hello and welcome to this lesson in this lesson we're going to be looking at the empirical and molecular formula we're just going to do three examples and by the end of it you will see exactly how these questions work i'm not going to go into all the details of how they use this in real life or anything like that you guys are in school you just want to know how to do it so you can get through your tests and exams and get some nice marks so what you do is the following they're going to give you a couple of elements and those elements are usually going to be given either in grams or they're going to be given as a percentage okay so it will be either in grams or as a percentage but even if they give it to you as a percentage it's still going to be based on mass now we know that chemistry is all about moles so we need to try convert those percentages into a mole okay so now how are we going to do that well what we do we do a little trick we know that if you add these percentages up it's going to give you 100 because you can't have um all percentages add up to 100 right so let's pretend that this sample that they've given us is 100 grams okay so if it's 100 grams and the percentages add up to 100 well now what we can do is we can say well that means we have 36.5 grams of n a and we've got 25.4 grams of sulfur and 38.1 grams of oxygen many of my students say kevin shouldn't the oxygen be o2 yes technically oxygen is a diatomic molecule but when you are busy with this section you need to treat oxygen as its own individual little thing you'll become more used to that as we go through this lesson now kevin why have we converted it to grams well we are one step closer to moles because we know that moles from our previous lesson is m over m r and so if i know the mass of each of them i can convert them into moles so for the sodium or the n a i can say that that's going to be its mass which is 36.5 divided by its molar mass which you would get from the periodic table but to save space i've already added their molar masses here from the periodic table and n a is 23. so if you go work that out the moles for n a is going to be about 1.59 you can just round to two decimal places for sulfur it's going to be its mass which is 25.4 over its molar mass which i've already found in the periodic table as 32 and so that's going to be 25.4 divided by 32 which is 0 comma 79 and then oxygen i'm going to just scroll down is 38.1 grams over its molar mass which i found earlier let me just scroll up on the periodic table as 16. and so then you can just type that in the calculator and that's going to give us 2 comma 3 8. all right all you do next is you divide each of those numbers that you have just found by the smallest one so the smallest one is going to be 0 comma 7 9 so you are going to divide each of them by zero comma seven nine so we're gonna say one point five nine divided by zero point seven nine for the sodium and that gives us two comma oh one okay then we're gonna do this one we're gonna say zero comma seven nine divided by the smallest one but it is zero comma seven nine so that's just going to give us one and then you're going to do two comma three eight over zero comma seven nine and that's gonna give you or three comma o one now what happens is that these numbers are all fairly close to being whole numbers and so we will just assume that this is going to be 2 this is going to be 1 and this is going to be 3. so the ratio of them will be 2 to 1 to 3. so the empirical formula for for n a s and o will just be we'll just write it out as n a to the two s we don't say to the one and then o to the three can you see what i've done there i've just taken the ratio number answers and i've put them into a formula n a is two s is one and oxygen is three that is called the empirical formula you don't need to understand the background they're not going to test you on that you just need to know how to do it get your marks finished okay the next part of this question says determine the molecular formula if the molar mass is 378. so this looks like a molecular formula but the empirical formula is not the proper molecular formula it's the simplified version of the molecular formula so all you need to do in a question like this and trust me they will ask this is we are going to just work out what the mass of this is and so that's going to be 2 times by 23 because sodium is 23 from the periodic table sulfur is 32 and then there will be three oxygens and each oxygen is 16 from the periodic table that's going to give us 2 times 23 plus 32 plus 3 times 16 and that's going to give us 126 grams so that means that the mass of this is 126 grams however if we look at the question they tell us that the proper mass is 378. so what we need to do is see how many of these 126s can fit into the 378. so we say 378 divided by 126 and you should always get a whole number if you don't you've done something wrong we get an answer of 3. so it means that the real molecule the molecular formula is three times as heavy as this so we can now update the answer although for the answer for b would be n a not to the 2 but to the 6 because we have to multiply everything by 3 s to the 3 and 0 to the 9. this is the molecular formula and then the one in red that is the empirical formula and they are going to ask you to do both in an exam and so here we go guys here's another question um they give you they'll always do it like this they'll give you a bunch of chemicals typically in grams or as a