Chemical reactions (stoichiometry)
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Molecular and Empirical Formulas
-
The Mole and Avogadro's Number
-
Formula from Mass Composition
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Another mass composition problem
-
Balancing Chemical Equations
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Balancing Chemical Equations Intuition
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Balancing chemical equations
-
Stoichiometry
-
Ideal stoichiometry
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Stoichiometry: Limiting Reagent
-
Limiting reagent stoichiometry
-
Spectrophotometry Introduction
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Spectrophotometry Example
Formula from Mass Composition Figuring out the empirical formula from a molecules mass composition
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- A couple of videos ago we figured out how to go from the
- empirical formula to the mass composition.
- And all I mean by that, mass composition, is that you could
- start with the empirical formula H2O, which is also its
- molecular formula.
- And then we were able to figure out what percentage of
- this was water and what
- percentage of this was hydrogen.
- The way we did it is we said, OK, oxygen's mass is 16 atomic
- mass units.
- Hydrogen is one.
- The mass of the entire molecule is 2 times hydrogen.
- 2 times 1.
- Plus 16 for that one oxygen, so it's 18.
- And then we said, the composition of
- oxygen is just 16/18.
- So oxygen is 16/18.
- I don't have my calculator anymore.
- I think the number was something on the order of 88%
- or 88.9% oxygen.
- So now let's see if we can go the other way.
- Let's see if we could start with mass composition, the
- percent composition of the different elements, and then
- go to empirical formula.
- And I think the first video I made, I
- spelled empirical wrong.
- Which is good reason for me not to do spelling videos.
- I think I spelled this with an e right here, because I
- pronounce it empricial.
- But it's spelled with an i.
- Empirical formula.
- So back to the empirical formula.
- So let's do some exercises here to see
- if we can get someplace.
- So let's say that I start off with-- I need space, my
- periodic table is right down there.
- Let's say I am told that I have a bag of stuff.
- And 75% of that bag, let's say it is mercury.
- Hg.
- And say that the other 25% of the bag is chlorine.
- So the question that I'm asked is, what is the empirical
- formula of the stuff that I have here, assuming that it's
- all one molecule or one type of molecule?
- So what is the empirical formula?
- So the way to think about this is, let's assume that you have
- 100 grams of this stuff.
- Right?
- And the reason why I'm picking 100 is because 100 is a useful
- number to deal with.
- So if I have 100 grams of the stuff, all right.
- So let's say I assume that you have 100 grams. If you have
- 100 grams, how many grams of mercury do you have?
- Well that means that 75% of that, that means you have 75
- grams of mercury, Hg.
- And that means you have 25 grams of chlorine.
- You might be saying, where did get these numbers from?
- Well, I got the numbers because I assumed 100 grams. I
- could have assumed 112 grams or 7 grams, but those would
- have been harder numbers to work with.
- So I'm just assuming 100 grams. So the question is, in
- 75 grams of mercury, how many moles of mercury do I have?
- So I need to convert from grams to moles.
- Let's go the other way.
- One mole of mercury-- this is at least how my brain thinks--
- let me pick a suitable color-- so one mole of mercury so 6.02
- times 10 to the 23 mercury atoms.
- What's the mass of that?
- Well it's equal to whatever mercury's mass number is in
- grams. So let's go look up mercury's mass number.
- So mercury is a transition metal.
- It's one of the few metals that are liquid at room
- temperature.
- Right there we have mercury.
- Its mass number, let's pick the 200.
- It has a mass number of 200.
- Obviously, there's some mercury out there that has a
- mass of 201.
- But for simplicity, let's say it's 200.
- So 1 mole of mercury is 200 grams. How did I get that?
- Mercury's mass number.
- One atom of mercury has a mass of 200 atomic mass units.
- So one mole of mercury-- 6.02 times 10 to the 23 mercury
- atoms-- has that many grams of mass.
- So instead of 200 atomic mass units per atom, we have 200
- grams per mole of that atom.
- It's an easy calculation.
- You just look up it atomic mass number and say it's that
- many grams. You have a mole of the substance.
- Fair enough.
- And what about one mole of chlorine?
- We do the same exercise.
- We go down to our periodic table.
- Chlorine has a mass number of, let's say, 35.
- It has various isotopes on this planet.
- But let's say the mass number is 35.
- So one chlorine atom weighs-- well now, we shouldn't say
- weighs, we have to be very careful.
- One chlorine atom has a mass of 35 atomic mass units.
- So 6.02 times 10 to the 23 chlorine atoms, or one mole of
- chlorine atoms will have a mass of 35.
- 35 grams.
- So how many moles of mercury do I have?
- Let me get my calculator going.
- So I have 75 grams of mercury.
