hello mr. Courtney here in this video we’re talking about mole concept and we’re going to be specifically looking at empirical and molecular formula over compound so we’ll be looking at how to calculate empirical formula how to calculate the molecular formula and all to get the molecular formula of a hydrate so we’d use the formula for compound to calculate percent composition in this lesson we’re going to see how we can use the percent composition to determine the formula of the compound so what is the empirical formula when perfect formula is the smallest whole number ratio of the elements in a compound we see the smallest whole number ratio mean the simplest form that the element the compound can be broken down into so when we look at the ratio of the elements in this element we have one carbon to oxygen we cannot break it up into any simpler form so that is our empirical formula in h2o you cannot break it on into a simple a whole number ratio so that’s our empirical formula but when we look at c4h10 we see that we have we it is able to be broken down into a simpler formula it is not the simplest whole number ratio we can divide the ratio of carbon to hydrogen by two we get a simpler ratio of c2h5 so c2h5 is the empirical formula for c4h10 so how do we calculate the empirical formula first we need to know the mass then we convert on mass into moles and how do we do that by dividing by the atomic mass of each other now if you give it a percent composition we’re going to assume it’s a hundred gram of the compound why can we assume it’s 100 grams of the compound you can do that because a percent is out of 100 so if you told the percentage of something is twenty-five percent that means out of 100 grams you have 25 grams of that element from the compound next we need to determine our mole ratio so was the mole ratio of each element in the combo what we’re going to do is look for the element that provides the lowest are the smallest number of moles and divide all the number of moles my element that number of mole and that will give us a whole number now if we do not get a whole number we need to multiply both the ratio by some number some factors that will give us a whole number once we’ve done that we’re going to use the ratios as a subscript in the formula so that’s the kind of example so compound is made up of 79 cuisine 79.8 percent carbon and 20.2 percent hydrogen or asked the fans empirical formula so first thing listia elements present in a compound then we determine what mass we have with a mass of carbon in the compound is 79.8 because we convert % 2 grams and the mass of hydrogen will be 20.2 we divide now by our atomic mass of these elements to get our number of moles over cotton we’ve had bagged 12.0 that gives us six point six five four hydrogen we / 1.01 that gives us 20.0 now after we have moles if you refer back to your steps we need to get more ratio so look at the number of moles we have for carbon and hydrogen you see that carbon has lowest number of moles which is 6 165 so we’re going to divide each number of moles that is six point six five four for carbon and 20.0 for hye-jin by 6.65 that gives us a mole ratio of 14 carbon and the mole ratio of three by hydrogen so empirical formula tells us we have one carbon and three hydrogen so what empirical formula will be CH three in the second example we have an oxide of aluminum which is formed by the complete reaction of 4.15 one grams of aluminum with 3.69 two grams of oxygen and we’re asked to calculate empirical formula of this come from so we’re given grams we’re not were current percentage so we give it a specific amount of grams of each element aluminum and oxygen again list our molds or math sorry now we get our molds x divided by the atomic masses of the element so aluminum will / 27.0 for oxygen will / is 16.0 so we look at our number of moles for each element and we use that to determine our moderation and this one we see that aluminum provides a smallest number of moles so we’ll divide the number of moles of aluminum and oxygen by each one guy point one five three seven we get one mole for aluminum and 1.5 models for oxygen now we didn’t get a whole number thank you go back to the steps we said if you don’t get a whole number you need to multiply it by

some factor that will give us a whole number so we’re going to multiply the 1.5 and the one multiplied both of them back to that gives us two aluminum’s and three oxygens so the empirical formula for this compound is al to go three so what this is showing us is basically how we move from the mass of the elements to empirical former so we start with mass percent of elements then we good we even mass percent we assume it’s out of 100 grams then we get grams of each element use the atomic weight so we divide by atomic weights to give us moles of each element so that gives us moles that we calculate our mole ratio by dividing by the smallest amount of moles then we can use those numbers to give us our subscripts in our company and that gives us the empirical formula so molecular formula molecular formula is active actual number each element or the actual month of each element in the compound so that’s what it tells us it gives us a ratio of the elements in the come from not the simplest phone number with just a ratio so it could be c2h4 it is not the simplest form but it is the correct ratio of the elements in the compound the empirical formula and the molecular formula can be the same let’s look at some examples in water each two is the empirical formula and the molecular formula because it is the actual ratio of the elements in the compound and it also represents the lowest whole number ratio of the elements in a compound for hydrogen peroxide h2o2 is the molecular formula but it is it is not the empirical formula because it is not the simplest whole number ratio of the elements the empirical formula will be H oh so calculated molecular formula to calculate molecular formula we need to know the ratio of the molecular mass to the ratio of the empirical mass we say in berk molecular formula Max and Virgo formula mass so we need to know these