In this lesson, I'm going to provide you a list of formulas that will be useful to you if you're studying regular stoichiometry or even solution stoichiometry. So here's the first one. N, which represents the number of moles, is equal to the mass in grams divided by the molar mass or the molecular weight of the compound in grams per mole. So I'm just going to put the units here.
This is in grams, this is in grams per mole, and you'll get this in moles. Now there are other variations of that formula. If you want to calculate the mass, the mass is equal to the moles times the molecular weight. So that's another variant of that formula. And of course, also, if you want to find the molar mass, the molecular weight, the formula weight, the process will be the same.
It's the mass divided by the moles. So anytime you need to go from grams to moles, or if you need to calculate the molar mass using those two, you can use these three formulas. Now, it's important to know that one mole, it specifies a quantity.
It's equal to Avogadro's number. which is 6.02 times 10 to the 23. In the same way that a dozen represents 12, a mole simply represents a large number of stuff, typically in the case of atoms and molecules. Now the next formula you need to know is percent composition. If you want to find out how much percent of nitrogen is in a compound, or the percent composition of nitrogen, you can use this formula. It's going to be the mass of...
the element divided by the total mass of the entire compound times 100%. By the way for those of you who want example problems on how to use this formula or any of these formulas feel free to check out the links in the description section below. I'm going to post a list of videos that will cover these topics these topics and how to use these formulas. Now the next formula is percent yield.
Percent yield is equal to the actual yield divided by the theoretical yield times 100 percent. The next formula, this one is not very common, but in the event for those of you who have covered this in your class, here it is. The percent error, instead of percent yield, is equal to the measured value minus the actual value divided by the actual value times 100%.
The error, not the percent error, but the absolute error, is basically the difference between the measured value and the actual value. Other textbooks may have a T here for true value, but true value, actual value, it means the same thing. Now, let's talk about concentration, specifically molarity.
There are different forms of concentration. There's molarity, there's molality, there's normality, which you'll probably cover in Gen Chem 2, but in Gen Chem 1, typically you'll focus on molarity. The molarity is equal to the moles divided by the volume. And the moles is molarity times volume.
And the volume is equal to moles divided by molarity. But now let's focus on this equation. Molarity is equal to n. n is the moles of solute.
v is the volume, but of the solution. You need to understand the difference between the solute and the solution. So let's say if we take sodium chloride, table salt, and we put it in water. The sodium chloride will be the solute.
It's the stuff that dissolves in the solution. Water is the solvent. The solvent dissolves the solute.
When you combine... And together, the mixture of these two forms the solution. So here, this is just the moles of the solute, in this case, certain chloride.
This is the volume of the entire solution, the volume of both certain chloride and water combined. So that's what you need to understand when you're dealing with this formula, m equal n over v. Now the next formula is the dilution equation, also known as the dilution formula. It's M1V1 is equal to M2V2.
This formula is useful, let's say, if you want to find the new concentration after adding water. So, for instance, let's say we have a solution of sodium hydroxide. Let's say this is a 4 molar solution.
That's the molarity. And the volume is 200 milliliters. Now let's say we add 600 milliliters of water. The volume of the solution is going to increase. So instead of being 200 milliliters, it's going to be 200 plus 400, which is 800 milliliters.
So notice that the volume of the solution went up by a factor of 4. This formula will help you to calculate the new milliliters. M1 would be 4, V1 will be 200, M2 we're trying to find, V2 is 800. But you could think of it conceptually. If you dilute it by a factor of 4, if you increase the volume by a factor of 4, the concentration will decrease by a factor of 4. So if we divide this by 4, it should be 1M.
That's how you could use that formula. Now keep in mind n is equal to m times v. The moles is equal to the molarity times the volume. So N is equal to M1V1. It's also equal to M2V2.
That's why this equation works because the moles of sodium hydroxide doesn't change. All we're doing is we're adding water to change the concentration. So that's why we can set M1V1 equal to M2V2 because in both cases, they equal N, the moles of the solute.
But that is the dilution formula for those of you who will encounter problems like this. So that's basically it for this video. By the way, for those of you who want a printout of these formulas, feel free to check out the links in the description section below.
I'm going to post a direct link to the formula sheet, which you can download for yourself. Thanks for watching.