all right now we start chapter three which is titled composition of substances in solutions but what you should be really thinking about with chapter three and chapter four to follow is this is where we start to get into the math gauntlet of general chemistry right three four and to a lesser extent five is really when you're going to be putting your math skills to the test in the context of chemistry so ideas from chapter one for example dimensional analysis make sure you're comfortable with how to use that because even in this video today we're going to continue to use those types of conversion factors and the big question we're focused on here in chapter three is how do we calculate what molecules are made of and then do math with their molecular compositions right you see their composition of substances that's what we'll start with and then solutions later on in 3.3 so we've got four subchapters in chapter three so we will have four videos and this one will cover 3.1 formula mass and the mole concept and then three videos to follow that the first two 3.1 and 3.2 will be a little bit on the longer side and then we wrap up with 3.3 molarity where we get a new conversion factor for solutions and then other units in the end for 3.4 which will be the smallest subchapter making sure that we continue to use dimensional analysis throughout so formula mass in the mole concept addressing that question up top what do we do with chemical formulas because in chapter two we learned how to put those formulas together in molecular or sorry in molecules and ionic compounds and we learned about atomic mass units and using atomic mass on the periodic table we used it for example to calculate the number of neutrons and understanding those atomic masses and the information that they give us as you'll see led to a lot of important discoveries on top of what we already saw from chapter two so what is formula mass right it is the sum of the average atomic masses of all the atoms in the formula okay so an atomic mass is the mass of one atom a formula mass is the mass of everything that's in a compound okay so it's kind of just an extension of that idea we can still keep the units the same and think about it as being in atomic mass units we are now just thinking about it as in for example on this slide a molecule so we take everything that's in the molecule add up their atomic masses and that is the formula mass of a covalent compound we can also call that a molecular mass so you can see those two terms used interchangeably right formula mass molecular mass because we're dealing specifically with molecules so how do we do that if we look at this slide here we're trying to calculate the mass of chloroform right if you're asked to calculate the formula mass well first we have to look at the formula right at this level i don't expect you to know right chloroform exactly what that molecular formula is so i look down here at the molecular formula it's chcl3 and that formula gives me important information it gives me the types of atoms that are present in the molecule and the quantity so i have one carbon one hydrogen and three chlorines so i break them apart one carbon one hydrogen three chlorines i get their average atomic mass from the periodic table okay so using that value of a you've got to have your periodic table and then i multiply them above you know buy one another so carbon and hydrogen that's easy chlorine 3 times 35.45 gives me 106.35 add all three together and that gives me the molecular mass so for every molecule of chloroform it has an average atomic mass of 119.37 amu atomic mass units and this is something for the chapters that follow right we've got seven more chapters to go three through nine you will continue to have to calculate formula math so this is an important idea you need to know how to quickly calculate that so here's an example that you can practice right calculating the formula mass of aspirin and again the formula will be provided to you anything that's not simple nomenclature if it's something we learned how to do in chapter two you know like take dinitrogen tetroxide and 204 right that i would expect you to know that means n2o4 but any common names like chloroform or aspirin i will give you the formula so i recommend you pause the video here and practice calculating the formula mass of aspirin knowing that that's c9h8o4 so the method is to break them apart carbon of which you have 9 hydrogen but you have eight oxygen four pull their atomic masses off of the periodic table so you'll have to look it up for carbon hydrogen and oxygen and multiply their quantity times their atomic mass and add everything together if you do that correctly you should get an answer of 180.15 amu if this is something you're struggling to see the method and exactly what's going on because it is shown for each one in this video how to do it then let me know that's something we can discuss in lecture or in review so these chloroform and aspirin these are molecules we can call that a molecular mass but what about ionic compounds where we have a metal and a non-metal we know that ionic compounds from chapter two we know how to name them we know that it's a cation and an anion coming together right electrostatically attracted to one another they're neutral overall right so they only come together in one ratio but ionic compounds don't exist as discrete units right we covered this before they don't float around just like molecules so we don't call it a molecular mass if we're dealing with an ionic compound if we just think about it as a formula mass but as long as you keep that in mind