Formula from Mass Composition
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Formula from Mass Composition

August 27, 2019


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.

Only registered users can comment.

  1. Very good video.
    I like the way you have an easy going manor which makes it easy to understand what you are teaching.
    This will be great for my son whom I am going to be home schooling this year.
    Thank you for the effort your expending on these videos. 5/5

  2. "some mercury out there has a mass of 201" the truth is that only some mercury has a mass of 200 most of it has more or less.

  3. needs to be an edit on the very last worked problem. easy mental slip up…but if there were .375 moles of Mg and .72 moles of I, then the empirical formula should be Mg(sub 2)I…since it is roughly 2 Mg to equal 1 I…. just wanted to make a note of that. This guy is an amazing teacher…no complaints from me πŸ™‚

  4. @xxkora NH3's molar mass is
    1×14.01g/mol
    3×1.01g/mol <<found on periodic table
    ———–
    17.04g/mol
    n=m/M
    M=17.04g/mol
    m=100g
    n=100g/17.04g/mol
    n=5.87mol
    i think thats what your asking for =)

  5. I'm preping for HL IB chem, having never taken chem before. I'm just going through these videos one after another and it's like taking a year of chem.

  6. @7RaPeKiNg7 From what I understnad goes like this, you knwo that the number of moles equals as the grams of substance divided per the molecular mass. And he did that and got the number of moles. So he knew in 100 g of substance you have 75 grams of Hg, using the equations you obtain number of moles equals the mass witch is 75 grams divided per the molecular mass of substance which is it 200, and you obtain the number of moles.

  7. omg.. this is the longest way to get the answer lolll if u have the periodic table u cud figure out its molecular formula and than empirical formula.. noo lets jus throw moles

  8. Actually, for the exam I sit it's paramount that you use the moles technique, it helps when you advance further into chemistry.

  9. Hey guys, I know this wont be liked, but I've created study guides on chemistry, organic chemistry, physics, and biology that I've made easy enough for anyone to understand, yet thorough enough for use for high school, college, and MCAT review. My goal is to change a scientific education for everyone regardless of income level, scientific background, etc. I'm also keeping the price at 9 bucks

    I can't post a link, but you can find all my books on amazon by searching "konstantinos papadopoulos"

  10. Other than looking and seeing that one amount is aprox. twice or 3 times, is there a more accurate way to deal with those weighted numbers? Like dividing by the smallest one so it is equal to 1, and hoping the others either clearly do round up to the next whole number or clearly don't?

  11. I Like all of your videos sir! But your vids would be a lot much more better if you brought your calculator. Cuz you don't have it like, all the time.

  12. can you explain why, even though the molcule contains more grams of Hg than Cl, there are more moles of Cl than Hg and therefore more atoms of Cl than Hg?Β 

  13. got an exam tomorrow… the best procrastinator in the world I think so!

    furthermore, the way my teacher helped us calculate empirical formulae is much more different than this… but this helps me fill in so many gaps in my knowledge i forgot about.

  14. If the word atom didn't get left out of the expressions this wouldn't be so difficult for so many students.

    It is very clear what one means with "a dozen atoms of mercury". We are counting atoms, and there are 12 of them (a simple substitution: "a dozen"=12).

    When it comes to moles, however, it becomes "1 mole of mercury", and the fact that we are counting atoms stops being obvious."[One mole of] atoms of mercury" = "[6.022 * 10^23] atoms of mercury" would make things a lot easier until the student gets familiar with the concept.

  15. When you say you have 25% or 9% of something "in a bag", you should be clear that you mean 25% or 9% by mass, not 25% or 9% by volume, because when we talk about having an "amount" of something, we often mean we have a certain amount by volume, such as one liter or one gallon, but in this case we should be clear that we're talking about percentages by mass or weight, such as 25% of a one kilogram or one pound bag of stuff.

  16. Just putting this out there – Mercury's most common isotope is 202, and it's second is 200. Because of the fact that it has 7 stable isotopes, it's average is 200.5.

  17. mass/molar mass = moles

    that's pretty much all that is required to solve molecular mass composition problems and of course SI unit conversion i've never seen amu used molgmol^-1 is used instead.

  18. Americans don't use this table?It's in Dutch, but you geet the point. https://upload.wikimedia.org/wikipedia/commons/thumb/f/f5/Scheikunde_chemisch_rekenen_overzicht.png/680px-Scheikunde_chemisch_rekenen_overzicht.png

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