Ch 6 Transcript PDF

SPEAKER 0 Both of you. Um, so the test where we happy about it? No, some of us were, and some of us weren't. Um, there were, like, five, eight still, which is good, but, like, the average was, like, lower than I expected it to be bad. Um, instead of, like, the 60 that we like to be around, it was a...

SPEAKER 0 Both of you. Um, so the test where we happy about it? No, some of us were, and some of us weren't. Um, there were, like, five, eight still, which is good, but, like, the average was, like, lower than I expected it to be bad. Um, instead of, like, the 60 that we like to be around, it was a 52 and so much lower. Um, so I just want to remind you that you need to be studying outside of spam. Okay? Um, like, I give you the resources. If you're struggling with the information, you should be in my office hours, right? Or in the tutoring center. Um, having help understanding, finding ways to write. That's what you get when you come to my office hours, right? We go through things. Maybe we figure out what, uh, you know, the the struggle is right. Um, with your understanding that information. Okay, so that we can maybe find better ways for you to study and understand the material and, um, be able to, like learning so that you can use it and obviously pass your exams with it, but also because you're going to use it as you move forward in Stem. Okay. So this isn't like a gen can we like, take it and we cross our fingers and pass it and then we never have to use it again. You will use this information that we have in this class throughout your entire Stem career. Okay. Um, it'll come back in biology and engineering and, uh, even physics. Okay. This all comes back in all of those fields. As you move forward. It'll be in your upper level courses. Okay. So it's really important that you're understanding and getting this information. I did say that this last test was the one with the least amount of math on it, right? Very little math on this last exam. The next exam will go back to, like a lot of the stuff we learned in chapters one, two and three. Right. Um, so it'll be much more math based. So if you're more comfortable with math like I am, then you'll be happy that we're getting back into math. Okay. Um, but that information on exam two will be on the final because the final is comprehensive. So if you did poorly for then you wanted to do on that test, make sure that you're going through and reviewing those, um, those answers that you got wrong so that you can know why. Okay. Um, I've already had three students sign up for office hours and come to my office to go over their exams for exam two, because they're like, I don't know why. Like, I thought I did better and then I didn't do as good as I thought. So can we go over the problems that I missed? And that's a wonderful way, right. To figure out where that disconnected and to think about it a little bit more. Okay. If you're taking an exam and then never looking at it again. That is not the way you're going to get through and have a positive outcome with your Stem courses. Okay. It's very important that you review that information. Again. You can come and review that information during office hours with me. Right. Um, or you can like do it on your own, which is fine too. But it's making sure you set aside that time. Um, you should be spending. Right. 2 to 3 hours, mostly three. Right? Really? New studies show 3 to 4 hours for Stem. Okay. Fields, 3 to 4 hours outside of class for every one hour that we have in class. So you should be setting those times aside to be making plans to be looking over material, to be doing your homework. That should be helping you with understanding the material. Okay. Um, and saying all of that, I will do an extra credit opportunity, which is this event next Wednesday evening? Um, from 6 to 7 p.m.. It's going to be in this room, um, from again, 6 to 7 on Wednesday, because next week is National Chemistry Week. Yay! Super excited. So the chemistry club on campus, which is, um, the American Chemical Society student chapter, they actually have gotten awards every year. They've been chapter. Uh, which is pretty awesome. They presented the national conference they're presenting this spring in San Diego. They presented last spring in Indianapolis. We take all the officers. If you're a member, you also get a cord. We graduate. You can be any major, which is great if I'm plugging the club, too. Um, but if you go to this event, um, and take a picture of yourself at the event and write a summary of what you learned at the events, and upload it in a Dropbox that I'll put on blackboard. Um, you can get extra credit for this course. I mean, sorry, the second, the first poorly done. The second exam. Okay. So that extra credit will specifically go to that second um, sad. If you can't attend this event because of work or something, you need to let me know, like ASAP, because I have something that is like a instead of this event, right. Um, which is tomorrow, um, which would be the same thing, taking a picture of yourself at the event, writing a summary of the event. And I can put up the flier for that one, which is this flier right here. Um, I think the more fun event will be that one on Wednesday next week. Um, but this is also going to be a good event for Stem majors, right? This one, uh, talks a lot about research, um, and things like that. So the other one is like a live webinar. It's going to be like thousands and thousands and thousands of people across the US in ACS that are going to be like attending that event on Wednesday next week because of National Chemistry Week. Um, they'll be like prizes and games. And they talk about like, um, the chemistry of photography, which is really kind of cool. So a fun applied chemistry lesson, but it's going to be like, I think the host is actually a comedian. So it'll be like very fun and interactive. Um, the one on Wednesday. So again, I highly want to push that one. But if you can't attend that one, this is the other one that you could go to, um, to uh, again, get that extra credit if you go to both. No, you do not get double the extra credit, right. You only get there's only going to be one Dropbox for extra credit, you need to upload the photo and the summary of what you learned at that event. Um, for that again, I will put that on blackboard as well, along with these two fliers. Um, so that you can do that okay. Any questions. Next week again is also like I said, National Chemistry Week. So the chemistry club will be doing several things on Monday. They have a demo they're going to be doing from noon to one for their general meeting. Um, it's a it's an lbw b to 88. But it's they're going to be doing elephant toothpaste and pumpkins which is really fun. And then they're going to be doing an invisible ink demo. So again super cool. Um, things that you can do hands on. Right. For, for that. Um, again, it's kind of going along with National Chemistry Week and then obviously Halloween theme, right? Um, if you look at their Instagram page. Right. Uh, what is it, T x and mu. Right. Because that's always the like beginning X is the ending right for American Chemical Society. Um, that'll pull up their information on Instagram, which is awesome. They're also going to be doing a week long scavenger hunt to find a mole, because a mole is 6.022 times