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The combustion of 0.4196 g of a hydrocarbon releases $17.55 \mathrm{kJ}$ of heat. The masses of the products are $\mathrm{CO}_{2}=1.419 \mathrm{g}$ and $\mathrm{H}_{2} \mathrm{O}=0.290 \mathrm{g} .$ (a) What is the empirical formula of the compound? (b) If the approximate molar mass of the compound is $76 \mathrm{g}$ calculate its standard enthalpy of formation.

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$49 k J / m$

Chemistry 101

Chapter 6

Thermochemistry

University of Central Florida

Rice University

University of Maryland - University College

Lectures

05:27

In chemistry, a chemical reaction is a process that leads to the transformation of one set of chemical substances to another. Both reactants and products are involved in the chemical reactions.

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In chemistry, energy is what is required to bring about a chemical reaction. The total energy of a system is the sum of the potential energy of its constituent particles and the kinetic energy of these particles. Chemical energy, also called bond energy, is the potential energy stored in the chemical bonds of a substance. Chemical energy is released when a bond is broken during chemical reactions.

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So this problem has two parts and it involves the combustion of a certain hydrocarbon. And party asked us to find the empirical formula for the compound. So another in general, for any hydrocarbon reaction that is the combustion reaction that we have Hydrocarbon. What I will call X is one of the reactions. You have some out of oxygen. We have CO two and liquid water. So if we know it's high hydrocarbon, the only elements in that compound or hydrogen and carbon and that can all be traced to the amount of carbon in CO. Two in the amount of hydrogen and h 20 So the way they were going to go about finding the empirical formula, the compound is finding how many moles, uh, carbon were there in co two relative to how many moles of hydrogen were in the hydrocarbon from the H 20 So we know that the masses of the products is 1.419 grams. That's for the CO two, and for the water it was point 290 grams. So what we can do is and we can multiply both of these by their respective more masses and we confined. How many moles of each were created? Let's first do the carbon dioxide. It's we have 4.419 grams FC or two. You knew it for every one mole of U two or the more massive sea or two that is 44 point. Oh, and only multiply that out. You get that approximately your 0.3 2 to 4 moles of CO. Two were produced and now we go to water. We know that there was 0.29 You're Grimm's water produced, and we know that the more mess of liquid water is approximately 18 grams. Then when will try this out? You get it? There was 0.1 six 11 moles of each 20 formed. So now we need to relate this to carbon and hydrogen. So in 0.3 to 2. For moles of CO two, there must be the exact same amount of moles of carbon because there's one mole of carbon in every mall of see you too. So that corresponds to your 0.3 2 to 4 mol of C. On the other hand, when we have 0.1611 mole of H 20 that that that doesn't necessarily correspond to the moles of H because in every Mol of H 20 there's two moles of hydrogen atoms. So we need to multiply this by two. And when he multiply that answered by two we get there's zero point zero 3222 moles of hydrogen. It's not to find the empirical formula. We need to find the ratio of carbon to hydrogen. And as you can very clearly see that the ratio of C two h it's just this and that is very, very close to one. So we know that our hydrogen carbon are hydrocarbon, has an equal amount of moles of carbon and hydrogen. So we know that the empirical formula must be ch. So that solves part a of this question. So now what we need to do is we need to find the Standard and Tapia formation for this hydrocarbon. Given that we know that it's approximate Moeller mess. So now that we actually know it's more mass, we confined the actual formula for the compound instead of simply using the empirical formula. So we know that our empirical formula is C H. And the more mass of ch is around 13 grams per mole. That's a rough approximation. And we know that the molar mass of the actual compound is 76 grams per mole 76. So we confined. How many? What, how maney ch is actually fit into our hydrocarbon to find the actual compound that we're dealing with. So all we do is we take 76 and then we would divide that by 13. And when you plug that into your calculator, you'll get that approximately five point 84 which it is roughly six. So now we know that we just have to multiply our empirical formula by six to get the actual compound we get that are actual compound C six h six. And so that matches both the empirical formula and the molar mass. I see six h six happens to be benzene. So now we know that the compound we're dealing with his benzene, we need to find a reaction that is suitable for finding the heat of formation of benzene. So what we're gonna do is we're gonna actually right. The combustion reaction for benzene and we're going to reverse engineer it so we can actually find the heat information of benzene. So you know that, as always, combustion reactions have oxygen as a reactant liquid water and co two as a products. Now we need to balance it first. Look at carbon six on the left, one on the right. We multiply basics to correct that hydrogen. Six on the left, two on the right, multiplied by three to fix this. And now we look at oxygen. So we have three oxygen from the water. 12 oxygen from this year. 2.5 6 we have zero extra. So this now on his ticket count for 15 so we can add 15/2, and that will correct it. However, this isn't the final reaction because there is a fraction as a coefficient for the 02 So to correct that need to multiply everything by two. And then you will get our final combustion reaction, which is and following. So this is the combustion reaction that that we wanted. So now how how do we reverse engineered this? So we know that the this reaction is found by taking the heats of formation of the products and then subtracting out the heats information of the reactions. But we can ignore the oxygen because it's an element in the heats. Informations of elements zero. So that's gonna be our strategy. And we know that two times, whatever the information of benzene happens to be is going to be subtracted out. But since are variable is in the heats of formation of reactions. We need to actually find what is the heat of this entire reaction. Well, we're given a step towards that answer in the problem, because we know that 5th 17 0.55 killed jewels of Pete are released whenever zero point 4196 grams of the hydrocarbon that we're dealing with her combusted. But we know what the hydrocarbon aids and we confined. How much is released whenever two moles of Benzino reacted like we have here in a reaction. So we just take you would just take the molar mass of benzene and has given to us in. The problem is around 76 grams who just multiplied by 76 to get one more of benzene and then you would multiply this by two moles event because That's how much we have here in a reaction. So that will give us how much heat is going to be released in one mole of this reaction. And you will get that that is approximately negative. 6534 killing jewels, promote of the reaction. So now we can solve this problem once and for all. So we have negative 6534 kill Egil from all that's the entropy of this reaction and that is found by multiplying heats, informations of the products and then subtracting out the information of benzene. So we have six moles of water and the heat information of water is negative. 285.8 killer jewels from all. And then we have carbon dioxide in the heat of that information that is negative. 200 and 93.5 Children from all and then finally subtract out two of the heat information of benzene. But we don't know that, So I will simply call that X. So now we have finally equation and we can solve for X. So we know that Chu X is equal to this expression and then we just divide both sides. Thank you. When you plug all of that into your calculator, you will get that The heat of formation of benzene is approximately positive. 49 killer jewels promote Rex, and that is the final answer.

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