it's calculated the same way it's just a different name formula mass for ionic compounds molecular mass for covalent compounds or molecules but after that if we take sodium chloride for example it's nacl so 1na 1cl add the two together 58.44 if you had something like you know cacl2 for example calcium chloride then you would need 2 times 35.45 because there's two chlorines but as you see it's calculated the same way i know how to calculate formula mass or molecular mass now everything we've talked about so far in this video is thinking about individual molecules or ionic compounds but you know if we're thinking about working in the lab for example we're working in the macroscopic realm of chemistry where we can actually hold things with our hands and see them with the unaided eye so we need some sort of conversion factor to bring us between these realms and this is the idea of the mole okay which is not the mole the animal which is in the top right corner here it comes from latin for bulk and this is a concept you've probably heard of having taken chemistry in the past all right we can't measure atomic mass units directly we can't we don't have a scale that can measure the individual amu right so we use mole to as kind of an aid to help us calculate the atoms in you know a whole quantity of something that we can measure in the lab for example yeah so what is the mole it is an amount of something okay so if you say that a dozen you know you get a dozen donuts or a dozen beverages right that means 12. mole is the same idea okay and the number comes from avogadro's number which we have on the next slide where did the idea come from right the definition of a mole it is the number of atoms in a sample of pure carbon 12 that weighs 12 grams and you're going to see this is really useful because it allows mole and amu to be directly related to one another and that gives us a link between the mass of the sample and the number of atoms molecules or ions depending on what we're dealing with specifically that's the key to relate those two worlds right what we're dealing with on the individual atomic scale and what we can measure in the lab and deal with the macroscopic realm so i alluded to it before here we go on slide 10 if a dozen is 12 a mole is 6.02214 times 10 to the 23rd and you don't have to know that whole number but i do expect you to know the first four significant figures okay so you absolutely 100 need to memorize avogadro's number right here it's represented by the symbol n sub a equals 6.022 times 10 to the 23rd and that would be units per mole right anything you want 6.022 times 10 to the 23rd atoms in a mole of atoms 6.022 times 10 to the 23rd molecules in a mole of molecules all right so it's a quantity right so what that means if i'm measuring out the quantity depending on what i'm measuring it's going to have different masses and that's what the last bullet point here is talking about okay the masses of one mole of different elements will have different masses because the mole is the quantity right if i take a dozen you know water bottles compared to a dozen i don't know bricks right the bricks are going to have a different mass even though there's 12 of each so here if i take exactly a mole of these different atoms they have different masses right 65.4 grams of zinc 12 grams of copper 207 grams of lead down here and you see other examples up there in the molecules the number of molecules for these guys is the same okay so all of those represent exactly one mole it has the same amount the same number of atoms in this case as the others so one mole of something has the same quantity as one mole of something else but as you see here the masses are very different because the atoms that make these things up have different masses themselves okay so atoms and molecules have different masses i hope this is making sense and you might notice something notable about those numbers right they correspond to the atomic masses on the periodic table which we're getting to okay so as you can see 6.022 times 10 to the 23rd right 6 02 and then 20 more zeros after that it is a crazy big number right a single mole right if you took one mole of red blood cells that's more than every human being in the world has in them which is crazy to think about because you yourself have 20 to 30 trillion red blood cells in just your body multiplied by every human on earth and it's still not a mole and if you took a mole of basketballs it would be the size of the earth but we need that huge number to make the connection between those crazy tiny atoms that we talked about in chapter one and the macroscopic domain where we can actually measure things out and that brings us to the idea of molar mass okay so this is tying back to what i referenced before and which is at the bottom of the slide here right the units for mole and atomic mass unit are both based on the same reference carbon 12 which is incredibly useful and then all other elements are based on that same reference so we get the same kind of relationship and that relationship is the fact that the molar mass of a compound which has the units of grams per mole is equal to the formula mass in atomic mass units which we just talked about how to calculate so that's what this middle bullet point is saying molar mass of any substance is numerically equivalent to its atomic or formula mass and amu difference just being right if it's an atom or a molecule or an ionic compound so what is the molar mass of an elementary compound it is the mass in grams of one mole of that substance okay so jumping back to this slide the mass in grams of one