A certain alcoholic beverage contains only ethanol (C2H6O) and water. When a sample of this beverage undergoes com-bustion, the ethanol burns but the water simply evaporates and is collected along with the water produced by combus-tion. The combustion reaction is When a 10.00 g sample of this beverage is burned, 11.27 g of water is collected. What is the mass in grams of ethanol, and what is the mass of water in the original sample?

Video transcript

Hi everyone for this problem. It reads a sample of vinegar containing acetic acid and water is burned vinegar burns and produces water while the water and the sample just evaporates the water produced by the combustion of vinegar as well as the water that evaporates is collected. The combustion reaction is the following. When five g of the sample is burned, 3.64 g of water is collected. How many grams of acetic acid and water are there in the original sample? Okay, so this is the question that we want to answer. All right. So, it's asking for the original sample, which means we're going to have two samples. So, our first reaction is just going to be the same one that's up above. So, we'll rewrite that and our second sample is going to be when it is burned again. And the question that we want to answer is how many grams of acetic acid and water are there in the original sample. So, let's go ahead and define variables. Here, we're going to let X equal the mass of water and the five g sample, and we're going to let y equal the mass of acetic acid In the five g sample. Okay, So what we're gonna have then is X plus y. Is 1, 2 equal five X plus Y. Is going to equal five g. So, if we rearrange this equation and we make it so that we're solving for Y. When we arrange this, we're going to get Y is equal to five minus X. So, let's remember that. Let's remember why is equal to five minus X. Okay. So we know that the mass of water collected is 3.64 g. Okay. We're told that in the problem. So, we're going to take 3.64 g and set this equal to X plus Y. All right. And what we're going to do here is solve or do dimensional analysis for the remaining part. Alright. So, what we're gonna do is we're going to take our molar mass of acetic acid. So, we have one mole of acetic acid. Its smaller mass is 60. g of acetic acid. And looking at our multiple ratio, we want to know our multiple ratio between acetic acid and water. All right. So, looking at our multiple ratio of acetic acid and water. We see that we have a for every one mole. Underline it in red. For every one mole of acetic acid consumed two moles of water is produced. So our multiple ratio is 1-2. Alright, So let's go ahead and write that. So every one mole of acetic acid consumed to move of water is produced. All right. So, let's look at what variables are what units cancel so far. So, we see our molds of acetic acid cancel. Alright, So now Let's look at our moles of water. So in one mole of water, what is our molar mass? Our Molar mass is 18.02 g of water. All right. So we have are moles of water cancel. And we're left with grams of water. Alright, So now that we wrote this out, what we can do is we're going to substitute for Y. Okay. So we defined why above as five minus X. So where we see why here? We're going to go ahead and plug in five minus X. All right. So we have 3.64 g is equal to X plus Y. Which we defined as five minus X. Okay? And we're going to copy and bring down everything we have in blue. Alright, So this whole part is going to we're gonna bring it down. All right. Okay. So now we're going to do the calculation for the blue part. So we're going to take one divided by 60.05 to multiply it by two And then multiply it by 18.02. So, this side here becomes 0.6001. And we're going to multiply it. Bye. We're going to bring down the five minus X plus X equals three point Let me move this over is equal to three 64 g. All right, So now what we're gonna do is we're going to solve for X. Okay, Because X is what we're going to solve for and figure out our mass of water in the sample. Okay, So let's go ahead and foil this part. Okay, So we'll take it step by step, we're going to get 3.64 g is equal to X plus 3. -0.6001 X. So we foiled that part out. So now we're going to subtract 3. from both sides. Okay? So we're going to get 0.6395 is equal to x minus 0.6001 X. Alright. So remember we're solving for X. So we have x minus 0.1 X. So remember there's a one here. So we're taking essentially one minus 10.6001. So that leaves us with on the right side. 0.3999 X. is equal to 0. Will divide both sides. Because remember we're solving for X. Well, divide both sides by 0.3999. And we get X is equal to 1.59. And this is the mass of water in the sample. Okay, So we're almost done here. We know that We defined why as the mass of acetic acid. And we said why is equal to five minus X. So that means y is equal to five minus 1.59. So y is equal to 3. g. All right. So, we just solve both X and Y. So we know that we'll write it in the corner here. We know that we have 3. g of acetic acid. Because we defined acetic acid as our variable y. And we know we have 1.59 g of water because we defined our variable water as variable X. So these are our final answers here, and that is the end of this problem. I hope this was helpful.