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Example of finding reactant empirical formula

Video transcript

This problem asks us, suppose you isolate an acid from clover leaves and know that it contains only the elements carbon, hydrogen, and oxygen. Heating 0.513 grams of the acid in oxygen produces points-- let me write these down or let me underline them-- so heating 0.513 grams of the acid in oxygen produces 0.501 grams of carbon dioxide and 0.103 grams of water. What is the empirical formula for the acid? So let's just write the equation here. And once again, you can't balance this equation because we don't know how many carbons or hydrogens or oxygens we have in the acid. But let's just write it down in general terms. So we have the acid. It has x carbons, y hydrogens, and z oxygens, and we're going to heat it in the presence of an abundance of oxygen. And when we do that the problem tells us that this produces-- let me do it in another color-- this produces some carbon dioxide and some water. And they tell us the amounts of carbon dioxide and water measured in grams, and they tell us the amount of our original substance. Let me write that here. So we're heating 0.513 grams of our original substance. And then that produces 0.501 grams of carbon dioxide, and 0.103 grams of water. Now we're going to do this problem very similar to the way that we've done the last one. Remember, we just have figure out the empirical formula. We just have to figure out the ratio, the simplest whole number ratios of carbon to hydrogens to oxygens. The only thing that makes this slightly more difficult than the last one is we have oxygens here, and the oxygens are kind of mixed in with the two products. So we can't just look at the moles of our two products and say oh, we must have this many moles of oxygen. What we're going to do here is we're going to figure out the moles of carbon dioxide using this information there. And that'll tell us the moles of carbon that must have come out of our mystery acid, because this was the only source of carbon that might have generated the carbon dioxide. Then we're going to use the water to figure out the moles of hydrogen in the product, which has to be the same as the moles of hydrogen in our mystery substance. And then we can figure out the mass of carbon and hydrogen out of this 0.513 grams, we'll be able to figure out because we'll know how many moles of carbon and how many moles of hydrogen we started off with. We'll be able to know their mass, and then we'll be able figure out how many grams of oxygen we must have had in our original substance. And then we can figure out the moles from there. And then figure out the ratios. If that confused you, hopefully just solving through it will make a lot more sense. So first let's just figure out the moles of carbon in our product, and the only product here that had carbon was the carbon dioxide. And that'll tell us how many moles of carbon we had in our original mystery acid over here. So we are end up with 0.501 grams of carbon dioxide-- that tells us right there. And we want to convert this into moles of carbon dioxide. So we want to multiply this times moles of carbon dioxide per grams of carbon dioxide. And if you care about dimensional analysis you know that because these grams of carbon dioxide will cancel with these grams of carbon dioxide. So how many grams is one mole of carbon dioxide? And we've done this multiple times. The molar mass of carbon dioxide-- you have 1 carbon plus 2 oxygens, so 16 times 2. That is 12 plus 32 or 44. So one mole of carbon dioxide has 44 grams of carbon dioxide, or 44 grams has 1 mole. but we wrote it moles per gram because we want to cancel out with the grams right there. So let's cancel that out, and you're going to get 0.501 divided by 44 moles of carbon dioxide. Let's get the calculator out. So we've got 0.501 divided by 44 is equal to 0.-- we have three significant digits here, and this is just a pure number or pretty close to a pure number-- so 0.11-- I'll round it up to 4-- let me put it on the side so I don't forget this-- so this tells us that we have 0.0114 moles of carbon dioxide in our product. Now that also tells us for every molecule of carbon dioxide we also have one atom of carbon. So however many carbon dioxide molecules we have, we have that same number of carbon. Remember, a mole is just a number. When I say 0.0114 moles I'm just telling you a number. So this tells me right there-- I'll stay in blue-- this tells me that we also have 0.0114 moles of carbon. That's what that tells me, because for every carbon dioxide I have a carbon. And the carbons in the product all had to come from my mystery substance. So how many grams of this mystery substance must have been carbon? Well I had this many moles of carbon in our original reactant. They all end up over here. Let's figure out their grams. So let's multiply-- so times-- now here we want to do grams per mole, because we want to end up in grams. We want to cancel out the moles. We want grams of carbon per mole of carbon. And the molar mass of carbon is 12-- it's a good one to memorize. So we have 12 grams of carbon per mole of carbon-- we've got 12 over 1 if you like. So what does this equal? How many grams of carbon are we dealing with in this mystery substance right there? Get the calculator back. So if we multiply-- it's actually this exact same number, let's just multiply it by 12-- so times 12 is 0.137. So let me write that down. So we have 0.137 grams of carbon in our original reactant. So of this 0.513 grams of this mystery acid, 0.137 grams are carbon. So that's what we got so far, so we have two good take-aways. How many moles of carbon we had over here, and how many grams of carbon that translates to. Now let's do the same thing for the hydrogen. Same drill. Let's do the hydrogen-- we're using that information right there. So we're starting off with 0.103 grams of water, and we want to turn this into moles of water. So let's multiply it times moles of water per grams of water. Once again, to cancel out the grams of water, numerator and denominator. And what is the molar mass of water? You have two hydrogens, so it's 2 times 1 plus the molar mass of oxygen, which is 16, so it equals 18. So 1 mole of water has 18 grams, or 18 grams per