ten to the 23rd, which 1023 is next Wednesday when we have our live event. Right? Okay. Okay. Um, so again, uh, next week is like a super fun week for chemists. Okay. Um, like, across the world. Really? Uh, did I tell you guys, like, one of my friends in grad school actually got married on 1023. And like, it's because she's got straight, like, awesomeness. Okay. Um, anyway, so lots of fun to be had next week. Again, there should be a scavenger hunt with a mole. You'll learn more about it if you go to their Instagram page. Um, but if you find a mole, you actually get a prize. And so they have little mole with lab coat plushies that are for like the grand prizes. And then they have some, like other prizes that will be for the like lower level level moles, like the ones that aren't gold or silver. Right? Um, so those will be hidden around campus and you'll search for them and find them, which will be super cool. It'll go on all week next week. Okay. Um. Any other questions now? We don't have any questions now, are we ready to start chapter six, which is, um, really nice because, uh, we've been kind of introduced to some things that are in chapter six already, which is really, um, beneficial for us to understand what's going to be going on here. So let me swap this. All right. So in chapter six we're going to talk about formula weights again. Where do we get those from. Periodic table. Right okay. The same thing as molar mass right. Um so we're going to also review what moles are 6.022 times ten to the 23rd is how many we have in a mole. Right. Uh, how many footballs we have in a mole. Right. Not 12 dozen. Right. It's it's 6.02. Two times ten to the 23rd. Moles. Okay. Uh, per mole. Right. Um. All right, so I'm just gonna. I didn't get my poor clicker out already. Here we go. And then we'll also talk about percent composition. I think we kind of started talking about this in chapter two, but we're going to go over it even more in depth now. And I know you guys have done this in lab, which is good to have that like preview of it. We'll also talk about empirical and molecular formulas. Again, we talked about empirical in chapter two. Right. It was like the lowest whole number. Right. Um factor for our chemical formula. Okay. Um so we're going to talk about going from empirical to molecular, and we're going to talk about how to get empirical from things like percent mass. Okay. Um, and then we'll also talk about molarity and dilution. Um, which are concentration units. Right. Molarity is concentration unit that you've already been using in lab. But we'll talk a little bit about how to calculate that. Um, in this chapter okay. So I totally lost my poor clicker, which is what it is. All right. So the first thing we're going to do is look at a chemical formula again on the test that we just had. Right. We talked about chemical formulas. Right. We had to like be able to name them. Right. And then we also had to be able to put them together if they were ionic compounds. Okay. Um, and so that's what we will begin by talking about here is just a review of that information. Okay. Um, so al two so four three. What is this called? Aluminum sulfate. Right. Again, that was something we should know from our test that we just had aluminum sulfate. We should have known aluminum is a plus three on the periodic table because of the column it's in on the periodic table. Right. Um, sulfate. That was one of those polyatomic ions we had to memorize. Okay. Um, it was great at student. Actually, I don't know which student is sorry, but, uh, student actually, uh, messaged me through through blackboard and said that they found a really cool, like mnemonic phrase that helped them, like, memorize their polyatomic ions. Um, and they put, like the screenshot of it and stuff, which I thought was a really wonderful because they used ChatGPT to help them study. They asked ChatGPT, right, how can I memorize polyatomic ions? And ChatGPT said, oh, here's a nice phrase if you remember this phrase right, the number of consonants is the I don't even remember what was the number of oxygens, I don't know. And then the number of vowels was the charge on that. Um, and so if they remembered that one phrase and they can remember their eight endings, right. And then it were just one less same charge, right? Um, so again, using ChatGPT to be like, wait, how do I memorize this? Or what's a good way, right, of understanding this better? Sometimes it actually gives you some really good information. You can use that as a tool now, like it's your personal tutor. Of course, if it sounds wrong and you're like, wait, I don't think that this is correct, then you should like, not use that particular information. Right. Um, but for the most part, right. It does pretty good. Sometimes it does fail, right. But for gen chem, I would say, um, it's usually pretty. Pretty. Okay. Right. Because we're not high enough up yet for it to have like super wrong information. Okay. Um, but definitely like make sure. Does that make sense to what we're talking about in class? Um, but you can be using that, right, to get some ideas on how to memorize things. Uh, another student asked me like, how they can, like, learn those polyatomic ions. And I said, flashcards are really good. And I even googled, like, um, like flashcards for polyatomic ions. And there's already ones made on a website online. You can just flip through them. Right. Um, so again, super nice resources to help you study. You should be spending time on studying. Um, all right. Anyway, so in this molecule, we can get the formula weight by taking the mass of each of these elements, multiplying it by how many of each of these atoms we have and then adding those together. Right. So again for aluminum sulfate we have two aluminum. So there's two of them times our mass that we get from the periodic table for aluminum is going to give us our mass of aluminum uh, in our overall compound. Okay. Um and then here this three outside the parentheses, it's important to remember that that distributes inside the parentheses. So we have three sulfur and we have three times four. So 12 of our oxygens. And so that's what we see here. Right. Three times our sulfur 12 times our oxygen. Again we add all those up and that gives us our overall molecular mass. And this is what the awesome picture looks like. Um, of this aluminum sulfate. So a really large molecule. Right. Because we have three of these sulfate ions, we can see them. The yellow yellow sulfur, by the way. Uh oxygens red right. Um, aluminum is obviously that, like it's like a brownie color, um, up there. Okay. So important to know how to get your formula weight or your molecular weight. Um, as we move forward. Okay. Uh, we obviously needed to know it on on exam one as well. Right. So, um, a mole just to review what a mole is. It's 6.02, two times ten to the 23rd mole days next week. Right? Um, Avogadro's number is what we call that. Okay. Um, and it is just a unit for a specific number. Right? It's like a dozen. How do you explain a dozen? Well, a dozen means 12, right? A mole means 6.02. Two times ten to the 23rd. Okay. Lots of these things. Um, so most provide a bridge to get us from a molecular scale, very small scale to a real world scale. So if we had just one molecule of H2O, that mass we calculated, we would say is an atomic mass. Units