welcome to Chapter six Part C still chemical composition learning objectives in this part we will discuss the mass percent composition we are going to give the definition and we are going to see calculations we are going to discuss molecular formula and the pinnacle formula we are going to give definition the definitions actually we are going to give a couple of examples and then we are going to compare the differences between the two we are going to see how we can calculate determine the empirical formula from empirical data and then how we can determine the molecular formula given the empirical formula and molar mass of the compound let's start mass percent mass percent composition an element in a compound is the percent of the total mass of the compound that it's due to that element in other words the mass percent the position of X is the mass of X that we have in the mass of the compound in the total mass of the compound times 100% let's see an example we have 1.9 1/2 grams of example that is decomposed in order to give zero point six nine zero grams of calcium and one point two to two grams of chlorine how to later mass percent of calcium and chlorine see what in 1.9 grams of sample we have 0.6 nine grams of calcium so we can calculate the mass percent for that the mass percent of calcium is going to be the mass of calcium over the mass of the compound or point six nine over one point nine one two the grounds of the calcium over the total mass right here we go 36.9% point zero nine percent I meant to say on the other hand for the chlorine we are going to have the mass of the chlorine over the total mass or if you want one point two to two over one point nine one two and that's the grams of the compound times 100 the final double is sixty three point ninety one percent notice that the sum of this two should be 100 that's a perfect way for you to check your calculations mass percent composition from chemical formula the mass percent proposition X is going to be the mass that we have for element X in one mole of the compound over the mass of one mole of the compound times 100 it's in example it's going to be much easier calculate the mass percent of iron and chlorine in iron 3 chloride the first thing that we need to do is to calculate the molar mass because we need to know the mass of one mole of the compound right well it's once the molar mass of iron plus three times the molar mass of chlorine so it's 162 point 20 in one mole of iron 3 chloride we have one mole of iron that means in one sixty two point twenty graph twenty grams of iron three chloride we have fifty five point eighty five grams of iron on the other hand in one mole of iron chloride iron three chloride we have three moles of chlorine or if you only have three times thirty five point forty five grams of chlorine in one sixty two point twenty grams of I don't leak law right let's start with the mass of iron with a mass composition of Iowa [Music] in one mole of iron trichloride we have one mole of iron so in 162 grams of compound we have only 55 point 85 grams of iron so the iron has a 34 point 43 mass percent composition for the chlorine in one mole we have 3 moles of chlorine right so in 162 we have 3 times 30 5.45 do the calculations and the mass percent composition for chlorine is 65 point 54 remember while we said before the sum should be equal to 100 and it is another one example calculate the mass percent of oxygen in the water pause the presentation and try to do it by yourself the first thing that we need to do is to calculate the molar mass of water we have done it before it's eighteen point zero two so in one mole of water we have one mole of oxygen that means in 18 point zero two grams of water we have 16 grams of oxygen the mass percent of oxygen is going to be 16 over 18 point zero 2 so the mass percent composition for the oxygen is 88 point 79 the remaining is going to be water another unknown problem for you we need to calculate the mass percent composition of oxygen in calcium phosphate all the presentations I'll try to do it by yourself and view we are going to check your work later on welcome back the first thing that we need to do is to calculate the molar mass of the calcium phosphate you can do it by yourself it's three times phosphorus I'm sorry three times calcium I wanted to say twice the molar mass of phosphorus and eight times the molar mass of oxygen three hundred ten point eighty in one mole of calcium phosphate we have eight moles of oxygen eight moles of oxygen so in 310 grams of the compound we have eight times 16 grams of oxygen we are going to use this one here in order to calculate the mass percent composition the mass percent for the oxygen is equal eight times the molar mass of the oxygen over the total mass 41-point 27% molecular versus empirical formula [Music] the molecular formula gives the actual number of each element in the molecule however the empirical formula gives only the relative number of each element in the molecule that's an example consider the molecule that we have here the thing we think consists of two carbons and four hydrogens so the molecular formula is going to be c2 h4 if I take if I take this one here the ratio between the carbon and the hydrogen is 2 2 4 in all if you want one to two if we further simplify now the empirical formula gives the simplest ratio between the elements that make up the molecule outside we are putting a variable and end basically the empirical formula corresponds to a family of compounds and not a sink a single compound like the molecular formula so the empirical formula disappear achill formula represents a compound that it has a ratio that it has axle numbers 1 2 2 4 3 6 and etc for another one time molecular formula corresponds to a single compound and it give us the actual number of atoms that we have from each element in the structure empirical formula corresponds to a family of compounds for this particular one that contains carbon and hydrogen and the ratio is one to two so specific compound family of compounds that have that consists of the same atoms with the same ratio another 1 propene here we have one two three four carbons hydrogen's 