mole. But we wrote 18 in the denominator, because we want the grams into the denominator. So these cancel out and we have 0.103 divided by 18 moles of water. Let me get the calculator back. So we have 0.103 divided by 18 moles of water. Let me put this on the side so I can still see it, very bad memory. So this tells us that we have 0.00572 moles of H2O, of water. We just converted these grams into the number of molecules that we actually have. Now if we have this many moles of H2O, how many moles of hydrogen do we have? For every molecule of water we have two molecules of hydrogen. So you multiply that times 2-- so let me do that right here-- so 0.00572 moles of H2O times, we have 2 moles of hydrogen for every one mole, I should say, of H2O. So we're just going to multiply this by 2 to get moles of hydrogen. So let me get back the-- let's multiply it by 2. So times 2 is equal to 0.114. So this is equal to-- let me get that back. I always forget how many zeros there are, so I remember it now. OK, so it is equal to 0.0114 moles of hydrogen. So so far we figured out the number of moles of carbon in our product and its mass that must have been in the original reactant, the mystery reactant. And we've done the moles for the hydrogen. We have this many moles of hydrogen. So we must have started off with that many moles over here. Let's figure out the mass of that. So what is the molar-- we want to figure out this in grams-- so times-- the molar mass of hydrogen, it is 1 gram per mole. So hydrogen's easy-- times 1 gram of hydrogen per mole of hydrogen. I wanted the moles in the denominator so it cancels out. Moles cancel with moles. So we also have-- I'll do it in that same orange color-- we also have 0.0114 grams of hydrogen. So that's another important take-away. Now we have moles of carbon, moles of hydrogen. If we can figure out moles of oxygen in our original acid then we can figure out the smallest whole number ratio between them, then we would be able to find our empirical formula. And to figure out moles of oxygen we need to first figure out how many grams of oxygen we have here. And to do that we have a total of 0.513 grams of this thing. We subtract out how much of it is carbon, how much of it is hydrogen, and that'll tell us how much is oxygen. Let me scroll down a little bit. So if we want to figure out how much is oxygen, we literally just take 0.5-- let me write 0.513 grams-- let me just write this-- minus the grams of carbon. So minus 0.137-- let me do in that same color so you know where I'm getting these numbers from-- minus our grams of carbon, 0.137 grams of carbon minus-- I'll do it in the same orange color-- minus the-- where's our grams of hydrogen? Minus 0.011-- let me, just so you get the whole idea. We started off with 513 grams of acid, whatever it is, minus-- let me write it this way even better-- CxHyOz minus our 0.137 grams of carbon, minus our 0.0114 grams of hydrogen, will give us our grams of oxygen that we're dealing with. So let's figure that out. Because everything in that substance, everything in this mystery acid right here has to be either carbon, hydrogen, or oxygen. We know how much is carbon and hydrogen, the leftover has to be oxygen. So we have 0.513 minus 0.137 minus 0.0114 is equal to 0.36-- I'll just round it-- 0.365. So this is equal to 0.365 grams of oxygen. Let me make sure that I got that right. Yeah, 0.365 grams of oxygen. So let's figure out how many moles this is. So we're starting off, or this mystery acid has 0.365 grams of oxygen, and we want to convert that to moles. So times-- we want moles of oxygen in the numerator, grams of oxygen in the denominator, so they cancel. And we know the molar mass for oxygen is 16. 16 grams per mole-- right, the 16 has to match up with the grams. So this tells us that we have 0.365 divided by 16 grams of oxygen. So that's the number right there. I'll just divide it by 16. So 0.22, I'll say 8 just to round it. Or 0.0228-- let me put it aside, I want to make sure I don't lose any zeros here. So this tells us that we have 0.0228-- these guys cancel out-- moles of oxygen. We're almost done. We now know, we've now been able to solve the moles of carbon in our original reactant because all of this carbon had to come from here. We have our moles of hydrogen. We were able to figure out from this information because all the hydrogen had to come from here. And now we know the moles of oxygen in our original reactant. Our moles, because the leftover out of this 0.513 grams had to be oxygen, so we were able to figure out the moles of oxygen. So let's see if we can figure out the smallest whole number ratio between the three. So the ratio-- and I'll write the ratio like this-- we have-- let me do it in the same colors-- we have 0.114 moles of carbon, to-- and then we have 0.0114 moles of hydrogen. These look nice, they're actually the same numbers. So we're going to have a 1:1 ratio, to-- then we have the oxygen, 0.0228 moles of oxygen. And this looks like exactly twice times that. So if we divide everything, if we divide-- this is a ratio, this is another way of writing the ratio-- if we divide everything by 0.114-- let me be careful here. This up here was 0.0114 moles of carbon. So this was a 0 over here. Let me clear it out. I want to be nice and neat here. So let me clear this. I'm always afraid that I'm going to lose a zero here. Our moles of carbon was 0.0114. So let's divide everything by this amount because that is kind of the smallest number of moles, and that's exactly that number there. Do that in white. So you divide by 0.0114, 0.0114, 0.0114, and what do you get? What does this turn into? This turns into-- this is just a 1-- so this is 1 mole of carbon for every-- this is just a 1 right here-- for every-- I'll do it in the orange color-- for every 1 mole of hydrogen-- for every-- this is exactly 2, right? 228 divided by 140, that's 2. This is exactly 2-- for every 2 moles of oxygen. So that tells us our empirical formula for our mystery acid. It must be-- we don't know the molecular formula, but we know the simplest ratios, whole number ratios-- it must be 1 carbon for every 1 hydrogen, and for every 1 carbon and 1 hydrogen we must have 2 oxygens. So it's CHO2. That is the empirical formula for our mystery acid. Hopefully you found that fun.