amaz. Okay. Atomic mass units. And that's really what we get from the periodic table. Um, but if we have a mole of these things. Right. Um, then that means that we have 6.022 times ten to the 23rd, right? Molecules, which is going to give us instead of 18 amps, it'll give us 18g of this substance. Right. Because we have a mole of them. So moles let us get from one molecule to a mole of molecules weight, which is something that we can actually measure in weigh and lab. Okay. Um all right. So again, in order to get from a molecule to, um, a mass, we can weigh in lab, we go from our molecule, We use Avogadro's number right to get to our number of moles. And then we're going to use molar mass from the periodic table to get to our number of grams. All right. So here's an example. It says our bodies synthesize protein from amino acids. One of these amino acids is glycine which is the molecular formula C2 H5O2N. It says how many moles of glycine molecules are contained in 28.35g of glycine. So we can go ahead and calculate this. Maybe what I really want to do is open up my visualizer here. So what are we going to do to start out with remembering our wonderful, um, great, wonderful exam one material. Chapter two. We're going to start with what number? What number are we going to write down first? What's given. 28.35g right of our glycine. So that's of c2 H502N. Right. And then it asks for how many moles right we have in this. So we want to get to moles. How do we get from grams to moles. To be at a table. Exactly. So we're going to use the molar mass because molar mass is grams per mole. Right. That's what the molar mass is. So we can go to the periodic table. And we can say that carbon has a mass of 12. I'll put it up here 12 right times two carbons. Plus hydrogen has a mass of one times five hydrogens. Plus oxygen has a mass of 16 times two oxygens. Plus nitrogen has a mass of 14. Right. Question mark. I think so. Thank you. I was like, that's the one that I'm going to get stuck on. 14 again, those come from the periodic table, right. The mass is on the periodic table. And that's going to give us a total of what. What's the molar mass. What'd you guys get? 78 is what you said. Aha I know. 75 okay, great. Perfect. So 75. What what are the units on molar mass? Grams per mole. So where are we going to put that 75 in our equation here. Denominator. Why. Because we want to cancel unit. So 75g for every one mole. That's what this means. 75g per mole means per one mole. Right? Okay. Um, so 75g per mole grams are going to cancel and we're going to be left with moles, which is what we want to be left with. So we should be at the end right already which is nice a nice one step problem to remind us how to get to moles from grants. What'd you guys get for this? 28.35 right, divided by anything under a line is divided by. So 75 is going to give us. Is this what you got equals 0.378 moles. Is that what you guys got. Okay okay. And that's again moles of glycine seen Glee is what I would probably put just for like, a, you know, abbreviation. Okay. Any questions about how to get from mass to mole? Very important that you know how to do this moving forward. All right. So now that we've talked about molar mass again, we can get the mass of a molecule. Right. We can get the mass of an atom. All of those come from the periodic table. We can also do composition of compounds. So a chemical formula is going to be just a combination of our molar masses of constituent elements. And it helps us indicate their quantities right of each element in our compound. So we can do, um, percents of each element in the compound. And we determine that by the formula. Okay. So we can look at chemical formula. And we can figure out our percent of a certain element in the compound by putting the mass of that element in our chemical formula over the mass of the entire chemical formula. Okay. Um, so of the compound. Right. Um, again, we should be able to multiply that by 100, um, to get a percentage value instead of a ratio value. Okay. Um, percentages might not always, uh, like, add up to 100 due to rounding, but they should, right? Because if we have whatever CO2, okay, there's going to be a percent of oxygen and a percent of carbon. And if we have all of that added together, carbon percentage and oxygen percentage, we should get pretty close to 100. Um, again, rounding issues might give us like a tiny bit of difference in our decimal places, but it should be 100. All right. Example for our percent iron in our iron oxide. Do you guys remember doing this problem in chapter two because we did. Yes. Um, but again iron and iron three oxide again, that three is not how many irons we have. It is the charge on our iron, right? That's what the Roman numerals mean. We learned that in chapter four, which was on this last test. Um, so our iron three oxide is Effie 203. So we have two irons. We need to look at the periodic table and add those up to get the mass of iron over our 4203. So three oxygens and two irons total mass. Write our molecular mass, um, in the denominator. And then we multiply that by 100 to change that to a percentage. Okay. Um, so again iron on the periodic table is 55.85. Right. Um, and we have two of those atoms. So we need to multiply that by two because we have two iron atoms over that 159.7 atomic mass units um for the total mass of our 203. So that was 16 times three plus 5.85 times two equals that 150 9.7. Okay. Um, so that was just the whole mass of that. I know you see Amos here, but we could have also used grams per mole. It doesn't really matter. Right. Because the grams per mole would also cancel giving us just a ratio. Right. Okay. So, um, it would just give us a ratio, which is what we're getting from this, a mass ratio. Um, and then we multiply that by 100 to get our percentage. Okay. So this would actually give us 69.94% of iron in iron three oxide. Okay. Um, so what is the percent of oxygen in iron three oxide. Again we can do the same thing with our oxygen. What would we put in the numerator for this one. Right. We would put um for oxygen in iron three not three three oxide. Right. Because it's a plus three plus three right. This one we know is a minus two. That's why this one's two and this one's a three okay. Because they come down like that. All right. Um anyway oxygen we would put in the numerator the mass of what, 4% of oxygen in our 2033 oxygen. So three times what's the mass of oxygen 16. Again we can put amaz here or grants per mole here. We're just doing a mass percentage. So it's either unit is fine okay. Because the units are going to cancel over. What's the total mass again. It was 159 I think. 