so we will be looking at percent composition empirical and molecular formulas we want to start with percent composition so when we say percent composition we're generally referring to percent composition on a mass basis so what percentage by mass are the different elements in this compound so percent composition shows the percentage by mass of each element in a specific compound we don't need to over complicate this idea of percent composition for some composition or it's just like every other percent out there whenever we're thinking about a percent it's always the part that you're interested in divided by the total that of course is going to give you a decimal to change it from a decimal into a percentage we multiply by a hundred percent and percent by mass is going to be no different than this basic percent calculation that we're used to doing all the time and I do a real quick example of a percent composition calculation so this one asks us to determine the percentage composition by mass of carbon and oxygen and carbon dioxide so for this we just need to think about parts and total and here we're going to sum up the molar mass just like we've done in a lot of other problems because that molar mass is ultimately going to be our total so looking at the periodic table we're grabbing these molar masses from pure table carbon twelve point O one oxygen 16 we have another oxygen at 16 so that the molar mass for carbon dioxide is 44 0.01 grams per mole very similar to any other molar mass calculation that you've been looking at so now we're just going back to this idea of park compared to a total so if the part we care about is carbon let's take the part that is carbon part that's carbon is 12 point o 1 so it's going to be the numerator the total is that total molar forty 4.01 that's going to give us a decimal to get from decimal to a percentage we multiply by a hundred percent to give us twenty seven point two nine percent carbon now being everything else that's not carbon would be oxygen we get kind of by process of elimination just say okay well 100 percent minus twenty seven point two nine percent would have to be the percentage of oxygen but just for added practice let's go do this kind of the long way we're looking for the percentage of oxygen let's take the parts that are oxygen I have two oxygens at 16 that's going to be 32 the total is still the same total from before 44 0.01 that's going to lead to a decimal times a hundred percent to change that decimal to a percentage and then we would get the percentage of oxygen that we would have so percent composition by mass pretty straightforward you're doing the same thing you would in any sort of percentage base calculation you're comparing a part to a total then changing the net decimal to a percentage now I want to start talking a little bit about empirical and molecular formulas and what we want to start with for this is kind of a real quick definition of each empirical formula this is our smallest whole number ratio of the elements to each other in a compound or molecule so it's that smallest whole number ratio that's kind of the key word you want to think about for empirical for molecular it's the actual ratio of the elements to each other in a compound or molecule so molecular formulas when you're doing nomenclature what you're writing were the molecular formulas and just by looking at these two definitions you can kind of say which one you would prefer you would definitely if given a choice of having one of these as opposed the other you would much rather have molecular formula you'd much rather have the molecular formula because they tell you exactly what you have if you have the molecular formula we're gonna see you could go back and determine the empirical formula but just with the empirical formula formula you couldn't determine the molecular formula without more information so even though we'd rather have the molecular formula as a probe as opposed to the empirical formula empirical fail formulas still have value and are still important and the reason for this is you can determine empirical formulas pretty quickly you don't need any sort of expensive instrumentation in order to do that trying to determine the molecular formula formula though requires much more work it's either takes more time or the use of very expensive instrumentation hundreds of thousands of dollars worth of instrumentation meanwhile empirical formulas you can figure out with any without any sort of that expensive instrumentation so empirical formulas are still useful in many ways one do some examples to talk about the relationship that exists between empirical and molecular formulas because often you need to use this relationship in the course of actually doing a problem so what I've done here is I've listed some relatively common compounds cyclist of three common compounds so one of the common compounds that list is benzene benzene you might not be as familiar with but benzene is kind of an industrial solvent it's used in a lot of things like paint thinners and other sorts of industrial solvents it's not used as much now as it has in the past because it is a mild carcinogen but even with that you still find it in quite a few industrial applications and industrial solvents it's a liquid it's molecular formula is C 6 H 6 then empirical form and remember the empirical formulas the smallest whole number relationship between the molecular formula and the empirical form sorry the molecular formula is that smallest whole number ratio of the elements to each other so C 6 H 6 we could actually simplify that down to just CH Oh empirical formulas CH acetylene you might be familiar with acetylene is sometimes used in some well some welding torches it's a gas it's pressurized this uses that fuel source for doing welding a lot of times suddenly have some molecular formulas c2h2 so it's a gas but when you look at the empirical formula C 2 H 2 well 2 2 2 can be reduced to 1 to 1 so it's empirical formula is CH and then the last example we want to look at is carbon dioxide pretty familiar with carbon dioxide molecular formulas co2 one carbon to two oxygens well 1 to 2 can't be simplified down and still have whole numbers anymore so it's empirical formula is still just co2 