The combustion of liquid ethanol (C₂H₅OH) produces carbon dioxide and water. After 4.62 mL of ethanol (density = 0.789 g/mL) is allowed to burn in the presence of 15.55 g of oxygen gas, 3.72 mL of water (density = 1.00 g/mL) is collected. Determine the percent yield for the reaction. (Hint: Write a balanced equation for the combustion of ethanol.)

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Welcome everyone. Our next problem says the combustion of phenol C six H 50 produces carbon dioxide and water. After 10.6 mL of phenol density equals 1.07 g per milliliter is allowed to burn in the presence of 52.9 g of oxygen gas. 5.32 mL of water density. 1.00 g for milliliters is collected, determine the percent yield for the reaction. Hint, write a balanced equation for the combustion of phenol. And our answer choices are a 92.1%. B 81.6% C 72.4% or D 68.5%. So let's think about the information we're given and the information we need and how we would plan to get from one to the other. So the information we're given is that we have a specific amount of phenol and a specific amount of oxygen participating in the reaction. So we have known amounts of phenol and oxygen gas. And that's somewhat unusual. Usually you don't have a specific amount of oxygen and we know that we have a known amount of water produced and specifically, we can guess that this is not the theoretical possible yield since we're supposed to calculate the percent yield. So our reaction can theoretically yield more water than it actually does. So with these known amounts of phenol and oxygen, we have to figure out which of these is going to be the limiting reagent. So unless there are exactly equal numbers of moles of both these substances, one of them is going to be completely combusted before the other one is used up. And then once we know which is the limiting reagent, we would say how much water is theoretically produced by the complete combustion of our limiting reagent. In other words, how much water is theoretically produced when the limiting reagent is completely used up. And then we will compare that theoretical amount of water produced to the actual amount in the form of percent yield. So that's our sort of road map to where we're going before we even start out on this road. As our problem says, with its hint, we need to come up with a balanced equation because we need to be able to compare how many moles of our limiting region are used up and how many moles of water are produced. So let's write out this equation for the combustion of phenol phenol C six H 50 in combustion, we know it reacts with oxygen gas. So I leave a space before the oxygen to give me room to write in coefficients as I balance my equation space there. And then 02 remember that oxygen gas always exist as a diatomic molecule and this produces carbon dioxide co2 plus water. So when I'm balancing this type of equation, my first step will be to balance the carbon atoms. So I look on the left, my source of carbon is the phenol and there's one carbon atom from that on the right, it's the CO2 and oh excuse me getting mixed up here, not one carbon atom, there's six carbon atoms in our phenol. So six carbons on the right side, those in red to keep track here. And then on the right side is one carbon atom from carbon dioxide, flipping those two round. So all we need to do is put a coefficient of six in front of our carbon dioxide and we end up with six carbons on both sides, then we want to balance our hydrogen. So on the left side, now be careful because notice that the hydrogens are listed separately from each other, the five in the benzene ring and the one on the alcohol group. So that phenol has a total of six hydrogens got six hydrogens on the left side. Now, on the right side, we have water. So that gives us two hydrogens. So to balance the six and the two, we put a coefficient of three in front of the water and it gives a six on both sides. Our final step is to balance oxygen. We always do that last because as you can see in this type of reaction, we have oxygen atoms in every single molecule. So as we've changed, put coefficients in um we've made that a more complicated issue. So you save that for last. So our phenol has one oxygen atom, oxygen gas has two. Remember that as we put coefficients there, we're adding two every time carbon dioxide has six, multiplied by two is 12 and water has three, multiplied by one is three. So a total of 15 oxygens on the right. So on the left, well, the one in fol, if we subtract that we have 14 to make up, we have 02 providing two each time. And therefore putting the coefficient of seven in front of our 02 gives us 14 plus one is 15. So we now have a balanced equation and I'm just going to erase these balancing numbers to leave room for the calculations here. I'll go ahead and highlight in blue, my balanced equation so I can find it easily. So let's take that first step along the road. Taking