156 59.7 thanks. Uh, one 59.7g per mole. Right. And that was for everything. So again, this was just our mass of our oxygen. And this was the mass of our Fe 203 times it by 100. And what should our percentage be for this one? You don't even really need to do this math, right. We could have just subtracted that other one from 100 because there's only oxygen and iron in this. But either way it gives you the same exact value. What is our value for this? It's going to be 30.06 okay 30.06% okay of oxygen in for 203 okay. Yay. Super nice. So far so good. Trust me, it'll get more fun. Okay. Because math is awesome. All right. Um, so now that we know how to do mass percent and what mass percent is, right, it's just the percent of that element in that compound. Okay, um, we can get empirical formulas from mass percents, uh, because when we're in a lab, we can actually calculate mass percent using things like mass spectrometry. Okay. So we have instrumentation that allows us to know, right, how much of different elements we have in our, um, compounds that we shove through that instrument. Okay. And so let's say we put something through the system and it told us that there was 69.94% of iron and 30.06% of oxygen. Right. Um, we're going to go ahead and go from those values, percent mass of our elements in our compounds to providing us with an empirical formula. Okay. So the first thing we're going to do is that we're going to assume that we're in 100g of iron three oxide. Okay. We're going to assume that we're in 100g of this compound. And so when we assume that we're in 100g, I think I have it actually come up here. That means that we're just going to change our percentages to grams, okay. Each of them because we're going to assume that we're in 100. So if we're in 100g, then our 69.94% changes to 69.94g. So that's always the first step when we're going from math percent to empirical formula is we're going to get rid of percentages and we're going to put grams in place of them. Okay. Um, so our 69.94g of iron for every 30.06 right grams of oxygen is what we're going to have. Um, and what I would do right on my paper here because this would be the better way to do it. I would put 69.94g of iron. Right. Because I've just changed my percent to grams and then 30 leave some space because we're going to do both of these guys to see how much of each of them we have in our system. 30.06g of oxygen. So once we're at grams, we can get to moles because chemical formulas actually give us like mole ratios, those numbers that are subscripts or ratios of each of those in our compound. Okay. So what we want to do is we want to get to moles. And then we're going to look at mole ratios. So how do we get from grams to moles. Molar mass. We just did it right. We went from grams to moles for that. This this compound up here. Right. Glycine okay. Um so we're just going to go from grams to moles using molar mass which comes from the periodic table okay. What goes in the denominator for iron. 5955 point. What was it 55.85 right. Grams of iron that again, we grabbed from the periodic table for every one mole. Right. And now that's going to give us our moles, which is 1.25 moles of iron okay. And then oxygen. We also want to change oxygen into moles because we're trying to get more ratios to get our chemical formula from this. Okay. Um, again specifically we're getting our empirical formula, the lowest ratio numbers. Okay. Um, so for oxygen what's our molar mass. 16. Right. And that 16 is going to go in the numerator or the denominator denominator because it's 16g for every one mole right grams per mole. And so that's of oxygen. And so that's going to give us I'm sure I have it right here. Um 1.88 when we put that into our calculator moles of oxygen okay. So this is always how I do empirical formulas. I put my percentages in a line so that I can continue with math to the right of them. Okay. It makes it really nice and easy. Uh, sometimes we have more than just two things, right? Two elements. Sometimes we have 3 or 4 elements. Okay. Um, anyway, so now that we have moles, we need to do mole ratio. So we're going to divide by the smallest number of moles for both of these because that's going to make one of them become one. And one becomes some multiple of that. Right. Which is our ratio okay. So we're going to divide both by what number. The smallest one which is 1.25 moles. Right. So this um iron becomes what one exactly. So nice. One of them's going to become one because we're dividing it by itself. And then the other one here particularly is going to become 1.5. Okay. Now we're in our mole ratio. This is our ratio okay. We have 1 to 1.5. Can we have half of an atom? No, we can't have half of an atom. So because we have a decimal here, we need to figure out the whole number that we can multiply this by. Right in order to make this a like whole number. Okay. Not a fraction number. Um, what number can we multiply this by to make it a whole number? Two. So multiply it by two. If we multiply this one by two, we have to multiply the other one by two. Okay. So we're going to have two. And what is 1.5 times two is three. Right. So that means we have three oxygens and two of our ions okay. So f e right. 203 is our empirical formula from mass. Yay. We're going to be able to do this because it's going to be on the next test. All right, um, questions. We're going to do more problems like this. Don't worry. Questions right this second. Okay. Um, we knew it was going to be 203 because we did the other part first. Right? And we saw those percentages. But you won't have that other part first. You'll just be given percentages and you'll have to calculate empirical formula. All right. Um, so here's another one. It says for thousands of years the mineral, uh, calcite, right, has been highly prized source of copper. Its chemical composition is 79.85% copper and 21.15% sulfur. What is the empirical formula? How do we start to solve this problem? Exactly. We're going to convert. Right. Um, our percentages to grams. Everybody say it with me. Convert percent to gram okay. Super easy. Okay? Not like rocket science, right? Super nice and easy. We're going to say that this is a nice easy math problem okay. So 79.85g. We just changed our percents to grams right. And then that is of copper because we need to know what element so that we can get to moles. Right. And then the other one was 20.15g of sulfur. And now we're going to use a periodic table in order to convert these to moles. Right. And our periodic table tells us what is our mass for each of these. Copper is 63.5 is that we said don't rat. 63.5 63.5 ish on both sides. Yeah. Say it again. Did I write it on the wrong one? Was that for copper? Not for sulfur. Thank you. 63.5g per mole for copper. And then what's sulfur? 32.07. Is that what you said? Grams for every one mole. Right. And so now. We're going to go ahead and do that math. Right. So we've got 79.85 divided by 63.5 gives us 1.2574. Right. I would just put it all out there for now. Uh that's moles. And then our other one was 20.15 times, not times divided by right times one, but divided by 32.07. And that's going to give us 60 0.62 83. Okay. So we changed percent into grams. We changed grams to moles. Now what do we do. Divide by the smallest number. Okay. Which one's the smallest. Yes this one is smaller than that one okay. So 0.62830.6283. And then we should get from that. What value. Here. This one we don't even need our calculator for right one. And again that's my sulfur I'll put my sulfur over here so that I can remember. That's for sulfur. And up here what are we going to get to. Probably right because it looks pretty darn close to two right 0.6. It gets us to point 12 okay. So two. And that's for copper. You guys did this already didn't you. Did you do a lab like this already. Yeah. All right. So that means that our empirical formula is what CU2S. Uh again empirical formulas like to start with the cation and then end with the anion. When we're writing those empirical formulas out