so with these 3 examples there's three key points we want to look at and make for how these empirical and molecular formulas are related to each other the first thing one bila state is that different compounds can have the same empirical formulas and our example for that was with the benzene and the assembly so with the benzene and the acetylene when we were looking at that these are very different compounds they have different chemical properties they have different physical properties one's a liquid one's a gas ok they have different molar masses they have different reactivities these are very very different compounds but even though these are completely different compounds they still end up with the same empirical formulas that's the first thing we want to kind of look at when we're going through and doing these the next thing we want to look at is that the molecular formula is some whole number multiple of the empirical formula so the molecular formula is some whole number multiple of the empirical formula and again we can kind of see this going from looking at benzene molecular formula compared to the empirical form though what's this relationship well this relationship here would be a six fold multiplication molecular formula is six times what the empirical formula is for acetylene we look here so if we're looking at the acetylene our whole number multiple comparing the molecular formula to the empirical formula is to the molecular formula is twice what the empirical formula was our last example our last point we want to make I guess kind of follows the same thing what we just said with the whole number multiples is kind of a special case that we want to give this own category an individual compound can have its molecular and empirical formula be the same one way of saying that is the whole number ratio would be 1 it kind of gets us back to that second point but for this we'd be looking at something like carbon dioxide we look at carbon dioxide or on the right we see that the empirical and the molecular formulas are the same if the empirical and the molecular formulas are the same that would make that whole number ratio 1 so these are three key things that kind of pop up when we're doing problems different compounds can have the same empirical formulas see this with benzene and settling molecular formula is always some whole number multiple an empirical formula for benzene is 6 for acetylene is 2 for carbon dioxide is 1 which would make identical we will come back and use some of these points in future problems so I want to spend some time talking about these right now on spring respect to this slide this was slide we did previously in this presentation where we had a compound where we knew what the formula was we knew what the formula was sio2 and we wanted to figure out what was the percent composition of that so same slide we did before I wanted to show this again because what we're going to do next is we're gonna do the complete opposite imagine you didn't know what the empirical or molecular formula was for our compound but you did know what the percent composition data was so what we're doing now is we're kind of doing the backwards of this we're in take percent composition information and work to figure out what empirical and molecular formulas are that's what our next couple problems are going to look like kind of the reverse of what we did earlier well the best way to do this is to just do an actual example of this this is determining an empirical formula from percent composition information so let's start the problem the compound is forty point nine two percent carbon four point five eight percent hydrogen and fifty four point five zero percent oxygen by mass from this information we're asked to determine the empirical formula of this compound that's our goal figure out what this empirical formula is going to be so what that really means is we're looking at what are these subscripts in this formula here I have them as XY and Z because we don't know what they are yet we need to determine what XY and z are before we actually go through the calculation there's some baseline things we kind of need to think about we usually think of subscripts as being red an atom quantity but could they also be run in mole quantities let's kind of look at an example of this carbon dioxide we usually would read this as one carbon atom for two oxygen atoms but could we also describe that relationship as one mole of carbon to two moles of oxygen and we could just like you could say one dozen carbon two dozen oxygen and still have that make sense so we could read this in atoms we could also read this in moles and that's an important thing to notice about this because being able to read this in moles what that does is that allows us to have a handier unit we don't have to go all the way down to that atom level with a whole bunch of crazy small numbers we can use mole amounts which are much easier to manipulate mathematically so some other things we want to think about this percent composition data that we are given the forty point nine two percent hydrogen the four point five eight percent hydrogen and the fifty four point five zero percent oxygen the percent composition data is an intensive property when it means when we have an intensive property is it doesn't matter how much of the substance we have whether we have the smallest speck of this compound that teeny tiniest little bit that you can see or you have a humongous pile of it a mountain of it it will always have the same composition data our percent composition data no matter how much weight you have that small aspect or that humongous pile carbon is always going to be forty point nine two percent of the compound hi ujin is always going to be four point five eight percent of the compound oxygen is always going to be fifty four point five zero percent of the compound now the problem doesn't tell us how much we start with and ultimately it doesn't matter now remember we were talking about using mole amounts here instead of going to atom amounts and one way we know of how we can get to mole amounts as if we have gram amount information so grams are easy to weigh measure and manipulate so that's going to be a key thing you can actually do this