mole so the units oop jump ahead there are grams per mole so the units are grams in one mole of that substance and as i just mentioned you know 30 seconds ago that is equivalent to the formula mass on atomic mass units which is why we started this section by figuring out how to calculate formula mass because if you can calculate the atomic mass the molar mass is the exact same number we just changed the units atomic mass versus grams per mole and they have the same number but remember that they are very different scales right 12 amu so it's 12 atomic mass units in one carbon atom it's 12 grams so that's something you can actually measure on a scale in the lab in exactly one mole of carbon keeping in mind of course that one mole is very different for different things right here we see one mole of some solid mole can also be a liquid or even a gas okay but remember mole is a quantity it can look different as we see on this side and it can have very different masses so now keeping that in mind what we've covered so far we can get to the math part of this chapter that i alluded to before because those relationships between formula mass the mole and avogadro's number right are huge they tie in a lot of things in chemistry together we can do all sorts of different calculations and compute different quantities relating to the composition of these different substances and what we are covering in this section 3.2 and 3.3 it's all these same type of calculations fundamental concept moving forward you have to know how to do this math as we proceed in general chemistry okay so we've got mass chemical composition and moles if we know two of them we could typically calculate the third and they're related in this triangle which i pulled from outside of your textbook okay some people like this slide some people don't right it shows you how to do all the different calculations to relate the mass of an element the number of moles of an element and the number of atoms of an element and we're going to look at a lot of these calculations for the rest of section 3.1 here so if you like this slide use it if you don't you don't have to consult it because we're going to see how to do these calculations and then as long as you're using dimensional analysis to track your units and putting the numbers in the right places you'll be all right but that's just an extra resource for you to have if you want it so now as i said let's look at the math right how do we do these things a bunch of example problems okay according to the nutritional guidelines from the us department of agriculture the estimated average requirement for dietary potassium is 4.7 grams right if you are working on this on a test for example you don't care about most of that sentence the only thing you need to pull out is 4.7 grams of potassium and know that potassium is k on the periodic table and it asks what's the estimated average requirement for potassium in moles okay so i was given grams of potassium right mass and i need to find moles so how do i relate those two well i'm starting with grams i want grams to go away so it was in the numerator and now the denominator how do i relate grams to moles well i can use the molar mass of potassium right so i look at potassium on the periodic table it's got a average molar mass of 39.10 that's grams in one mole of potassium it's not moles in one gram so 39 goes down here 39.1 grams in one mole and then i take 4.7 divided by 39.1 with significant figures i get 0.12 moles of potassium yep it's a unit conversion similar to what we saw in chapter one now just you know continuing to use dimensional analysis to calculate different quantities so there we're relating grams to moles okay what about going this way on slide 19. okay liter of air has 9.2 times 10 to the negative fourth moles of argon and then we're asked to calculate the mass of argon in a liter of air so this is the part of the video right they're all broken down in the slides but you're going to want to pause for each of these and try to find the answer so here all right i've got 9.2 times 10 to the negative 4 moles of argon okay i'm asked to convert that to mass and we just saw i can convert grams to moles right but how do i do it okay i have moles and i want to convert to grams so if i set my conversion factor up right i want moles to go away so it started in the numerator now it's in the denominator and i want to convert it to grams i can do that i can always relate grams to moles now i just have to put my numbers in the right place so this was 9.2 times 10 to the negative fourth if you haven't ever seen that e before it represents times 10 to and then i can relate grams to moles that's no problem but i have to do it for argon specifically so i go to argon on the periodic table i look up its molar mass 39.95 grams per mole now right moles cancels out because it's on the top and the bottom i'm left with grams i multiply 9.2 times 10 to the negative fourth multiplied by 39.95 and that gives me my final answer which is shown on the next slide here 0.037 grams of argon now i've related moles to grams i can go grams to moles i can do moles to grams no big deal what about a problem like this this help tells me how many or asked me rather not tells me you're gonna tell the answer how many copper atoms are in five grams of copper wire now i just said you can always go grams to moles or moles to grams what you should commit to memory now is you can never go directly from grams to atoms or atoms to grams you cannot do that you always have to make a pit stop at moles or another way i've taught it in the past is you have to go over