starting with the metal ending with the nonmetal. Right. Like reading the periodic table our metals and then our nonmetals okay. How are we doing? Super good. Right. Empirical formula. That's what that gives us. Okay. So here's another one. Asbestos was used for years in insulating material and buildings until prolonged exposure to asbestos was demonstrated to cause lung cancer. Um, asbestos contains magnesium. contains magnesium. I'm going to write this down so that I know which elements I'm thinking of while I'm going through here, okay. It contains magnesium and silicon and oxygen and hydrogen. Right. Those are the ones we're going to need to deal with. It says one form of asbestos crystal tile, right, is 520. Um, is the mass of that okay? It says the composition is 28.03. Is that percents or should I go ahead and write it in grams? Grams. Because that's what we're doing here. Right? Um, and 21.6g of silicon. Right. And 1.16g of hydrogen. So that means that they left out which element? Oxygen. How much oxygen should we have? We sum those up and subtract them from 100 okay. Because it should add up to 100. So 100 which is our total percent by mass. Right. Minus our 28.03. Because we get counted for that already 21.6. We accounted for silicon -1.16 to account for our hydrogen. That leaves us with 49.21 right as our percent of oxygen. Okay. We could see here, right. This is like about 30, 40, 51%. So we had another 40 something percent, right that was unaccounted for. But that's all going to be from oxygen okay. So 49.21 is that what you guys got to I did that right. Right. Math. Mhm. Okay. Um so now we can go ahead and determine the empirical formula for this. Are you ready? Yes. Uh, how do we get from what we were given to empirical formula? We first changed to grams and then went. Divide it by the molar mass to get to moles. Right. And then what? Divide by the smallest number right to get ratios. And then make sure that they're all whole numbers okay. To get empirical formula. So we're going to go ahead and do that I know this one 16g per mole right. This one's one gram per mole right. 49.21 divided by 16 is going to give me 3.0756 moles of oxygen. And then silicon. If I had a periodic table it would be super helpful. UNKNOWN But I didn't. Say it one more time. SPEAKER 0 And then what is it for? Magnesium. Thank you so much. Okay, so 21.6 divided by 28.09 right. Gives us 0.7689. And then this one up here, 28.03 divided by 24.31 is going to give us 1.1530 grant. And starting out grams moles moles moles right. Moles of magnesium moles of silicon okay. All those moles moles. Now we're going to do what what's the next step. Again divide by the smallest number. Which one is the smallest number. This guy right? Divided by 0.7689 is going to give us one for silicon. Right. Um, divided by 0.7689 divided by 0.768 and. Okay I'm going to do that. So I still have my number that I had up there already up there. So I'm going to divide that by 0.07689. Gives me like ish 15 right. So what you guys got. No that doesn't even sound right. I don't know why. Why did it give. Oh because I look I put I put an extra zero. I was like that's such a big number. I put an extra zero. Right. So. So I have to do it again. 1.153 divided by 0.7689. Right. Because we know that seven and seven gives us 1.4, so it's not going to be 15 times, right? We knew right away that was wrong. Okay. Yay critical thinking skills. We should know when our calculator was messed up. Um all right. So that's going to give us some 1.5 okay. Which makes sense. Decimal over I added 01.5 of magnesium. And then we have 3.075 uh six divided by our 0.7689. That gives us four for our oxygen. Right. And then hydrogen 1.16 divided by 0.0 not zero. I don't know why I keep it's because I have A00.7689. Right. And that gives us 1.5. Again it was really close to this one. So we should have kind of known it was going to be the same right 1.5 hydrogens. So far, so good. Yes. Can we just take these and make our our empirical formula already know. What do we need to do. Well it's by everything right by two. So we're going to have two magnesium and two sorry 331.5 times two is three three magnesium and two of our silicones. Right. And eight because four times two is eight oxygens and three hydrogens. So our empirical formula should be magnesium three Silicon two, oxygen eight, hydrogen three. This is our empirical formula. Okay. Let's. Yes. UNKNOWN This is it. Okay. SPEAKER 0 Um, again, we want to start out with the, like the positively charged one. So magnesium, we know it's going to be out front and then usually like our hydrogens coming in. So it's usually like that hydrogens only out front if it's an acid okay. Just like it likes to be off. It's in um but outside of that it goes to the end. Um, but yeah, it's, it's, uh, cations first and then those other ones. I mean, that was the order they gave them. So we're going to go ahead and put them in the same order, um, because our magnesium was out front already anyway. But yes. Kind of arbitrary ish. It's it's fine. Um, so we can actually calculate our molar mass for this right compound from the periodic table. So three times one for hydrogen plus eight times 16 for oxygen. Right. Plus um what two times our silicon value was 28.09. Thank you. Plus our three times magnesium which was 24.31. And that's going to give us 260. So our empirical mass equals 260.11 right grams per mole. Okay. That that's our mass of our empirical formula. Does that mass match the mass that we were given for this compound. No. So our molecular mass. Right. Our molecular mass was given as 520.27g per mole. And we got our empirical mass which was 260.11g per mole. If we divide these two right, we're going to get the ratio between them. And that's the number that we'll divide our empirical formula or multiply our empirical formula by our subscripts to get our actual molecular formula. Okay. Because empirical formula is the lowest ratio we can have. But obviously since our empirical weight and our molecular weight did not match, we're not at the lowest whole numbers. We're not at the empirical formula. Our molecular formula should be wet. This division gives us two. We can see that hopefully right away. Right. 26 to 52. Okay. Um, so if we have two here, that means that all of these subscripts need to be doubled okay. So magnesium six. Silicon four. Oxygen 16. Hydrogen six. Right. This is our this is our empirical formula. And this is our molecular formula. Super easy right? Questions? UNKNOWN Okay, so if we're asked for. SPEAKER 0 Empirical formula, we get it directly from our percentages. If we're asked for molecular formula and we're given percentages we're going to get our empirical formula first. And then our molecular mass would have had to be given for us to get from empirical to molecular. So we would have to know the molecular mass, right, in order to get to our molecular formula. Right. We would have had to been given that information once we were in the problem. So it was nice. Nice to see. Okay. All right. Um, so again, just to reiterate, right, because we talked about empirical versus molecular in chapter two as well. But empirical is the simplest whole number. And our molecular formula is the actual ratio. Um so again the simplest whole number These would be empirical formulas, but a, c two, H two, O