Rutger's University
M.Ed., Columbia Teachers College
Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.
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Rutger's University
M.Ed., Columbia Teachers College
Kendal founded an academic coaching company in Washington D.C. and teaches in local area schools. In her spare time she loves to explore new places.
An empirical formula is the lowest ratio of the atoms within a molecule. The empirical formula is accurate when describing ionic compounds, which cannot be broken into a single molecule unit. But when describing covalent compounds we use a molecular formula which describes the atoms within a single molecule. The ratio of atoms within a molecular formula is the same as that in the empirical formula, but it is not reduced. Molecules with the same empirical formula have the same percent composition.
So we're going to talk about and molecular formulas. Empirical formulas are smallest whole number ratio of a compound. Ionic compounds that are already in its formula because it's already in its lowest ratio when we bring two ions together its actually in its lowest ration already so ionic compounds don't have an empirical molecular formula. However, covalent compounds or molecules do have empirical molecular formulas which are what that will illustrate up here, so we have CH4 and CH4 actually is an empirical formula already in its lowest lowest ratio it has the basically at the subscripts of 1 and 4 and we can't reduce those anymore so this actually is an empirical formula. However, it's also is a molecular formula there is a compound up there that has the formula CH4 methane a very common compound so this is empirical formula and a molecular formula.
Let's go to this guy we have C6H24, these numbers can actually be reduced to lower to a lower ratio 6 can go into 6 and 6 can also go into 24 making it CH4. Notice they have the same empirical formulas however the molecular formu- formula there are very different they have different chemical chemical properties but its empirical formula is the same it actually comes in handy later on.
Let's look at this guy C18H72 when you, this guy is also its molecular formula but it can be reduced too 18 can go into itself and 72 making it's empirical formula also CH4 so any time you have it's lowest ratio that's an empirical formula if it's not in it's lowest ratio, we're going to call that a molecular formula okay.
Let's go into percent composition and that will help us in determining molecular and empirical form- empirical formulas okay? So if you know the chemical formula you can also find the percent composition okay let's take methane CH4 alright so in order to find percent, percent is part of the whole so if we find the mass of the whole thing then find the mass of each part so the mass of the whole thing for CH4 is carb- looking at our periodic table, the mass of carbon is 12 grams the mass of hydrogen is 1 gram times 4 its 4 grams so 12+4 is 16 grams this whole thing has a mass of 16 grams for every mol okay so what we're just going to do the percentage of each element so let's look at carbon. Carbon has a mass of 12 grams like we had mentioned from the periodic table the whole thing has a mass of 16 grams and since it's a percentage we're going to multiply by 100 and we know that of this whole thing carbon has a mass of seven occupies 75% of this compound while our hydrogen, hydrogen you can you can always use obviously is the same subtract 75 from 100 and you get 25% let's actually calculate that to make sure. Hydrogen has a mass of 4 and this case 1 for every 1 we need 4 of them so we have a mass of 4 grams I have a total mass and the whole thing is 16 grams multiply that by 100 and indeed you do get 25% so in this case carbon 25, 75% sorry 75% of methane and hydrogen is 25% of methane.
Let's look at this C6H24, C6H24 if this guy says 6 carbons and 24 hydrogens give us a total mass of 96 grams for every mol, okay so let's do the part again C6 6 carbons gives us a mass of 72 grams over a total of 96 grams the whole thing multiplied by 100 and low and behold we should get 75% carbon which makes complete sense because they have the exact same empirical formula they should be a ratios of each other which makes complete sense which then I'm not going to considerate calculate but we can also assume that its 25% hydrogen okay so this is very helpful in determining empirical formula is very helpful when determining percent composition as well.
But let's say instead of having percent composition if I need percent composition let's say they give us percent composition let's say we analyze the substances which we didn't know what it was and we found it to be 36.84% nitrogen and 63.16% oxygen they gave us the percent composition. What are we going to do with this information? We have to figure out the compound so in order to do this I'm going to change this I'm actually going to assume I have a 100 grams of the substance so I can change this percentage to grams because if I if I have 36.84% nitrogen I have a 100 grams of it, that means I have 36.84 grams of nitrogen and I want to figure out how many moles that is so that I'm going to divide it by its molar mass and the molar mass of hydrogen is 14 approximately 14 grams and I get, what do I get? 2.63 that's how many mols of hydrogen I have in the substance. I'm engaging into the same with oxygen I get 63.16 grams divided by its molar mass in this case it's 16 grams and I get 3.95 mols this is in mols okay so essentially if I just stopped I can say I have N2.63 O3.95 but this one ugly looking empirical formula it is one ugly looking compound we don't like having decimals in the compound so what we're going to do? We're going then to divide it, make it trying making it into whole number so we're going to divide it by the smallest, the smallest one is 2.63 so we're going to divide everything by 2.63 because that is the smallest one and then I get N1 I don't indicate the 1 O1.5 oh oh it's still in decimal point a lot of it won't be it would be nice to round for you but this case it's still in decimal point so what I'm going to do, I can make this a round number or just multiplying everything by 2 N2O3 and it's in its empirical formula yes in deed it is it does follow everything we just talked of to this empirical formula and we just discovered it, awesome great.
But what if we go a step further it's it is known that the molar mass of a substance is 228 grams per mol what is the molecular formula? Well this is empirical formula what is the mass of the empirical formula? 2 times 14 because that's the mass of nitrogen, I'll not put the units, that's for nitrogen plus 3 times 16 give me a molar mass of 76 grams per mol okay? So then if you we want to see how many times is this to get 2 I will have to multiply 76 by to get 228 so I can just say 76 grams per mol times x equals 228 grams per mol and I solve this problem and I get x to be 3 so basically I'm going to have to multiply N2O3 these subscripts by 3 to get the molecular formula so I'm going to say 2 N6O9 that is my molecular formula and if I would have check it and figure out the molar mass of it which I'm not going to do right now but if I would check the molar mass of this and multiply everything 6 times 14 plus 9 times 16 which is the molar mass I will get 228 grams which will make sure I check it and this is correct so that is basically the mass of difference between empirical and molecular formulas and being able to calculate them from their percent composition.
Unit
Chemical Reactions