two values so to get n that is how many times is the molecular formula larger than our empirical formula we take the molecular mass and / the empirical mass once we get that ratio and we’re going to take that and multiply by the empirical formula to get our molecular formula when we get that we’re going to multiply each subscript in the empirical formula bag and so that’s very important we get a multiply each subscript by the emperor in empirical formula by that ratio n so that’s the kind of example here we’re told a white powder is analyzed and fund to have the empirical formula p 2 o 5 the compound has a molar mass of twenty eight point three nine to eight to eighty two point nine grams and we need to find a molecular formula so first we need to determine the ratio how many times is our molecular formula greater than 0 empirical formula so we need to find the empirical mass of p2o5 once you do that you get the empirical mass to be 140 2.0 so now we’re going to divide our molar mass so we take our molar mass and / the zone molar mass here and / or empirical mass and gives us two so now we have to as a ratio we multiply our empirical formula bag that ratio to give us our molecular formula so p 2 o 5 times 2 number 1 we’re going to multiply the subscript so 2 times 2 and 2 times 5 so I gives us p 4 4 10 and as our molecular formula now a compound used as an additive for gasoline to help prevent engine knock shows a falling percent composition so we have twenty four point seven percent carbon four point zero seven percent hydrogen 71.6 five percent chlorine we’re told the molar mass is known to be 98.9 six and we’re asked to determine the empirical formula and the molecular formula of this come from so we we’re incorporating both steps in this problem so we start by calculating empirical formula so we have carbon hydrogen and chlorine and since we’re given percent we’re gonna see you miss out of 100 grams so the masses will be the same as a percentage except we changes a grandson then we divide by the atomic masses of each of these elements

to get our number of moles so we / 12.0 to give us 2.0 24 carbon / 1.01 or hydrogen and that gives us 4.03 / 35.5 for chlorine that gives us 2.0 to we now need to know Oh a mole ratio so we look for the smallest number of moles that were we’d be calculated before we calculated 2.02 4.03 and 2.0 tooth so 2.02 is the lowest number of moles are the smallest number of mole so we divide the number of moles II calculated for each element like 2.02 so get one for carbon 2.0 for hydrogen and one for chlorine so that means our empirical formula is ch ch 2 CL now we need to final empirical mass or the empirical formula mask which find out to be forty nine point five grams once we follow a miracle formula mass that we need to find a ratio of the empirical mass to sorry of the formula mass to empirical mass so we divide formula mass by the empirical mass and then we’re already given the molar mass or formula mass which is ninety-eight point ninety six so we take 1 98.9 6 and / 49.5 that gives us approximately 2 so that means I’m molla molecular formula sorry is two times larger than our empirical formula so we multiply our empirical formula by two so i’ll give a c2h4 co2 is again we’re multiplying our subscripts so we multiply our subscripts the two times one it gives us to two times two is four two times one is to hydrate a hydrant is a compound that has water molecules chemically bonded to it then ink component has water molecules chemically modded to it it’s considered a hydric how do we know if something is a hydrate here we have a compound assault followed by a dot and then set amount of water molecules attached to it this number tells us how many water molecules and this h2o the watermark is attached to it is what leave lets us know that this is a hydric you can also look at the name of the compound the name of the compound will come contain the term hydrate so we have carpet to solve it tetrahydrate and tetra tells us we have four water molecules in it now to name the hydrate the name of the salt when in the salt and then we follow it with the term hydrate and with some numerical prefix attached to it the numerical prefix tells us how many hydrates we have in the compound so for this one if you were to name this company we named the salt plus sodium carbonate we have sodium carbonate and we have water I have any water molecule we have been out of 10 so that’s sodium carbonate and since we have hydrate you know it has water markets in the end of the term hydrate coming out 10 of them it’s decahydrate so the name of the compound is sodium carbonate decahydrate so what’s the formula for copper to sulfate tetrahydrate so copper to solve it copies plus to solve it is minus two so that’s cuso4 tip we put a dot and tetrahydrate will be 4h to go now the formula mass we do the formal matches as how we do we’ve done for any of any compound before just by adding the individual masses up together so we do for copper sulphur oxygen now we can do water as 18.0 15 or 18 point 0 2 x 4 or if you went to each one individually to 10 1.0 1 plus 16 and all that x for which is the same thing and that gives us 2 30 1.5 grams per mole to determine the formula for hydric now the moles of water present is going to be a whole number multiple of the moles of salt present and we saw that before we have one more one mole of the salt and four moles of the water in caco-2 sulphate tetrahydrate so let’s see how we determine that we’re gonna work it out simple some are similar to how we did empirical formula except we’re not going to be using elements we’re going to be using the compounds whatever the salt is and the water monkey so we look at this problem we’re given a sample of a

