moles of water cancel out now have moles of H For every one mole of H, it's mass is 1.8 g h. So now we're gonna get 0. g of h. So we found out how many grams of h grams of carbon within which exists within our sample. So that takes us to step three in Step three, We say, if necessary, subtract the grams of step one and two in the grams of the sample to determine the third element in this case, we're told that our sample has in it carbon, hydrogen and oxygen, and we're told that the sample ways 0.2500. So it has carbon, hydrogen and oxygen in it. So subtract out the grams of carbon, plus the grams of hydrogen, and what will be left would have to be my grams of oxygen. All right, so when we do that, we're gonna get 4 g of oxygen 40.1332 to 5 g oxygen. Now, this is important, because now that we know the grams of each of the elements, the steps kind of become familiar to us, because at this point, we could determine the empirical formula. So here we're going to step four, we're gonna convert all the masses that we've just isolated from our three elements into moles. So take each one of those sample amounts that you just got each one of these element amounts that you just got and convert them all into molds. So remember, we divide them all by their atomic masses, and we'll have the moles of each of them so we can see that this process is not a quick process. It does require a good deal of steps, so make sure you follow each of these steps thoroughly to get your final answer. And then we have 0.1 332 to 5 g of oxygen. And that's one mold. 0 16 g of Oh, so when we do that, we're gonna get for the moles. For each we're gonna get 0.833 moles of C We're gonna get 0167 moles of H We're gonna get 833 moles of O That leads us into step five. We're going to divide each bowl, answer by the smallest small value in order to obtain whole numbers for each element. The smallest mall answer that we got was 0. So each of them gets divided by that. So when we do that, we're going to get one C to H. One of now. This will give us our empirical formula of C H. 20 Now let's say we didn't get whole numbers. Then we go into step six and Step six says if you get a value meaning these numbers here and they had a 0.1 or nine, then you can round to the nearest whole number. So if you got 1.1, you go round down to one. If you've got 2.9, you could round up to three. Now, if you can't round that, we'd multiply them all by an integer a number in order to create whole numbers. In this case, we didn't have to do that. Now, in this case, based on the steps that we've taken, the empirical formula for our compound would be C H 20
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Problem

Valproic acid, used to treat seizures, is composed of carbon, hydrogen and oxygen. A 0.165 g sample is combusted to produce 0.166 grams of water and 0.403 grams of carbon dioxide. If the molar mass is 144 g/mol, what is the molecular formula?

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Problem

When hydrocarbons are burned in a limited amount of air, both CO and CO2 form. When 0.460 g of a hydrocarbon was burned in air, 0.477 g of CO, 0.749 g CO2, and 0.460 g of H2O were formed. What is the empirical formula of the compound?

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Combustion Analysis

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We've talked about the combustion of hydrocarbons compounds that possess Onley, hydrogen and carbon. But now we're gonna take a look at non hydrocarbons. We're going to stay here. A non hydrocarbon represents a compound containing not Onley carbon and hydrogen, but also sulfur, nitrogen or a halogen. Now, remember, your halogen are the elements from group seven A. So we're talking about flooring, chlorine, bromine or iodine. We're gonna say through combustion analysis they can create gashes products such as eso to which is called sulfur dioxide N 02 which is called nitrogen dioxide as well as die atomic molecules. Remember, the halogen in their natural forms exist as f to a C L to be our to and I to. So this is a little bit different from what we're accustomed to seeing with combustion reactions where we typically just see seal to and water being formed but realize sealed twin water being formed is strictly for hydrocarbons and compounds with carbon, hydrogen and oxygen. Now, when we introduced these new elements, expect to make new types of final products, So just keep that in mind when we're dealing with non hydro non hydrocarbons within our question
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Combustion Analysis Example 2