experimentally with a very small amount of material you can do this with way less than a gram and still get those data out but if you make one assumption at the start it makes your calculations a lot easier the assumption you can make is that if you assume we have 100 grams we're gonna find that assuming 100 grams will make our calculations easier why is I can make it easier well if we assume 100 grams then these percentages could go directly to gram amounts if we started with a hundred grams then forty point nine two percent of a hundred grams is going to be forty point nine two grams you know up here for carbon if we start with a hundred grams and four point five eight percent of its hydrogen well that means we would have four point five eight grams of hydrogen fifty four point five zero percent of its oxygen and if we have a hundred grams or if we assume we have a hundred grams fifty four point five zero grams of oxygen you don't have to assume 100 grams in fact you can assume any amount you want and this will still work out completely fine and will work it's just if you assume 100 grams you take that first kind of mathematical complication simplify the percentages would become grams right away and that's the assumption we're going to make we're going to start with the assumption that we have 100 grams of this material now remember what we are thinking about we're thinking about getting this empirical formula those subscripts and mole amounts right now wearing grams you can Ricci the forty point nine two grams the four point five eight grams of hydrogen the fifty four point five zero grams of oxygen but we want to convert them to mole amounts because our subscripts can be read and mole amounts they cannot be read in gram amounts so the first step is going to be a one step in a conversion problem same stuff you've been seeing before converting the grams to moles by using the molar mass so forty point nine two grams of carbon change that to moles one mole of carbon over 12.01 grams of carbon your grams of carbon of course are going to cancel your and we left with moles of carbon that's going to leave you with three point four one moles of carbon from that one step calculation hydrogen four point five eight grams of hydrogen same thing use the molar mass of hydrogen to convert the grams to moles same thing we've been doing kind of previously in this chapter your grams of hydrogen well they're going to cancel what are you going to be left with moles of hydrogen that numerical answer right time you do that calculation for 0.53 moles of hydrogen same process for the oxygen change the grams of oxygen two moles of oxygen by using the molar mass of oxygen we look up oxygen on the periodic table it's 16 grams per mole and on the grams cancellous why the grams are in the denominator grams of oxygen cancel with grams of oxygen we're left with moles of oxygen so I'm get three point four one moles of oxygen so right now at this stage when we are at three point four one moles of carbon four point five three moles of hydrogen three point four one moles of oxygen we're closer to our ultimate goal but we're not there yet we're closer because now we're in mole amounts and we know that we can read the subscripts in an empirical formula in mole amounts but we're not done yet because those subscripts have to be in whole number amounts and these are not in a whole number amount so this is not the smallest whole number ratio we could make out of these values that's gonna be our next sequence of steps whenever you have numbers and you're trying to set them in the whole number ratios the easiest thing to do is to take them all and divide by that smallest value so that's gonna be our next step we want to look at these three values of three point four one moles of carbon the four point five three moles of hydrogen the three point four moles of oxygen pick the smallest and divide everything by that well I didn't show the work here but if you take three point four one moles of carbon and divide it by three point four one what are you gonna get you're gonna get one mole if you take the four point five three moles of hydrogen and divide it by the three point four one you're gonna get one point three three moles of hydrogen when you take the three point four one moles of oxygen and divide it by three point four four one you're going to get one mole of oxygen so now we're a little bit better spot than we were now we are at one mole of carbon one point three three moles of hydrogen and one mole of oxygen we're not quite at the smallest whole number ratios but it's looking better than we were we have some whole numbers here but we don't have it for everything yet so we still have a little bit of work to do here so we're not at or extremely close to whole numbers yet you can't round that down that one point three three down to one that's not going to work doesn't know we're close to one but what we can't recognize here's where you can start recognizing common decimals and common fractions what we want to recognize is this point 3 3 what's that really close to that point 3 3 is really close to 1 thirds that's what we're doing we're recognizing the common fraction or the decimal of 1/3 now we have to think about how are we going to get rid of that third well you just can't toss it out it's there and we're looking for whole numbers so how are we going to deal with that well anytime you have a third of something and you want to make that whole what would you do you'd multiply it by 3 well at this stage if you multiply it by 3 whatever you do to the hydrogen you also have to do to the other compound so if we want to get rid of 1/3 we would multiply everything by 3 to get rid of that third if you were close to 1/2 meaning you were at 0.5 or 0.49 or 0.5 1 somewhere close to 1/2 you'd multiply everything by 2 to get rid of that half if you are near a quarter like 0.25 point 2 4 0.26 or you know three quarters 0.75 point seven six point seven four you'd recognize you're at a quarter to get rid of a quarter you'd multiply everything through by 4 to get rid of that fraction that's our next step you recognize their common fraction or decimal recognize the third in order to get rid of a third we have to multiply by 3 what we do to