- One mole of mercury would have a mass of 200 grams. So I
- could just take 75 divided by 200 and it tells me how many
- moles I have. 75 divided by 200 is equal to
- 0.375 moles of mercury.
- How did I figure that?
- One mole of mercury would be 200 grams. I only have 75
- grams of mercury, so I have a small fraction of it, roughly
- 0.375 moles of mercury.
- I've just expressed the number of atoms of mercury I have.
- Moles tells me it's 6.02 times 10 to the 23 atoms of mercury.
- So I have 0.375 times Avogadro's number of mercury
- atoms. Fair enough.
- Now let's do the chlorine.
- I have 25 grams of chlorine.
- One mole of chlorine is 35 grams. So I
- have 25 over 35 moles.
- So what's that?
- So 25 divided by 35 is equal to 0.714 moles of chlorine.
- Now these numbers, they're not exact.
- Because if you take twice this number up here, you don't get
- twice this number right here.
- If you get twice this, you don't get this.
- But this is roughly twice as many moles of chlorine as I do
- of mercury.
- So that tells me that for every mercury atom-- remember,
- moles is just a number-- for every mercury atom I have, I
- have two chlorine atoms. So the empirical formula here is
- for every mercury I have two chlorines.
- I have two chlorines, right there.
- And the numbers didn't work out almost exactly right.
- Especially in the real world, when you're actually trying to
- figure out things empirically, your numbers seldom will.
- And maybe there's some random other things running out there
- in terms of other things that are contributing to the mass.
- But this is close enough to know that the ratio of mercury
- to chlorine is roughly 1:2.
- Right?
- This is a number.
- For every one chlorine atom, you have roughly two.
- You have twice as many.
- I guess the other way of thinking about it is you have
- twice as many chlorine atoms in the
- bag as you have mercury.
- Roughly.
- Although this is a little bit more.
- This is close to 0.75.
- But it's close enough for you to know that you're dealing
- with mercury chloride.
- Right there.
- Let's do another one of these.
- Let's see how the numbers turn out for this one.
- Let's say you have another bag that is 9% magnesium.
- And let's say the remainder of the bag, 91%, is iodine.
- 91% is iodine.
- So the way to do all of these is, you do the same thing.
- Assume you have 100 grams. So if you have 100 grams, of
- which you have 9 grams of magnesium.
- And you have 91 grams of iodine.
- And then figure out how much a mole of magnesium-- what would
- be the mass of a mole of magnesium and
- then a mole of iodine.
- So let me write here.
- One mole of magnesium.
- And we want to figure out one mole of iodine.
- Let's figure those out.
- So magnesium's mass number is, let's just go with 24.
- Let's say we have the isotope that's 24.
- Magnesium is 24 and since we're already down here,
- what's iodine.
- 127.
- It's 127.
- Iodine is one of the halogens.
- So let's see.
- You have 127 and you have 24.
- Let me write those down.
- So you have 127 and you have 24.
- So one atom of iodine has a mass of 127 atomic mass units.
- So a whole mole all of it, 6.02 times 10 to the 23 iodine
- atoms will have a mass of 127 grams. You just take the
- atomic weight, or the atomic mass, and the mass will be
- that many grams when you have this many of the atom.
- So then one mole of magnesium will be 24 grams.
- So now we just figure out how many moles of each of these we
- have. We have 9 grams of magnesium.
- So what fraction of a mole is that?
- A mole is 24 grams. So this is equal to
- 9/24 moles of magnesium.
- And what is 9 over 24?
- 9 divided by 24 is equal to 0.375.
- So this is equal to 0.375, which is similar to what we
- had in the last problem when that 0.375 showed up.
- And 91 grams of iodine is what?
- 91 grams of iodine.
- Well, in a whole mole of iodine you're going to have
- 127 grams. So we have 91/127 moles of iodine.
- And what's 91 over 127?
- I have a feeling we're going to have very similar numbers.
- 121 divided by 127 is equal to 0.716.
- I should do another problem with better ratios than this.
- These have the same ratios as the last problem, 0.72,
- roughly, moles of iodine.
- And this is moles of magnesium.
- So we have roughly twice the number of iodine atoms as we
- do of magnesium atoms. Right?
- For every one magnesium atom we have roughly two iodine
- atoms. I know this isn't exactly 1:2,
- but it's pretty close.
- So the formula here is magnesium iodide.
- Right there.
- And that's the empirical formula.
- We don't know, maybe in every atom maybe you had two
- magnesiums and four iodines.
- We don't know.
- All we know is that the ratio of magnesium to
- iodine here is 1:2.
- In the next video, I'm going to try to look for ones that
- have more interesting ratios, because I don't want to do two
- problems that both have the same ratio.
- But hopefully you found this a little bit helpful.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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