Learn about Advanced Placement Chemistry, Stoichiometry 3/ Solutions 1, in this comprehensive video by bannanaiscool.
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Male Speaker: So here the masses of each of those compounds again and now what do we do we when we got the mass of each compound we need to find the moles of each. You find moles and then you can figure out the empirical formula, so, I take the grams of the carbon here, and I divide by the molar mass, that's how many grams there are in one mole, so the grams cancel here and that with the moles of carbon. I do the same for the hydrogen but see the hydrogen mole mass is 1.01 right, because that's mole mass of 1 H and for oxygen its 16, and I'm going to divide that into that, we get these numbers here, and those represents have the moles of each of those elements that we have, in that original unknown compound. What you do now? Oh its quite simple really, you take the lowest number of moles here and divided it into each to get the empirical formula, So the lowest number here is from 0.0174 I mean you divide that into itself, we get O as 1, if we divide this number into this number, you know what you get, look at you get 2, so its C2, now how many H's are there in this formula you divide this into this, I guarantee it this is what you get, 6 and that's the empirical formula compound that's all you are asked to find Its C2H6O or C2H5OH. Asked for I was getting that, you know what that is. Don't You? Two carbons and a Hydroxyl group at the end that organic molecule called ethanol. Here I just burns and booms. Advanced solutions In this solutions you are going to perform, the one thing that we concentrate it on was concentration, a concentration of course is moles of solute divided by the liters of solution, okay, so you know that formula and you are very comfortable with it and now that there is 3 extra ones that you really have to know, that are going to be very important, especially the last one of them that has a very weird unit in the end when you do math, okay so concentration moles per liter, moles solute divide by liters in solution here is mass percent, now mass percent we have been going through percentages that's expression pretty easy to understand. The mass percent of the solute in a solution is the mass the of the solute divided by the total mass of the solution, I remember the mass of the solution can be determined by knowing the mass of the solute which you should now appear, and that's going to be added to the mass of the solvent. Whatever the solvent is add those 2 mass to get the solute solvent to get solution right. Okay, so that's mass percent, you got now how to do that, what else is there is mole fraction, okay so mole fraction is going to be moles of solute divided by moles of solution which just means the moles of the solute plus the mole of the solvent here in the denominator divided into moles of solute. They get, that gives you in the end. No units, all that is, is just a fraction, so its called the mole fraction but in this formula here for mass percent all of these gets by multiplied by 100 that's going to be of course in mass percent. So you put the % sign, as the unit in behind. But then there is molality which is abbreviated by written M, molality is actually the moles of solute divided by the kilograms of solvent, not solution. Oh, that's sounds just odd, but waits till it comes our later it's beautiful.