two would not be an empirical formula. The empirical formula for that compound would actually be H0, which is what we would see if we had percentages, mass percents, and we change those to a formula, we would get empirical formula. And then we would see, well the mass is actually double of that. So we would need to get to our molecular formula that way okay. By multiplying it. Okay. So again molecular mass is the blue equation. Here. Molecular mass over empirical mass is going to be equal to the molecule that we're going to multiply those subscripts by. Okay. Um so make sure you know like if we have just one of something right. It would go to two obviously because that one we don't right there. Um it's like not okay. Um so here is another example for combustion. Combustion analysis of an unknown compound indicates that we have 92.2% carbon and 7.8% hydrogen. It says the mass spectra indicated that the molar mass. So it gives us a molar mass um, is 78. So we're probably going to be asked for the molecular formula instead of the empirical formula okay. Because it's giving us a molar mass. Um it says what is the molecular mass of this formula okay. So how are we going to start to do this one. What are we going to do first to get to molecular formula convert to grams percents go to grams right. So we're going to have 92.2g of carbon. And we're going to have 7.8g of hydrogen. Right. And then what do we do. Right. We're going to change to moles. Right. Because really what we're doing is we're first figuring out our empirical formula. And then we're going to use our molecular weight right. Our weight of our molecule molar mass. Right. To get to the molecular formula. Okay. Um so 12g per mole. Right. And then hydrogen is one gram per mole from the periodic table. So we end up with 7.8g moles of hydrogen. Right. And what is 92.2 divided by 12 should be 7.68. 3333. So divided by 0.7.8 divided by 7.8. We know that this one's one. We could calculate the other 17. whatever. I'm just going to take the number that I have because it's already there. So divided by 7.8. Right. And that's going to give me Go to. It's fine. 7.68333 divided by 7.8. It gives me.985. This is close enough to one right to to push it to one. Okay. So it's a 1 to 1 ratio. So what's our empirical formula. C h right. Because this was for carbon. So one for carbon and a one for hydrogen. So 0.98 was close enough to one okay. Um 0.75. Not close enough to one. Right. We can multiply that by four. Right. In order to get to a whole number. Okay. Uh, 0.33. If we had a 1.33. Right. Well, we'd multiply that by three to get to a whole number. Okay. Um, so those ones, those ones that we can think of fractions, right, that we can think of. Um. don't just round right. But 0.98 is really close to one. So we would just round. All right. Um, yes. We did. Oh, I see what you're saying. It doesn't matter. Yes, it should have. I don't know why I thought it was going to be one. Um, yes. It should have been divided by the 7.68. The smallest number. Okay. But it would still give us 1 in 1. Okay. Give us, like, a little higher than. Right. Did you guys do the other way? Which is what you should have done. 1.0, 1.0 something. Yeah. So this one was one. Thank you for pointing that out because it is the smallest number. I just they were just so close that I moved on with life. And I shouldn't have, uh, 7.8 divided by 7.6833 gave us 1.01. Again, close enough to one to just round it. So it would be the same outcome, right? Because because it was like that. But it should always be. And when I got something lower than one, I should have been like, oh, I divide it by the wrong number, obviously. Right. Okay. Anyway, close enough to one. Thank you very much I appreciate it. I'm sure other people were like, we've been doing the smallest this whole time. Now it's still to this moment. Um, all right. Thanks for catching that. Okay. So c h is our empirical formula. Is that our molecular formula. No. How do we know that. Because we have a different molar mass. The molar mass that we have for our molecular molar mass. Right. Molecular molar mass is what, 78g per mole. We are given that right in the problem that's given 78 given okay. And our empirical mass. What is the mass of this 1312 plus one from the periodic table, right. So right away we should have been like, oh, it's 13, that's not 78. So it's not the same formula. Right. We need to do a multiple. So we're going to need to do this division. We divide by the smaller number empirical formulas always going to be the smaller number okay. The empirical weight. Um and what is that going to give us. 78 divided by 13 six. So what is our this is our empirical formula. What is our molecular formula. C6 h6 okay. Because those are invisible ones. They're right for each of those. All right. Do we like these I like these empirical formula. They're super fun. Okay. Um, so now we're going to talk about solution composition. If there's no more questions for that one. Yes. Always divide by the smallest number when we're getting from our moral values to our moral issues. Okay. Um, so solutions, um, our solutions. Right. We have homogeneous mixtures, right, of two or more substances that are going to be our, um, our solution. So we have two parts of a solution. Any solution. Right. We're going to have the solute and the solvent. Um the solute is going to be the one in the smaller amount. So if we have like salt water boiling on the stove, if we put salt into our water, right. Not now because we don't use our water right now. Right. Okay. Um, but once the city figures that out, um, if we can, I guess we boil it. Are you guys boiling water? We're boiling it to wash all of our dishes, and it's no fun. So we've been trying to use less dishes, which is hard to do. Yeah. Um, anyway, So. The salute is going to be the one in smaller amounts. So it'd be the salts in that salt water okay. And the solvent is going to be the one in greater amounts. So it's going to be the water in that salt water. All right. The H2O okay. Um, so water is a solvent that we often use, right. That's very common water being our solvent, when we have water as our solvent, we call it aqueous. So sometimes you'll see a cue next to a chemical formula when we have chemical reaction okay. And that AQ means that it's aqueous. All that means is that it's dissolved in water okay. So NaCl with an AQ next to it means that it's dissolved in water. Okay. Aqueous. Um, so we can change concentration by using different amounts of solutes and solvents. There's actually a fat Phet simulation that does this in the chapter six folder. I've already opened it up right before class today so that you can see that fat simulation. If we increase our solute right then our concentration increases. Okay. Um, so if our concentration is really high, right. If we had something that was colorful in our solution, it would be a really dark color, right? If our concentration is really low right then it would be a light color. Okay. Um, so change in concentration again, to change that concentration we're going to use different amounts of solutes over solvents. Um, and so here are concentration units. Again we're going to talk a lot about molarity. That's what we're going to use mostly in this class moving forward is molarity as our concentration unit. There are other units