Write the balanced chemical equation for the complete combusti...

Question

Write the balanced chemical equation for the complete combustion of each of the following compounds:

b. 3-hexanol

Accepted Answer

Hi, everybody. Welcome back. Our next problem says, provide the balanced chemical equation. When liquid to Piny undergoes complete combustion, we have four different answer choices labeled A through D, each one is a different equation. So we'll look over these and describe them in more detail after we work out kind of the generalities of what our balanced equation needs to look like. So let's think about what's happening in a combustion reaction. Our pinol will be reacting with oxygen gas. So it will be the other reactant will be 02. Since it's always in diatomic form gas, we'll just have a single arrow. It's a complete combustion doesn't go the other way it generates so much heat and our fentaNYL will be completely gone. Our only products will be carbon dioxide gas. So co2 gas plus water vapor. So plus h2o gas. So the next thing we need to be able to balance this equation is the molecular formula of our two fentaNYL, excuse me. So we'll start by thinking about our base compound here. Pinol, that OL ending will mean that we have pentane. So a five carbon chain with an O group. So let's start with the molecular formula of pentane. When we have an alkane like this, the general format of the molecular formula will be C subscript N H subscript N plus two. So in the case of pentane, it will be C five H 12. And then with that oh group from that base molecular formula, we'll need to subtract one hydrogen because the oh group will take the place of one of the hydrogens in this case on a hydrogen on carbon two. But we don't need to know that for our molecular formula. So we kind of just ignore the two. So I've said minus H and then we'll add one oxygen and one hydrogen from our alcohol group. So the number of hydrogen stays the same, we've subtracted one and added one and we just add an oxygen. So the molecular formula of our tiny will be C five H 12 0. So let's move on to writing out our equation and working on balancing it. So we'll have C five H 12 0, you know, it's in a liquid form plus and then I'm going to leave a space since I'll be adding coefficients here. So then 02 gas, my single arrow leave a space co2 gas plus leave a space H2O gas. Now, before we go any further and work through balancing this equation that takes some time. And one of the benefits of a multiple choice style question is that often I don't need to work through the entire problem to be able to eliminate enough choices to get to my answer. So in a problem like this, it's worthwhile, checking those answers over really quickly before working there, something like balancing the whole equation. So now is the time to look back and think about our answer choices. So in choice, a first, I'd look at, do I have the correct molecular formula for my fentaNYL? And do I have the correct reactants and products? So in choice, A, I do have C five H 12 0 plus 02 becomes co2 plus H2O. So I've got all the correct reactants and products. However, when I look at this equation, there are no coefficients. And of course, I know even without calculating what they are, I've got five carbons in the Piny and there's just one molecule CO2 on the right hand side. So this definitely is not a balanced equation. So without going any further, I can eliminate choice a right away. Choice B I have C five H 12 0. That's correct plus 1002. So I correctly, I have my +02 becomes a co two and then I have plus two co carbon monoxide. So I have an incorrect product here that then goes on to say 8 H2O, I'm not missing anything but I have a product I shouldn't have, there shouldn't be any carbon monoxide there that would only occur if I had incompletely combusted my hydrocarbon which is not the case. So choice B eliminate that right away. Now lets look at choice C it says two C five H 12 0 plus 1502 plus 10 CO2 plus 12 H2O. So I've got all the correct reactants and products and my molecular formula of fentaNYL is correct. So nothing to eliminate it upon first glance. And then in choice D well, I don't even need to look at are things correct in terms of reacting the products. My first thing here is 1.5 C five H 12 0 that gives me a fractional coefficient. And my equation could be balanced in terms of having matching numbers of atoms. But we just don't list balanced equations with fractional coefficients. It doesn't make sense to have fractional molecules. So choice D is ruled out just by being, having that fractional coefficient there. So this is a great example of how we don't even need to balance the equation. If we're on a test with this type of question, I, all I have left is choice C, I'll pick that as correct and move on. This is a practice problem. However, so to be thorough, we'll work our way through balancing the equation and just make sure it matches choice C. So I'm going to write my coefficients in red to make them stand out a little here. Let's work our way through the steps of balancing the equation. So my first step in a combustion reaction is to balance the carbon atom. So balance C. So on the left hand side, I've got the five carbons in my fentaNYL. On the right hand side, one carbon per molecule of CO2, that's the only source of carbon in the products. So I just put a five in front of my co two to balance that quite simple step two will be to balance my hydrogen atoms. So balance H on the left side, I have 12 hydrogens in my piny. On the right side, the only source of hydrogen is the two hydrogens per molecule of water. So I've got 12 hydrogens on the left. If I put a six in front of my h2o, I've got 12 on the right. So those two are pretty straightforward. Last of all, I balance the oxygen we see right away why we do this last? Because I have two sources of oxygen in my products and two sources in my reactants. So that's a little more complicated. On the left side, I have that single oxygen from my fentaNYL and then from oxygen, I have the possibility of two oxygen atoms per molecule. And you'll see why I write that in that form in just a moment. On the right side, for every co two, I have two oxygen atoms and there's five co2 molecules. So that gives me five multiplied by 2, 10 oxygen atoms coming from my carbon dioxide plus my water. There's one oxygen atom, my coefficient is six. So six oxygen atoms coming from the water, 10 plus six equals 16. So I have to balance the 16 on the right. So let's take away one that comes from the hydrocarbon and then I've got 15 left that I'll balance with oxygen. Why do I separate it like that? Because that oxygen has only oxygen in it. If I start adding coefficients to the fentaNYL, I'm changing the number of carbons and hydrogens on the left and I have to rebalance them so much simpler just to use the oxygen. So if I still need to make up 15 and I have two oxygen atoms per molecule, then the coefficient I need here is 15 halves. But now, as we mentioned, can't have fractional coefficients. I need to get rid of the fractional coefficients. So step four, eliminate fractional coefficients. Well, since my fraction has a two in the denominator multiplying the entire equation by two, we get rid of those and leave it still balanced. So I have two multiplied by the whole thing which will give me a final equation of two C five H 12 0 liquid plus I have two multiplied by 15 halves. So 1502 gas goes to two multiplied by 5, 10 co2 gas and then two multiplied by six, gives me 12 h2o gas. Does that match my equation in choice? C choice C says two C five H 12 0 liquid plus 1502 gas becomes 10 co2 gas plus 12 H2O gas. So choice C is indeed my correct answer for the balanced chemical equation. When liquid two fentaNYL undergoes complete combustion. See you in the next video.

Write the balanced chemical equation for the complete combustion of each of the following compounds:
b. 3-hexanol

Key Concepts

Combustion Reaction

Balanced Chemical Equation

3-Hexanol Structure

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