one thing we have to do to everything so we would take our one mole of carbon multiply it by three what's that going to give us that's going to give us 3 moles of carbon wherein take our one point 3 3 moles of hydrogen what are we going to do we're gonna multiply it by 3 we're gonna get 3 point 9 9 moles of hydrogen we're going to take our one mole of oxygen and what are we going to do we're going to multiply it by three whatever we hear to one wing up two to all and yet our three moles of oxygen it's not worth three moles of carbon three point nine nine moles of hydrogen and three moles of oxygen now technically we're not at whole numbers but here's where we're using our mathematical and chemical intuition knowing we're expecting whole numbers these have to be read out in the whole numbers it's three point nine nine there's probably some other digits in there if we carry through more digits through this we're really looking at that being a whole number three point nine nine we can legitimately round that to four without any problem so we would get three moles of carbon four moles of hydrogen three moles of oxygen remember what we were looking for in our empirical formula in our empirical formula we were looking for the ratio of the carbon to the hydrogen to the oxygen and we said that we could read those out in mole amounts with the X align the Zr as long as they are the smallest whole number ratio well we're in what amounts three moles for moles three moles or aren't will not so we're good is this the smallest whole number ratio this information can be displayed it 3 to 4 to 3 the answer is yes you can't reduce 3 to 4 to 3 down any more so we're able to come up with our actual empirical formula it's going to be C 3 where we got that from H 4 Oh three we have our empirical formula now that's where that information came from we worked hard on getting everything to that smallest whole number ratio now before we go on to the next problem I want to review the key points to this because you see this pattern every time you're trying to take percent composition data and make a molecular formula from that we served by assuming 100 grams what'll that do for us that allowed our percent composition data to just be read in grams then we went through and converted eath grams to the respective moles once we had everything in mole amounts and we divided by the smallest that's where we got the one mole of carbon 1.3 3 moles of hydrogen and the one mole of the oxygen here we saw we weren't up whole numbers but we recognized a key fraction or a key decimal to 0.33 that's really a third or really close to a third so how do we get rid of a third we multiply everything by 3 once we did that we hit whole numbers and we saw that these had the smallest whole number ratio so that's how we were able to grab and do this empirical formula next up we wanna do a problem it says to determine the molecular formula in the previous problem we didn't have enough information to determine the molecular formula you need more than percent composition data you need something else in order to go that last step in order to figure out what the molecular formula is going to be so in this problem we're going to do that so we're going to determine the molecular formula of a compound this problem reads determine the molecular formula of a compound that's forty nine point four seven percent carbon five point two percent hydrogen sixteen point four eight percent oxygen and twenty eight point eight five percent nitrogen the molar mass of the compound is between 180 and 200 grams per mole it's critical that we have some information about what that molar mass is and we have a little bit information on it we know the range that it's in without that information on the molar mass we could not determine the molecular formula instead we'd be stuck back where we were in the previous problem where as far as we could go would be to determine that empirical formula so to determine the molecular formula starts off the exact same way as it does when you're determine an empirical formula your first sequence of steps is to solve for the empirical formula just like we did in the previous problems or even go through that process again so kind of continue the first step is we assumed 100 grams assuming 100 grams allowed us to take the percent composition data and move it directly into the grams of each of our compounds where we got each of the gram amounts that I just highlighted now we know that the subscripts cannot be read in gram miles they have to be read in mole amounts so we need to change for each element the grams to moles by using the molar mass we'd find the molar mass on the periodic table it would be a one-step conversion going from the grams to the moles we go through and do that grams of carbon cancel what are we left with moles of carbon get our four point one two moles of carbon grams of hydrogen cancel what are we left with moles of hydrogen that's where we're getting our moles of hydrogen from grams of oxygen would cancel what we'd be left with moles of oxygen so we get the one point O three moles of oxygen from grams of nitrogen numerator/denominator those are going to cancel what are we going to be left with moles of nitrogen that's where we're getting the 2.06 moles of nitrogen now that we have all this information in mole amounts it's at this stage where we are dividing each of these by the small so we're going to take each of these values and divide it by one point O three so when you take and divide the four point one two moles of carbon by the smallest value than one point O 3 what are you going to get you're going to get four moles of carbon when you divide the five point one five moles of hydrogen by that smallest value that 1.03 what are you going to get you'll get five moles of hydrogen when you divide the one point O 3 moles of oxygen divided by that one point O 3 you're going to get one mole of oxygen when you take your two point O six moles of nitrogen divided by the smalls which is the one point O 3 what will you get you'll get two moles of nitrogen when you look at those values you just generated these are already at or extremely close to whole numbers there is no need to multiply to get rid of a fraction or a decimal