um, which are these uh mass units. And that's what our city is using to determine the safety of our water. right? Okay. Um, they're using mass units. So if you're doing environmental chemistry, typically it's in mass units. And the reason why is because it's in such low concentrations that these chemicals are allowed to be in our water supply. Okay. Um, so very low concentrations are allowed by the EPA, right? Environmental Protection Agency to be in our water that we consume. Okay. And so because they're in such low concentrations, we don't want to say it's 0.00000005. Right. Molarity okay. We want to say that it's something parts per million right. And a million of these right mass. Then we're going to have just one allowed okay. Um and so again parts per million is the grams of solute that are allowed per ten to the sixth which is per million parts per million, exactly the way it sounds. Right. Per ten to the six grams of our solution of that water. Okay. Um, and I say of that water. But it's really the whole thing. Solution means everything, right? Solvent and so use okay. So grams per hour, million of total. Okay. Um, and then parts per billion is grams of solute per ten to the nine grams of solution. Right. What is what do we know what the E.coli. I don't even know. What's the E coli EPA standard. It's probably not even in parts per billion. I would think it would be in parts per trillion. Anybody want to Google that real fast? EPA allowable E coli in drinking water citywide? I don't know, something like that. It would be fun to see and hear since it's like current topic in our city right now because, you know, it's not I'm going to Google it to, uh, EPA. Right. Say it again. UNKNOWN It says you can set a maximum cabinet level goal for you for life. SPEAKER 1 In drinking water at zero, meaning that no. SPEAKER 0 I00. There's not even a parts per trillion. It's just zero. Okay. Uh, there's other things in our water that's allowed to be in there, which is nice. Nice to know, because I read something the other day that was like, oh, it's lower than the EPA standard allowed. And I was thinking like, maybe they had a peaked, like maybe they're allowed a tiny bit, but no, they don't allow any. This is a really good to know. Um, but there's other things in our waters like metals, right. And like, we know fluorines in our water because we put that in there for our teeth to be better so that it wasn't crumbling like it used to be. Right. In the olden days or in, I want to say, Europe. Right. Okay. Um, but zero. That's nice to know. So other things though, typically parts per trillion, um, allowed like things that we know are contaminants like lead and, um, other types of things. Okay. Um. Oh. See? Ten parts per million or billion. Okay, so arsenic is ten parts per billion is allowed to be in our water, right? Um, and this is really cool because, like, we actually have a research that goes on over at like Casablanca and they test that water, and they've actually found that there's quite a bit of arsenic in that lake. Um, and other like, heavy metals that are dangerous and higher than EPA standards. But then people fish and then eat those fish. And so are you like, really? I mean, I've seen the studies because like, there's a class that does it, environmental chemistry. Right. Uh, that takes those samples and then they present on it and I get to go to those, which is nice. Um, and then they, like, test the fish and see how much is in the fish, and it's like over the values. But really my question is if we cook them though, does it help? I wonder because that would help. I would hope so. Okay. Anyway. Parts per billion, million, trillion. That just means mass over total mass. Mass of the thing we're looking at over the total mass, obviously the total mass and billions or the total mass and millions. Right. Um, so billions is ten to the ninth, right? Trillions is ten to the 12th, I believe. Right. Um, but those are the units for that. Okay. Um, anyway, uh, moles. So when we're dealing with instead of mass units, we're dealing with mole units, which is what we're going to do in here and what we do in lab. Right. Typically, um, our labs have a molarity labeled on the bottle. Okay. Um, so molarity is going to be the moles of solute instead of the mass of solute right over the leaders of solution. Okay. Moles per liter is how we're going to deal with concentration for the most part in this class okay. The other one's nice to know, especially if you're a biology major. You're going to be doing mass percents for the rest of your life. Okay. Um, which is, you know, this is this can be coming from a mass percent as well. Okay. Uh, because that's where we get grams, right? We just convert. All right. Um, so moles of solute, specifically moles of the small part over moles of not the big part, but moles of the, uh, sorry, liters of the total solution. Okay. Um, so let's go ahead and calculate this. It says, what is the molarity of an aqueous solution, uh, prepared by adding 36.5 of barium chloride to enough water to make 750ml of solution. So molarity, what are the units for molarity again. Right. Moles of what solute per liter of solution. Right. I swear I'm doing it on my slide here. Okay. Moles of solute per liters of solution. Right. That's what we want to put together. Do we have moles of solute? No. What are we given. Grams. Grams of our solute. Right. Because it says we have an aqueous solution. What does that mean about our solvent? It's water okay. Uh, aqueous means it's water. Water is our solvent. Right. And we have 36.5g of barium chloride. What's the chemical formula for barium chloride? Berkeley two. We knew that because we did it on our last test. So barium is a charge of plus two because of where it's at. On the periodic table in the second column, chlorine likes to be a negative one because it's a halogen. So it wants one extra electron to be a full octet. Right. Barium wanted to lose those two electrons. So two and one come down right to make this neutral okay. So there's our barium chloride. We don't want our barium chloride to be in gram units though. We want it to be in mole units because we're going to molarity moles. Okay. Um so molarity how do we get from grams to moles. Divide by the molar mass okay. So the molar mass of barium chloride is what. Again chlorine is what 34.5 or 35.5 35.45. I knew there was some fours and fives in there. Um times two. Right. Plus barium, which is what say say it again. 187. Okay. Mass of barium. Yes. Okay, so we get 208.23. Is our total 208.23g for every one mole. And that's going to give us our moles, which is going to be 36.5 divided by the answer right gives us 0.175. Whatever. 