hydrate which is 10.40 seven grams and is heated to drive off the water and hydrous this is important anhydrous or the dry sample as a mass of 99.5 20 grams we want the format of the hydrate and then the name of the hydrate so first of all we know we have barium iodine that’s our salt and we have water because it’s a hydrate we need we need to know the mass of the hydrate of water sorry and the mass of the area highlight now since we’re told in the beginning the sample is the hydrate then we dragged it off so the mass of the dry sample will be the mass of the salt and the difference between the two masses will be the mass of water let me calculate moles we divide 9.5 20 by the formula mass of the area mega dive that gives us the number of moles we do the same thing for water / 18.02 and gives us number of moles and again we’re going to calculate or mole ratio this here is a smallest number of moles so you divide this by 1 by itself we’re going to divide this by point zero two four three four also we do that we get one for barium iodine and two for water so that tells us in every mole of the hydrate there’s one very mad i and two water molecules so bai to dot 2 h2o so that’s very Amanda died died hydrate because we have two water molecules attached so we’re going to talk about combustion analysis in combustion analysis it’s using organic or and analytical chemistry to determine the empirical formula for pure organic compound and as you can see here the sample is passed through the furnace then you’re going to have and absorb it something that absorbs the water molecules and then something that absorbs our carbon dioxide oxygen and other gases will be given out at the end so once the substance is combusted the products and usually is going to be carbon carbon dioxide water and oxygen gas can be analyzed to determine the empirical formula so we get mass you go to moles and then empirical formula so it’s a kind of example set a complete combustion of a sample of propane gas we started off we sorry we ended up with 2.6 21 grams of carbon outside 1.1 full 2 grams of water as only products find the empirical formula of propane so we know propane is going to contain carbon and hydrogen so that means from carbon dioxide we need to find the moles of carbon and from Waterloo to find the moles of hydrogen so we start with this we start with 2.4 61 grams of carbon dioxide we know in 1 mole of carbon dioxide we have 44.0 grams so that gives us moles now look I’ve got the formula for carbon dioxide co2 we have one carbon atom in one mole of carbon dioxide one mole of carbon atoms in one mole of carbon out there so that gives us point 0 600 moles of carbon for water we start with 1.4 for two grams in one mole of water we have 18.02 grams the formula mass and in one mole of water that we have two moles of hydrogen because we have h2o so here we have two moles of hydrogen that gives us 10 16 same thing as we did before going to calculate our mole ratio this here we have a smallest number of moles from carbon so we do our ratio of carbon to hydrogen and that gives us 1 06 / point 06 or carbon point 16 / point 06 for hydrogen and gives us a ratio of one to two point six seven so now we gonna we don’t have a whole number 40 mole ratio and we need whole numbers so i’m going to multiply everything by 3 so I gives us three carbon atoms 8 hydrogen atoms so that means of empirical formula is c3h8 let’s look at one more no suppose you were able to isolate them acid from two leaves and know that it contains only the elements carbon hydrogen and oxygen when you heat it when you hit 8.5 13 grams of acid we get point 501 grams

carbon out 7.10 three grams water we need to know the empirical formula and the molecular formula if you’re also told that the molar mass was determined to be 90 point zero four grams per mole the notice we’re given that is made bit Elm the compounds are is made from the elements carbon hydrogen and oxygen but we’re only given the mass of carbon dioxide and the mass of water produced so from carbon dioxide we can determine grams of carbon from water we can determine grams of hydrogen then we can find the grams of oxygen from the as being the room in subtract those values point from point 5 13 to give us grams of oxygen so we do it similar to what we did the first time in previous examples of point 501 grams of carbon in one mole of carbon dioxide sorry and in one mole of carbon dioxide we have 40 4.01 grams we know that we have one mole of carbon dioxide one mole of carbon in one mole of carbon dioxide then we’re going to convert our moles to grams so we have 12.01 grams of carbon in one mole of common so in the compound we have point one three seven grams of carbon we do the same thing for water point 103 grams in one mole of what that we have 18.02 grams then we look at the mole ratio in hydrogen in water molecules we have two moles of hydrogen in every mole of water then we multiply by 101 grams in each mole of hydrogen that gives us 0 point 0 1 16 grams once we’ve done that we can calculate the mass of oxygen by subtracting these two values from point 5 13 so now that we have the mass of each element in the compound we can go ahead and cut empirical formula so we start with our elements we start with the mask which we just calculated you already know the moles of carbon and hydrogen is we we could have determined that already so we do that by their atomic masses to give us the moves we do the same thing for oxygen now we look at the the values that produces the smallest number of moles so let’s go with this one these two are approximately the same okay so we go here so for more Moorish of column we get one hydrogen we get one oxygen we get 1.98 that tells us it’s approximately one two one two two so empirical formula is CH 0 2 but we still need to find our molecular formula so to find molecular formula phone from empirical formula we must know our empirical formula maths or are the empirical mass that’s 45.0 and we will give in the molecular formula of the compound previously in the question so we have 90 point zero four grams per mole as a molecular formula we divided by our empirical formula mass sorry Lloyd molecular masses 90.0 for we divided by the empirical mass of 40.0 that gives us a ratio of two once we have the ratio of two we’re going to multiply this by our empirical formula which is CH 0 2 to give us our molecular formula so that will give us c 2 h 2 o 4 mr. that takes us to the end of this until the next time i’m out blessings