percentage but the percentage is also based on mass it's never going to be based on moles so here they give us a percentage so what we said is that percentages always add up to 100 so let's pretend that this is 100 grams there's nothing wrong with that so that means that the carbon is actually 49.5 grams the nitrogen is 28.8 grams hydrogen is 5.2 grams and then oxygen is 16.6 grams now that we have each of those in grams we can use n equals to m over mr and we could go get the moles of each one so the carbon moles would be we're using the formula n equals m over mr so the mass is 49.5 and then you would need the mr or the the molar mass of carbon but i've gone and found those on the periodic table already just to clear up a bit of space and so you can go work that out as 4.125 i'm just going to keep three decimal places because 1 2 5 is really nice but if there were a whole lot of decimals then i would round it up for nitrogen it's going to be 28.8 over its molar mass which is 14. and if we work that out you end up with 2 comma 06 grams sorry i mean moles okay that's mole and that's mole then we can do hydrogen so we know that hydrogen is going to be 5.2 over its molar mass let me scroll up for you its molar mass is one and so that's just going to give us 5.2 moles then we're going to look at oxygen which is going to be 16.6 over its molar mass which is 16 and that's going to give us 1.0375 moles next step we divide by all of those by the lowest one so the lowest one is going to be the oxygen and so we're going to say 4.125 it's quite a long process but extremely easy and that's going to give us 3.98 for the nitrogen it's going to give us 2.06 divided by 1.0375 and that's going to give us 1.99 and then for the hydrogen it will be 5.2 over 1.0375 and that will give us 5.01 and then for the hydrogen it's going to divide by itself because it is the smallest one and that's going to give us 1. then these numbers should always well they'll be fairly close to whole numbers and so we can obviously round this up to four this one up to two this one to five and this one to one so now when we put the formula down the empirical formula it will be c four n two h five o that is called the empirical formula the next question b says determine the molecular formula if the molar mass is 194 so what we do in a question like this remember is we go work out the mass of this one and so that's going to be 4 carbons which will be 4 times 12 plus 2 nitrogens which is 2 times 14 plus 5 hydrogens plus an oxygen and if we go do that we get 97 but now that they tell us that the molar mass is actually 194. so we see how many of these can fit into 194 so we divide it and if we've done it correct it should always be a whole number yes we get the number two so what that means is that the actual formula is double this one and so the molecular formula will have to double and so it's going to be c8 n4 h10 and then o2 and so we'll end this lesson but with quite an interesting one notice in this question they don't give us percentages they give us grams but that is actually better because if you remember in our previous ones they gave us percentages we then had to convert those into grams and then into moles now they've given us the grams already so we can go straight to the moles so for carbon it's going to be 19.95 over its molar mass now the molar mass of carbon is 12 and that will be 1.6625 remember i'm using the formula n equals to m over mr and then for hydrogen it's going to be 3.35 grams over its molar mass which is one and so that's just going to be three five and then for oxygen it would be 26.7 over its molar mass which is 16 and that's going to be 1.66875 if there's not a lot of decimals on the calculator then i would just write all of them then that we are we've already calculated the moles so at this step you divide by the smallest one now guys i have some students they'll look at these two and they'll be like hmm which one's the smallest guys they are the same they're both one comma six six 6 it's not going to change your answer much by being so specific with this type of question so just take these two as both being the smallest so in fact you can just chop off the end of that trust me it's okay so that's 1.66 so then that is going to be the smallest number so to find the carbon ratio you're going to say 1.66 divided by the smallest one which is 1.66 for hydrogen it's going to be 3.35 divided by 1.66 and that's going to give us and then for oxygen it's going to be 16 i mean 1.66 divided by 1.66 and that's going to be 1. so when you put this formula next to each other and remember the order doesn't matter it will be c h two o can you see because the c was one hydrogen was two and oxygen is one perfect that is called the empirical formula the order as i said does not matter moving on to question b they now tell us determine the molecular formula which is the real one if the molar mass is 60. okay so what we do is we work out the mass of this from the periodic table so that's going to be 12 plus 2 hydrogens which is 2 and then 16 for oxygen and if you go add that up you're going to end up with 30. so the mass of the empirical one is 30 but the real one is 60. well that obviously means we're going to have to double everything right and so the molecular formula which is the real formula you're going to have to double everything so it'll be c2 h4 o2 guys that is how you do empirical and that is also how you do the molecular formula thank you very much for watching