286 I'm going to just keep them all for now moles. And then I want to shove that in the top. And then what do I put underneath it. My liters of solution. How many liters of solution do we have. Yeah exactly. Somebody said 750. Right. But oh wait a minute. That's milliliters. That's exactly what your brain should be doing okay. So instead of 750ml. What's it going to be. 0.75. Because we move our decimal three places for milli right. Um.75 liters of our solution because it told us that's how many milliliters of solution we have. Um, again we could have also just converted right seven 50ml. We know that there's 1000ml for every one liter. Right. And that's going to give us our 0.75. That's the math we did okay. Moving the decimal over three. We know Millie right. It's smaller unit than liters. So we would have a small number of liters for 750ml okay. All right. So what's our answer then. Divided by 0.75 we get Seven. What's our unit? A capital M, which is for molar okay. A capital M. Okay. Uh, a lowercase m is for meters, right? Mole is for moles. Right? Okay. I like to do Molek for molecules, but a capital M is molarity, which means moles per liter. Okay, so you could have written a capital M, or you could have written a moles per liter, which is really the units we end up getting out of this. Right. Okay. Um, but that is what we have 0.23. And really, we should have gone to what, this decimal place. Right. Because we have three sig figs with that 36.5. Okay. So 0.234 molar is our concentration of barium chloride in our solution. Okay. All right. Another problem. How many grams of aluminum nitrate are required to make 500ml of a 0.0525 molar aqueous solution. So now we're given molarity and we're trying to find write our grams okay. So again molarity is equal to what. Write moles of solute over liters of solution. And we know how much solution we have. And we know right. These two are given. So we're going to be calculating this one moles of solute. But then we need to convert that one into grants because we don't want moles. We want grams. So we can do moles to grams by multiplying by the molar mass okay. Periodic table moles of grams. We multiply by the mass to get from moles to grams, we divide to get from grams to moles. Okay. And we know all of that even without remembering it. Right. Because of the units. So we follow units and cancel out units. All right. So we're going to have our molarity times our volume right. So I'm going to rewrite this molar molarity right is equal to moles in is typically the unit I mean the variable that we use for moles is n a lowercase n okay. And then over um volume right. Is our leaders volume in liters. It has to be in liters okay. Because that's the definition of molarity is moles of solute per liters of solution. So this has to be our moles of just our solute. And this volume has to be liters of our solution okay. So if we rearrange this because we want to solve for n up here moles of solute we're going to multiply both sides by Volume. Right. So we're going to have volume which was what 0.5l. Because we need to go from milliliters to liters. Yes. We're going to move that decimal place over 1 to 3 places. So 0.5 right. Liters times our molarity which was 0.0525 molar. I'm going to go ahead and put my moles per liter so that I can cancel out my liters so that I know what my unit is when I get done with this math. Um, again this should give us right equals n it should give us our n value. Um, again we just rearranged right. And so v times m usually you'll see it written the other way MV equals n. Okay. Uh, anyway 0.5 times 0.0525 gives us 0.02625. Right. Moles of what? Aluminum nitrate. How do we write the formula for that? Al? In oh three. Right. There's three of those nitrates. Right. H and oh three is nitric acid. So it's just a negative one charge. So it should be al aluminum nitrate right. Just al in oh three okay. All right. Um. At the beginning of class, we didn't have aluminum nitrate. We had aluminum sulfate, which was a negative two charge, which is why we had two of those aluminum at that point. Okay. I heard somebody say two because probably they were just remembering back and they were like, was it the same? But the iron was the iron was different. All right. So we're going to go ahead and start out with 0.026 to 5. Right. Uh moles of our Al in 033. And then we need to change that to grams. How do we get to grams for moles. Molar mass. Molar mass is going to go in the numerator right. What is the molar mass for aluminum nitrate we're going to have three times 14. That's our nitrogens right. Plus um three times three is nine times our 16 for oxygen plus whatever aluminum is. What's aluminum molar mass. Wait say it. The first part again. 2626.98 I'm sorry. I'm also going to f okay. And there's only one of those. So I'm just going to leave it like that. Right. All right. So. when we add all these up. SPEAKER 2 Um. SPEAKER 0 26.98 plus three times 14 plus nine times 16 gives me 212.98. Is that what you guys got? And that's grams for every one more. And so we're going to multiply that by 0.02625. Right. Because we're going to multiply everything in the numerator divided by one. It doesn't matter. That's is itself right. Multiplying by one doesn't matter. It is itself. We can ignore the ones okay. Um so that's going to give us 5.59 is that you guys got grams. Five nine grams of aluminum nitrate okay. So 335.59 is what that number says. Fire okay. The same number on here. All right. How are we doing? So far? So good. Yes. All right. Um, so the next one, it says, what is the molarity for the nitrate ion in the this 0.0525 molar solution of aluminum nitrate. So if we want molarity of the nitrate ion right. Then we can go back to moles of our aluminum nitrate. Right. And for this many moles of aluminum nitrate, how many of how many ions do we have of our nitrates. We have three for every one. Right. Aluminum. So we're going to go ahead and say 0.02625 moles of our aluminum nitrate. And then for every one mole of aluminum nitrate. We have three moles of nitrate ions. Okay. Because of this three here. This one outside the parentheses okay. Um, so for every one of these we have three moles of ions. Nitrate ions. Right. And so that's going to give us whatever that number is times 3.02625 times three. That gives us 0.07875. So we've got 0.07875 moles of night because this moles cancels with this moles. Right. So we left with moles of nitrate ions. And we want not moles. We weren't asked for moles. If we were that would be the answer. But we weren't asked for moles. We would ask for molarity. So what do we need to divide that by. The leaders of solution. the leaders of solution, which is 0.5 again, right. Leaders of the entire solution. And so divide that by 0.5 gives us um, our molarity for in oh three ion is going to be equal to 0.1575 okay. Molar. Or we could put moles per liter. Right. Either one of those units should work. Okay. Um of nitrate ions okay. That's our molar of nitrate ions. All right. We're going to start with dilutions on Tuesday. But what I do want you to do is a problem in Top Hat so that you can get credit for being here today. Yay. Um, and it's going to be on present composition. Are we good with that? Yes. Empirical formula is what I want you to do. And then I think that'll be good. So here is a question. And then once you're done with this question, feel free to go, okay? Because it'll take a minute because it's a molecular formula with several atoms. And it's asking for the molecular formula. So we're going to first find our empirical formula. Get our empirical back right. So back into our molecular formula mass. UNKNOWN And then get our whole number ratio. And let's say and then. SPEAKER 0 It is what it is. UNKNOWN Right. Oops sorry. SPEAKER 0 Those of you leaving, did you just Google the formula that I'm calculating it? You should know how to calculate it so that you can do it on a test. UNKNOWN Right? Now's the time. It's open for any. Day. Now. Okay.