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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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example
Combustion Analysis
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Hey, everyone. So in this video, we're gonna take a look at the combustion of carbon compounds here in this example, it says 23 dihydroxy succinate acid is responsible for the assiduous flavor of some candies and is composed of carbon, hydrogen and oxygen. If the combustion of 12.01 g of this acid creates 14.08 g of CO2 and 4.32 g of water. What is its empirical formula? Right? So for a question like this, the first thing we're gonna do is we're gonna convert the grams of carbon dioxide into grams of just carbon, right? So to do this, we're gonna say we have 14.08 g of CO2 and we're gonna convert grams of Co2 into moles of CO2, 1 mole of CO2. The combined weight of the one carbon and the two oxygens is 44.01 g grams cancel out. Then we're going to say that for every one mole of this compound of CO2, we see that there's one carbon within the formula. So there's one mole of carbon, the CO2 Coor. Now that we have moles of carbon, we can just change those to grams of carbon and say for every one mole of carbon, its atomic mass from the periodic table is 12.01 g. When we work all this out, this gives me 3.8423 g. Carbon. Next step two, we're gonna convert the grams of water into just grams of hydrogen. Now, here we're focusing on finding carbon and hydrogen from these two compounds because that's the only place they were located. Carbon dioxide is the only place for carbon. Water is the only place for hydrogen, right? So we have here 4.32 g, water for every one mole of water. So water has two hydrogens and one oxygen. Their combined mass together is 18.016 g rams cancel up. We're gonna stay here for every one mole of H2O we can see from the formula that there are two hydrogens. So there's two moles of each. And then we say for every one mole of H its atomic mass from the periodic table is 1.008 g. So this is 0.4834 g each. All right. Now that we found the grams of carbon, the grams of H, we go to step three if necessary. So track the grams of step one and two from the grams of the sample to determine the third element. Remember here we have a third element in the form of oxygen. So our substance weighs 12.01 g. And that's the carbon, hydrogen and oxygen altogether. If we subtract that what we just found out for carbon and for hydrogen, the difference of what's left would be the grams of oxygen. So 7.6843 grams of oxygen. Now that we've done that step four, just be, convert all the masses into molds. So we're just gonna continue doing that over here since we already have all their grams here. Let's just change them to moles here. One mole of carbon is 12.01 g. One mole of hydrogen is 1.008 g, one mole of oxygen is 16 g. So when we do that, we're gonna get 0.3199 moles of carbon. Remember when we find the moles, we gotta make sure we have at least four decimal places to avoid rounding errors. This comes out to be 0.4796 moles of hydrogen. And this is 0.4803 moles of oxygen. So we found the most for everyone. Next, we're gonna divide each more answer by the smallest mole value in order to obtain whole numbers uh for each element. So we come back up here. So we're gonna write down each of these moles and divide them by the smallest mole answer. So we have 0.3199 moles of C. We got 0.4796 moles of H and then we had 0.4803 moles of O. The smallest mole answer from all of these was the 0.3199. So when we do that, that's gonna give me one carbon. This is gonna give me approximately 1.5 hydrogen and, and this is gonna give me 1.5 oxygen. Remember, we can only divide to the nearest whole number if we got a 0.1 or 0.9. Well, here we have point fives. That means we have to multiply in order to create whole numbers. So think about it. What number can I multiply the 1.5 by to make them whole numbers? It's right too. And if I multiply them by two, I have to multiply everyone by two. So it's two carbons, three hydrogens, three oxygens. So here my PCO formula becomes C two H 303. So this would be the empirical formula based on this given question.
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Problem
Problem
An unknown compound is composed of carbon, hydrogen and oxygen. A 4.30 g sample is ignited and creates 8.59 g CO2 and 3.52 g H2O. If the molar mass is 176.22 g/mol, what is the molecular formula?
A
C3H5O3
B
C2H4O4
C
C8H16O4
D
C12H16O4
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Problem
Problem
In the presence of a small amount of oxygen a combustion reaction will not only produce carbon dioxide, but also carbon monoxide. The incomplete combustion of naphthalene, a hydrocarbon used in many dyes, produced 2.80 g CO, 4.40 g CO2 and 1.44 g H2O. Determine its empirical formula.