the first step along the road, take the next step along the road, which is the limiting reagent. Well, how do I determine that I look at the amounts I've been given and determine how much water can be theoretically produced by each of those reagents. The one that, that produces the least amount of water is the limiting reagent. So let's start with phenol, right? Phenol, I know that I'm starting with 10.6 mL of phenol. So in order to get to moles so that I can get to moles of water, I've been wanting to use the molar mass which is in grams per mole. So I need to get 2 g. Well, I use density to do that 1.07 g per milliliter. So I set up my conversion factor. So I multiply that volume by I put the grams on top 1.0 0 7 g of phenol divided by 1 mL phenol. So now I'm on grams of phenol. Now I'm going to use the molar mass to go to moles of phenol. So the molar mass ofen is 94.11 g per mole. So I want grams on the denominator. I want them to cancel out. So 94 wait 1 1 g of phenol, the denominator and the numerator, one mole pino. Well, now that I'm in moles of phenol, I can easily use my coefficients of my balanced equation to tell my, tell me how many moles of water this would theoretically produce. So now I just look at the ratio of moles of phenol, two moles of water. So my balanced equation has one as an implied coefficient in front of phenol. So one mole of phenol will theoretically produce. So the Mosen on the denominator three moles of water. So now I have all of my conversion vectors set up, check that my units cancel milliliters of phenol, milliliters of phenol grams of phenol grams of phenol moles of phenol moles of phenol. And I'm going to end up with moles of water. So let's plug those numbers into my calculator. I'll have 10.6 multiplied by 1.07 divided by 94.11 multiplied by three equals 0.36 one 56 moles of water. So let's highlight. This is for phenol and that's how much water that would produce. Now, we just need to compare it to how many moles of water would be generated theoretically from the amount of oxygen that I have. So 02, now my amount of oxygen is already presented in grams. So 52.9 g of 02. So I just need to go to moles. So my molar mass of 02 is in the denominator 32.00 grams of oxygen. And then my numerator there is one mole of oxygen and then I just need to convert two moles of water. So looking at my coefficients of my balanced equation in the denominator, seven moles of oxygen theoretically produces three moles of what? So again, I'll enter 59.2 divided by 32 multiplied by three, divided by seven. And I get 0.70 85 moles of water and we don't necessarily need to look at sig figs here because we're just looking to compare which one would theoretically generate more or less of our water product. So if we compare these two numbers, we have 0.36156 moles of water from the phenol, 0.7085 moles of water from the oxygen. So our final is going to be our limiting reagent. So we know what our limiting region is the phenol. And we know how many moles of water have been theoretically produced when it's completely consumed. So we just need to convert that to grams of water and compare it to how many grams of water were actually produced to get our percent yield. I've just scrolled up here to have a little more room for my math. So my theoretical yield was 0.36156 moles of water and the molecular weight like a mass, excuse me, of water is 18.02 grams per mole. So 18.02 g of water in the numerator, one mole of water in the denominator. And when I do that math, that will yield 6.52 grams of water. So that's our theoretical yield. And we have our actual yield given to us in the problem which is 5.32 mL. So we need to go to grams for that. So actual yield. So 5.32 mL of water, then we use its density to convert that to grams. Water is an easy one. We have 1.00 the top grams per milliliter. So 1.00 g per 1 mL. So milliliters on the bottom is going to equal of course, 5.32 grams of water. So that would be our actual yield. And we just need to get a percent well, for a percent yield, we're going to end up dividing our actual yield. So 5.3 ramps of our actual yield divided by our theoretical yield of 6.52 g multiplied by 100%. Yeah. And will we do that math? And that's going to end up equaling 86.2%. We have three significant figures or excuse me, I misread those numbers there. It yield 81.6%. That's better as our percent yield. We have three significant figures given by the values we're given in our answer in our question. I mean, so our answer here, 81.6%. We're up to our answer choices and we see that choice B is 81.6%. So we have this known quantity of phenol burned in the presence of a known amount of oxygen gas. We're given the amount of water produced and need to cal calculate the percent yield. We have choice B 81.6%. See you in the next video

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Key Concepts

Combustion Reaction

Stoichiometry

Percent Yield