3 the last 3 6 plus 2 8 so the molecular formula is C 4 8 8 on the other hand the ratio is 4 to 8 or 1 to 2 so the empirical formula for this compound as well its carbon 1 hydrogen to put it in a parenthesis and put a variable outside your books doesn't have the variable outside but I would like you to get used to that so put the variable outside and end in alpha and X whatever you want let's see a couple of compound this is the molecular formula for a compound it can also be written like this right this is another one compound we can write it as the US like this as well another one compound it can be written like ways another one example here we go notice something that all these molecules are using ch2 as a building unit the first one has two building units this one has three building units for building units five six and etc so this six compounds that I have here have exactly the same empirical formula after the first case n is equal to two at the second n is equal to three four five six and etc all these compounds that we have here all six compounds that we have here belong to the same family it's very useful in space especially in organic chemistry to write the empirical formula because in that way we can categorize compounds and study the chemical properties as a group let's see how we can determine the empirical formula of a compound from experimental data based of the analysis of a compound we know that it contains only nitrogen and oxygen it contains 30 point 45% nitrogen and 69 point 55% oxygen we have to calculate to determine the empirical formula of the compound let's start knowing that the compound contains only nitrogen and oxygen here they are nitrogen and oxygen we know that the empirical formula will look something like that where the x and y are represent a simple small ratio for the nitrogen and oxygen the N is a variable it's going to be there so what we need to do is to calculate to determine based on the numbers that we have here to determine the simplest model ratio between them sorry between the nitrogen and the oxygen based on the data that we have in 100 grams okay in 100 grams of the sample we have 30 point 45 grams of nitrogen and 69 point 59 grams of oxygen so we have a ratio between these two between the nitrogen and the oxygen but it is in grams it's not in moles so I'm going to take this one here the 30 point 45 multiplied by a conversion factor fourteen point zero one is the molar mass of nitrogen in order to convert the grams in two moles of nitrogen so thirty point five thirty point forty five grams of nitrogen corresponds to two point one seven thirty four moles of nitrogen I will do exactly the same thing for the oxygen I will take the sixty-nine point for fifty five grams of oxygen and I will convert it in to moles the sixty-nine point 55 corresponds to 4.30 for 69 moles of oxygen so the empirical formula is going to look something like this however these two need to be whole numbers so I'm going to take its number and divide it by the smallest number the 2 point 17 34 and it gives me nitrogen one oxygen two this is equal to 1 this is equal to two and I need to put a variable and outside and that's the empirical formula for the compound that we have here based on elemental analysis we know that the compound contains carbon and chlorine only carbon seven point eighty-one percent chlorine here it is ninety two point nineteen percent we need to calculate the empirical formula at this point this example is very similar to one we saw before I want you to show the presentation and try to solve it by yourself welcome back so the first thing that we are going to do is to assume that we have 100 grams of the compound at that case we are going to have 781 grams of carbon and ninety-two point 19 grams of chlorine let's convert those quantities into moles here we are converting the carbon this is the molar mass for carbon twelve point zero one so seven point 81 grams corresponds of carbon corresponds to point sixty five zero three moles of carbon let's do the same thing for chlorine 35 35 point forty five is the molar mass that we have for chlorine and ninety two point nineteen grams of chlorine corresponds to two point six zero zero six moles of chlorine so the empirical formula is going to look something like that however these are not whole numbers we are going to take each number and divide it by the smallest number this is the smallest number this is equal to one this is equal to four that's the empirical formula we need the variable outside don't forget get the variable your book doesn't have it ionic versus covalent bonding or if you want ionic versus covalent compounds consider sodium chloride and I only compound a water and two water a covalent compound or molecular compound same thing let's start with a sodium chloride in the sodium chloride we have sodium cations and chloride anions decide the sodium is interacting with all the chlorides around it and each chloride is interacting with all the sodium's around it here we do not have exclusive bonds this sodium interacts with all these chlorides as well as the chlorides of air behind it as well as the chloride that it's in front of it so when it comes play on the compound the only thing that we can write is the simplest ratio that we have between the sodium cations and the chloride anions we do not have exclusive distinct building units let's see the molecular compounds now let's see the case of water here is a little different the water molecule consists of an oxygen atom and two hydrogen atoms this is an exclusive molecule this is an exclusive bond that we have raised oxygen does not interact with any other hydrogens other than these two riffs one here is the building unit that we have for water let's you took a couple of compounds we are going to see two molecular compounds into an ionic molecular water and hydrogen peroxide molecular formula this is for the water this is for the hydrogen peroxide the empirical formula for base state where X is a variable here and the molecular and the empirical formula for the ASIS hydrogen oxygen ratio one to one and an exit side on the other hand I only compound for