Learn Advanced Placement Chemistry: Stoichiometry 2 Video

Learn about Advanced Placement Chemistry, Stoichiometry 2, in this comprehensive video by bannanaiscool.
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Rob Lederer: Finding empirical and then molecular formulas can be very straight forward as long as you remember that those percentages, those mass percents really are just how many grams of compound you have, out of a hundred percent, out of a hundred grams, so what? Here is the thing, when you are given, when you are told you have 71.65% chlorine in a compound, always pretend that you have 100 grams of the compound because if you did then you have 71.65 grams of it. So, we are dealing with everything out of a 100%. So, what we are going to come up with as a formula when we do this is a formula that well, is based on our observation of a 100%. So, we call it an empirical formulas based on our observations. So, here is what we do. Step number one is to take all the percents and just change them into masses of the compound. So, that's what I've done here. 71.65, 24.27, 4.08 those are all percentages. Turn them all into grams of that compound and then divide by the molar mass of those compounds. So, take that mass of chlorine, divided by its molar mass, its 35.45 grams for every one mole of chlorine, and then do the same for the carbon and the hydrogen and you get yourself these numbers of moles because, grams cancel in every one of these. They are now the moles of the chlorine, carbon and the hydrogen that are found in the compound. And remember, any formula really tells us how many of those units are in a compound or how many moles of each of those units are in one mole of a compound. So therefore, if you've got these as your mole ratios, step number three would be to divide the lowest number here into each one of the numbers and that would you get you your empirical formula. So, if you take the lowest number here, which is either one of these two. So, let's say this one here. Divide that by itself and you get one mole of Cl. So, I guess in this formula then, we've got Cl acting as a single unit. Now, if you divide this number into this, you also get one mole of carbon. And then, if you divide this number into 4.04, you get two moles of hydrogen. And so, ladies and gentlemen, your empirical formula for this compound is CH2CL, you know what you always do. We'll always put the carbons at the beginning of the formula, and then everything else gets ordered in terms of its increasing electronegativity. So, hydrogen does not have an electronegativity that's high in chlorine, chlorine is higher. So, we put it last hydrogen there, but, we put the carbons first because we just put the carbons first. So, CH2CL is the empirical formula.

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