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in this example question. It says, Ah, solve aging agent that contains carbon, hydrogen and chlorine is used for spectroscopic processes. Work told also that it has a molar mass of 1 47 g per mole. Now here, it says determined its molecular formula when a 0.250 g sample contains 0.451 g of carbon dioxide and 0.617 g of water upon combustion. Now here we're dealing with a non hydrocarbon, so we're gonna have an element that's different from just carbon and hydrogen. Here, the element is chlorine. This is gonna be our third element. Some of the steps you will be similar to what we've seen in the past. But because we're dealing with chlorine is just gonna be a little bit different at one point now, if we take a look at Step One, it says, in present, convert the grams of carbon dioxide, 2 g of carbon. So this is what we've done before with combustion analysis. We take our grams of carbon dioxide. We change those grants into moles. So one mole off carbon dioxide, remember, it's possesses one carbon and two oxygen when you look up there, masses on the periodic table and add them together. That gives you 44.1 g of carbon dioxide grams Here, cancel out. Next, we're gonna convert moles of carbon dioxide to just moles of carbon. For every one mole of carbon dioxide we see in the formula that there's just exactly one carbon. So that's one more of carbon. Then finally, we're going to convert. Finally, we're going to convert our moles of carbon into grams of carbon. So one mole of carbon weighs 12.1 g. When we multiply everything out together, we're gonna get here. 0.1231 grams of carbon. Step two, we're gonna convert the grams of water into grams of just hydrogen. So here we're gonna take our 0.617 g of water. Water possesses two. Hydrogen is and one oxygen. When you add their masses together from the periodic table, you'll see that waterways 18.16 g of water for every one mole of water. Then we're gonna stay here for every one mole of water we can see in the formula. We have two hydrogen. So for every one mole of water. There are two moles of just H and then for everyone, Mol of H. It weighs 1.8 g h. Moles cancel out and then we'll get the mass of hydrogen as 0.69 g of H now for step three. We're accustomed to subtracting out the amount of carbon and hydrogen from our sample to get our oxygen. But here we're dealing with a non hydrocarbon. So our third element is not oxygen, but instead chlorine. So step three, if necessary, to attract the grams of steps one and two from the grams of the sample to determine the third element, we're told that our sample is 20.250 g that contains carbon, hydrogen and glory. Subtract out the grams off the carbon and the grams off the hydrogen, and that will give us our grams of chlorine. So when we do that, we're going to get as our grams of chlorine 0.1 to g of chlorine. So at this point we have the grams of carbon, grams of hydrogen and grams of chlorine. So Step four tells us to convert all the masses into moles. So we're gonna convert each of those grams that we got into moles. So remember to do that, you just divide them by their atomic masses from the periodic table. So one mole of carbon, one mole of H one mole of chlorine Here are their atomic masses. From the periodic table grams he will cancel out, and we'll be left with moles of each one of these elements. So when we get that, we're gonna write their moles here, down here. So 0.102 moles of C, we're gonna have here 0.68 Moles of H. And then finally, we're gonna have our moles of chlorine, which is 0.0 34 Moles of chlorine. Now, Step five divide each Milan to buy the smallest more value in order to obtain whole numbers for each element. So we're gonna divide all of them by the smallest mall. Answer that we got, which is 0.34 and again 0.34 So that's going to give me three carbons, two hydrogen and then here, one CEO. That means that my empirical formula at this point is C three H two C L Now, step six. If you get a value, that was 60.1 or 0.9. We could just round to a whole number here. We don't have to worry about that because all of these numbers are whole numbers. Now, if we go back up here, notice that they're asking what is the molecular formula? All we've done at this point is determined the empirical formula. So we come back down here now, remember, it is R N factor R n factor Times are empirical formula equals are molecular formula. Right. So here we need to determine what our end factor is. So n equals R. Moller Mass divided by our empirical mass. Remember, your empirical mass comes from the empirical formula the C three h to seal that we isolated. So here, if we add up the three carbons, the two hydrogen and one cl, it will give us the empirical mass. So when we do that so we're gonna say we have three carbons. Two hydrogen is one cl. You multiply them by their atomic masses on the periodic table. Okay, so then that's gonna give us 36.3 g 2. g and 35.45 g. Add them all together that's going to give us our empirical mass are empirical mass. When we add them all together, get comes out to 73.496 g per mole. And on top, we just have the molar mass that give us in the very beginning, which is 1 47 g per mole. When you divide, those two gives us approximately two. So that tells me that my in factor is too. So I'd say to timed my empirical formula of C three h two c l all the subsequent get multiplied by two. So now come out to C six h four c l two so that there would represent our molecular formula. So realize that this is slightly different from our previous calculations in combustion analysis when we're dealing with typical carbon compounds. But now we're dealing with non hydrocarbons, so we're gonna have a different element that we're not normally going to see such as chlorine. In this case,
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Problem

If a compound made of calcium, carbon, nitrogen undergoes combustion in oxygen to create 2.389 g CaO, 1.876 g CO2, and 3.921 g NO­2, determine its empirical formula.

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The combustion of 4.16 grams of a compound which contains only C,H,O and F yields 7.7 g CO2 and 2.52 g H2O. Another sample of the compound with a mass of 3.63 g is found to contain 0.58 g F. What is the empirical formula of the compound?

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