ionic compounds the only thing that we can write is the empirical formula we do not have any other formula because there the only thing that we know is the relative ratio between the two nothing else so when it comes to the ionic compounds given that we do not have something like this in a molecular formula we do not put that variable that X outside the only thing that we can get is the empirical formula so the random analysis of a compound reveals that the compound contains iron and chlorine 34 point 43% I donate sixty five point 57 percent chlorine we need to determine the formula of the compound see something because this one here is an ionic compound consists of iron and chlorine the only thing that we can determine is the empirical formula we do not have an other formula right assume that we have 100 grams of the compound then 34 point 43 grams will be iron and 65 point fifty seven cards will be chlorine we have a ratio but we need to convert it into moles right we are going to use the molar mass for iron in order to convert that into volts and we are going to use the molar mass of chlorine in order to convert to that into moles 35 point four to five I got it from the periodic table so in 100 grams of sample we have that many moles of iron and that many moles of chlorine so the empirical formula will look something like that however these numbers here the subscripts need to be whole numbers so I'm going to take each number divided by the smallest by the point six one six five this one here is equal to one and this one is equal to three this is the compound we don't need a variable outside why don't we because lots in ionic compound the empirical formula is the only formula that we have it's the iron 3 chloride molecular formula for empirical formula and mass molar mass remember this table we saw it earlier this compound can be written also like this and this is the empirical formula let's try to calculate the molar mass the molar mass would for this one here is going to be twice the mass of this unit inside here right so it's going to be twice 12.01 plus twice one point zero one one what I have here in the parenthesis basically is the mass of this building unit and then I multiplied this one the whole thing by two because I have two of those units right let's see this compound here the molar mass of this compound is going to be three times the molar mass of the building unit the molar mass of this is going to be four times the molar mass of the building unit that's what I have here five times it doesn't unit six times the building unit so so the molar mass of any of these compounds is equal to the number n times the molar mass of the building unit of the repeating unit right so I can find the number n as long as I know the molar mass and the molar mass of the repeated unit the molar mass of this one here let's say an example compound contains carbon and hydrogen and we know that this one here is the empirical formula we have the molar mass so we need to determine the empirical formula basically the only thing that we have to do the molecular formula that's what we have to determine we know the empirical we have to determine the molecular formula so the only thing that we have to figure out is the N remember what we had from before that the molar mass that they give it to us is equal to the N times the molar mass of the repeated unit so if I solve for n here I have n is equal to the molar mass over the mass of the C AIDS the molar mass is seventy eight point twelve they give it to us and then the mass of this one here is twelve point zero one plus one point zero one grams per mole divided Wister this feeds six times into the molar mass so the molecular formula is this or if you want base car versus hydrogen six another one example elemental analysis we have carbon and hydrogen that much carbon that much hydrogen we have to determine the empirical formula and then as soon as we have the empirical formula we will have to determine the molecular formula of the compound knowing the molar mass basically here what we are going to do we are going to see how many times the empirical formula feeds into the molecular formula but let's deal with part a first in 100 grams we have that much carbon that much hydrogen we know it from here convert that into moles by using the molar mass for carbon so eighty five point sixty grams of carbon is equal to seven point twelve seventy six moles of carbon do the same for the hydrogen this is the molar mass of the hydrogen fourteen point forty grams of hydrogen is equal to fourteen point twenty-five fifty two moles of hydrogen the empirical formula is going to look something like that however the seven point twelve and fourteen point twenty-five need to be whole numbers so I will take each number and divide it by the smallest number by 7 point twelve seventy six here we go this is one this is two so the empirical formulas licks the ratio between the carbon and the hydrogen is 1 to 2 this is the repeated unit and in the actual formula we have n of those repeated units it can be written like this or like this now knowing the empirical formula and knowing the molar mass we have to determine the molecular formula we have basically to find how many times this fits into 56 point Club let's all friend plug in the numbers that's the molar mass of the repeated unit we have one carbon and two hydrogen so the molecular formula is this or if you want this one here at this point we'll have finished with Chapter six Part C and with a whole chapter six basically we discussed with we discuss mass percent composition we gave definition and we saw a couple of examples some calculations we spoke about the molecular formula and the empirical formula what are the differences and we saw a couple of examples as well we saw how we can calculate the empirical formula for experimental data and then after having the empirical formula and the molar mass how we can determine the molecular formula at this point we have finished with chapter six make sure that you answer all the problems on the worksheet and homework problems I will see you with Chapter seven thank you very much for your attention