A
C3H3O3
B
C5H4
C
C5H4O
D
C10H8
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concept
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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example
Combustion Analysis
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You were told that chloral methane, which has a molar mass of 50.084 g per mole. It's a numbing agent composed of carbon, hydrogen and chlorine and is used for sports injuries. Now, here it says, what is the empirical formula? When 0.9004 g sample produces 0.5811 g of carbon dioxide and 0.3568 g of water upon combustion. All right. So here we're gonna utilize the following steps in order to solve the empirical formula for this given example, question step one says if present convert the grams of carbon dioxide into just grams of carbon. So here we're gonna have 0.5811 g of carbon dioxide. We're gonna say here, first, one mole of carbon dioxide which is composed of one carbon and two oxygens. You're their molar masses from the periodic table. When we add those totals together, we get 44.01 g for carbon dioxide. Next, I'm going to say that for every one mole of carbon dioxide, we have one mole of carbon and for every one mole of carbon, we know that its atomic mass according to the periodic table is 12.01 g. When we solve this, we get 0.1586 g of carbon. Next, we have to convert the grams of water to grams of hydrogen. So here we're gonna say we have 0.3568 g of water. We have two hydrogens and one oxygen. So remember here's the atomic mass of hydrogen and oxygen. We 16. So their combined mass with the two hydrogens and one oxygen is 18.016 g. And that's for every one mole of water. Now, we're going to say that for every one mole of water, we see that there are two hydrogens in the formula. So it's two moles of H for every one mole of H, we have 1.008 g for that hydrogen. So that comes out to be 0.0399 g of H. So we found, we found carbon, we find, found hydrogen. Remember this sample has chlorine within it. So if necessary, subtract the grams of steps, one and two from the grams of the sample to determine the third element. So here we're going to say that the sample of the element is 0.9004 g. And that's composed of carbon hydrogen and chlorine. Together, we're gonna subtract out the amount of carbon and hydrogen that we found in order to isolate the grants of chlorine. So we're gonna do 0.1586 g, carbon plus 0.0399 g of each, what's left would be the grams of chlorine. So that's 0.4786 g of chlorine. We have the grams of all three. Now, so step force has convert all the masses into moles. So we're gonna take each of these grams that we found and change them into moles. So we have 0.158 6 g. Carbon 0.0399 g H and 0.4786 g. Chlorine divide them all by their atomic masses from the periodic table to isolate our molds, one mole of carbon, one mole H one mole. Chlorine. Here are their masses from the periodic table. And chlorine is 35.45 according to periodic table grams, all cancel out and we'll have moles of each. When I do this, I'm gonna get as my totals. 0.0132 moles of carbon, 0.0396 moles of H and 0.0135 moles of chlorine. Step five, divide each mole answer by the smallest mole value in order to obtain whole numbers for each element. All right. So here we're gonna say we have what we have 0.0132 moles of carbon, 0.0396 moles of H 0.0135 moles of chlorine. The smallest little answer we got was this 0.0132. So all of them get divided by that number. So here that's gonna give me one carbon three hydrogens and one chlorine. We got whole numbers for each of them. So we don't need to do this last step six. But if you get a value of 0.1 or 0.9 then you can round to the nearest whole number. We got whole numbers here. Um If you put in your calculator, you might get 0.1 for one of these, but it's gonna, you can round up no round down to the closest whole number. If you can't round we multiply by a factor to create whole numbers here. We didn't have to do that part. We already got whole numbers for everything. So that will mean that my empirical formula for chloral meth um for chloral methane is ch three. See you, this would be my final answer.
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Problem
Problem
A compound composed of carbon, hydrogen and nitrogen undergoes a combustion reaction to produce 264.21 g CO2, 63.06 g H2O and 46.4 g NO2. Determine its empirical formula.
A
C2H11N3
B
C6H7N
C
C6H12N4
D
CHN
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Problem
Problem
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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