remember previous problem we were close to a third we had to somehow get rid of that 0.33 here we have whole numbers so being we hit whole numbers at this phase we didn't have to do anything we didn't have to multiply by three or four to to get rid of a third or order our half because we already hit whole numbers we're now ready to write this empirical formula remember our empirical formulas what are these subscripts so these subscripts were able to deal with I got four five one and two whereat whole numbers there's no way of reducing these down to smaller amounts those are the smallest whole number ratios you can make but those so for moles of carbon you get our c4 five moles of hydrogen so where H five is coming in at one mole of oxygen one oxygen in there two moles of nitrogen two moles of nitrogen so now we're able to get our empirical formulas our empirical formulas c4 h5 o n2 and we did this the same way we did in the previous problem but now we're going to be able to go one more step we're going to be able to take this empirical formula and figure out what the molecular formula is so we're going to do that on the next slide so now what we need to do is now we need to convert to an empirical formula to the molecular formula and we're gonna use what we learned about the relationship between these two formulas earlier in the presentation and here are the key things that we're using the molecular formula is always some whole number multiple of the empirical form remember we talked about that earlier in the presentation we need to use that because that's true that the empirical formula is always some whole number multiple of the empirical formula what that means is that the molar mass which comes from electing the formula must be some whole number multiple of the empirical mass which is tied directly to the empirical formula we just need to figure out what exactly is that relationship between the empirical and the molecular formulas so we're going to look at that we just determined what the empirical formula was on the previous slide that was our c4 h5 o n2 now what you can do is you can figure out the mass of that well the mass of that if you add up all the parts is ninety seven point one one grams per mole we're trying to figure out what the molecular formula is now what we're trying to figure out what we do know about the molecular mass is that it's between 180 and 200 so right away we can figure out some things so right away is the what what is the molecular formula gonna be the same as the empirical formula no what can a formula cannot be the same as the empirical formula because we know the molar mass is gonna be between 180 and 200 the empirical mass is not between 180 and 200 so we know the molecular formula will not match the empirical formula so now we got to figure out is what whole number multiples back up here at this point that I'm coming out drawing your attention to in the rut molecular formulas some whole number multiple of the empirical formula we need to figure out what that relationship is so what is that relationship between the empirical formula and this molecular formula what's this relationship here how are these going to be related well they're going to be related in the exact same manner that the empirical mass is related to the molecular mass shown in green we have to have that exact same relationship here so the easiest thing to do is when you're given range of the molar mass or molecular mass easiest thing to do is just start trying different integers to see where you hit that range at well let's see the range can't be 197 is not between there so let's try two if we double 97.1 one with that value be between 180 and 200 well if you double 97.1 one you're going to get let's just do that real quick pull out a calculator ninety seven point one one times two what are we gonna get one ninety four point two two well is that between 180 and 200 yes so what does that mean that means because the molar mass is twice what the empirical mass is that means our whole number multiple is really two we're looking at times doing this we have to multiply by two to go from the empirical mass to the molar mass that means we need to multiply by two to get from the empirical formula to the molecular form that means our whole number multiple that we're looking at here bottom right like in our formula will be twice what the empirical formula is because our whole number multiple is that times two so what do we have to do we have to double that empirical formula so doubling that empirical formula is going to give us C eight each 1002 and four for our molecular formula so that's the process you need to go through in order to determine what that molecular formula is remember you have to have some information about the molar mass if you don't have any information about the molar mass you can't determine what that molecular formula is here we had some information we were able to use that information what we do is we compare our empirical mass what we knew about the molar mass and we were looking for that whole number multiple and that whole number multiple and love being two that means going from the molecular formula to the empirical formula that molecular formula had to be twice what the empirical formula was if it didn't work with two if that didn't get you in range then you try three you try four you just keep going up till you hit something in that range sometimes they may give you the molar mass not an arranged but is an exact number well then it would be a lot easier then you could just divide the molar mass by the empirical mass get that whole number ratio this one made it a little bit harder they gave us a range so we had to try different integer values to realize that it would be a times two for this so that's how you take percent composition data and a little bit of information about the molar mass and then use that information to first figure out what the empirical formula is and then kind of go that last step use that information about the molar mass to take that empirical formula and then kind of go through the process and figure out exactly what this molecular formula turned out to be we can see very clearly that the molecular formula is twice what that empirical formula was