the mole hill to go from grams to atoms or atoms to grams you always have to convert to moles first and so this is a two step conversion i have a mass of copper atoms i need to divide by the molar mass first to go to moles of copper atoms and then multiply by avogadro's number that's 6.022 times 10 to the 23rd to get the number of carbon atoms so i recommend that you try that one on your own i will upload a separate video showing how to solve this problem on paper okay a little cleaner than these annotations how about another one here example 3.6 our bodies synthesize protein from amino acids one of these amino acids is glycine which has the molecular formula c2h5o2n how many moles of glycine are contained in 28.35 grams of glycine okay this is a good question because we know we can go from grams to moles which is what this question is asking us to do go from grams to moles i just said we can always do that in a one step conversion but that's using a formula mass previously when we were doing atoms for argon and potassium we could look them up on the periodic table but glycine isn't on the periodic table because it's a periodic table of elements and glycine is a molecule so before i can convert from grams to moles i need to calculate the molecular mass of glycine using this formula up here two carbons five hydrogens two oxygens in a nitrogen multiplied by their respective atomic mass so first and i recommend you pause the video and do this you calculate the atomic or sorry the formula mass of glycine then when you have that that tells you how many grams are in one mole then you take that 28.35 divide by that new molar mass and that will give you the final answer in moles of glycine i recommend you pause the video and try that and then after you get an answer you can check yourself with this slide right here final answer of 0.378 moles of glycine using the correctly calculated molecular mass of 75.07 grams per mole and always double check that you're putting your units in the right place grams you start with so you need 75 grams on the bottom that's why i recommend you use dimensional analysis to track your units and make sure you're doing your math the right way because if you multiplied by 75 in this case you would not get the right answer another one okay example 3.7 this tells us vitamin c is a covalent compound with the molecular formula c6h806 okay recommended daily dietary allowance vitamin c for children in 48 years is 1.42 times 10 to the negative fourth moles again test taking strategy don't worry about a big word problem pull out the information that you need you need that formula c6h8o6 you need this amount 1.42 times 10 to the negative fourth moles great that's all i need i can ignore the rest what is the mass in grams that's what it's asking me to solve for so can i convert from moles to grams i can right we said we can always go from grams to moles or moles of grams and this question is even easier because it tells me the molar mass for vitamin c 176.124 so that's great okay i have what i'm starting with right 1.42 times 10 to the negative fourth moles of vitamin c i set up my conversion factor i know i want moles to go away and i'm converting to grams and because i know i want those to disappear and get my final answer like it asked me to do in grams and then it's just a matter of making sure i put the number in the right place right it's 176 grams in one mole right so i can't put the 176 down here because that would be 176 moles in one gram which doesn't make sense because a mole is huge so i do 176 0.124 in one mole so this question is asking me to do multiplication to get my final answer because those two are both on the top and then again the final answer is shown with sig figs 0.0250 grams of vitamin c or you could do it in scientific notation 2.50 times 10 to the negative second grams of vitamin c you should definitely try these on your own for practice if you're not understanding what's going on and you'll have more on your sapling one last example right a big one to tie all these ideas together okay packet of sweetener contains 40 milligrams of saccharin and we do need that formula c7h5no3s which has the structural formula shown below how many carbon atoms are in the sample okay so this is another one that i will upload a separate video for because it's a multi-step conversion but before you watch that video i do recommend you try this on your own you need to pay attention to the fact that it's 40 milligrams not grams which is significant you'll need to calculate a molecular mass for saccharin and then think about the fact that it's not asking just for it molecules of saccharin but rather the fact that it's asking for carbon atoms that are in there which is notable as well the approach is also shown on this slide 27 yep so i do recommend you try it on your own i'll have a separate video showing the total approach right explaining step by step what's going on and i've mentioned it as we've gone through this video right but these are all examples of questions you can see on a test so you should know how to do these practice from the slide practice from the book practice from sapling you have to know how to do all these types of conversions to relate grams to relate molecules to relate atoms to relate moles any which way like you saw with that triangle several sides ago you can relate all of these quantities it's just a matter of tracking the units and doing the right types of conversions so i encourage you to continue with that practice and we'll pick up with these ideas in 3.2 when we learn about calculating empirical and molecular formulas