Molecular and Empirical Forumlas from Percent Composition
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Molecular and Empirical Forumlas from Percent Composition

September 21, 2019


Using all of the major
components in oil of cloves. It has a molar mass of 164.2
grams per mole, and is 73.14% carbon, and 7.37% hydrogen. The remainder is oxygen. What are the empirical
and molecular formulas for eugenol? And just as a refresher,
empirical formula is the simplest ratio of the atoms
in the molecule. And we’ll see that
in a second. And the molecular formula is the
actual number of atoms in the molecule. So let’s see if we can
figure this out. So the first piece of
information they gave us is that, the molar mass
of eugenol is 164.2 grams per mole. So just to simplify things in
our head, let’s just assume that we have 1 mole
of eugenol. And just as a bit of a
refresher, a mole is just a huge number. It’s like saying we have a
million molecules of eugenol, but a mole is just an
even bigger number. And we know that the molar mass
of eugenol– if we have that many molecules of eugenol,
then since we have 1 mole, we know that, that is
going to have a mass of 164.2 grams. So another way to
say it is, we have 164.2 grams of eugenol. Now they gave us the
composition. They tell us what percentage
is carbon, what’s hydrogen, and the remainder is oxygen. So if we assume we have a mole
of eugenol and we know that mole will have a mass of 164.2
grams, we can just use the percentages to figure out what
is our mass of carbon, hydrogen, and oxygen. So let’s figure this out. So let’s start off with this. So 164.2 grams of eugenol
times– let me do it in another color– times 73.14%. This will tell us what is our
mass of carbon that we’re dealing with. So let’s figure this out. So our mass of carbon– get
the calculator up here. All right, so we have 164.2
times– so 73.14% is the same thing as 0.7314. So we have 120.1. Let’s go with that. We have four significant
digits in each of these numbers, so let’s just go with
120.1 grams of carbon. So this is 120.1 grams
of carbon. Now let’s do the same
thing for the hydrogen and the oxygen. So we have 164.2 grams of
eugenol times– what’s our percent composition
of hydrogen? It’s right here, 7.37%–
so times 7.37%. What is this going
to be equal to? Get the calculator out. So we have 164.2 times–
7.37% is 0.0737. And we will get 12.1. So we have 12.1 grams
of carbon. And they don’t give us,
explicitly, the percent composition of oxygen. But we could figure it out
fairly easily, because we know everything left over
is oxygen. So we could take the mass of
carbon, the mass of hydrogen– oh sorry, this isn’t
grams of carbon. That’s grams of hydrogen. We could take the mass of
hydrogen, the mass of carbon, subtract them both from 164.2,
and whatever is left over will be oxygen. So let’s do it that way. Or we could figure out the
percentage that’s oxygen, because these all have to add up
to 100%, and then multiply. Either way would work. So let’s do that. So if we have 1 mole of eugenol,
it’s 164.2 grams. We just figured out that 120.1 of
those grams are carbon, and that 12.1 of those grams
are hydrogen. So what’s left over? We have exactly 32
grams of oxygen. So what’s left over when you
subtract these two guys out? I’ll do that in green. We are left with 32
grams of oxygen. So if you have a mole of
eugenol, it has a mass of 164.2 grams. And if you break
it down by the masses of the constituent molecules, 120 grams
of carbon, 12 grams of hydrogen, 32 grams of oxygen. Now what we need to do is figure
out how many moles of carbon this is, how many moles
of hydrogen this is, and how many moles of oxygen. And this is maybe something that
you should have memorized at some point, because these
are the three most common elements you’ll ever
deal with. But just as a refresher, you
could look up on any periodic table for carbon, you would see
in the periodic table the atomis– well, most periodic
tables will give you the atomic weight, which is the
weighted average of the different isotopes
of carbon around. But in general, the atomic
weight, or the weighted average of the atomic masses
of the different isotopes. But carbon, you just remember
that the atomic weight is pretty close to 12 atomic
mass units. Which means, that carbon
has a molar mass of 12 grams per mole. Or if you have a mole of carbon
molecules, it will have a mass of 12 grams. We do the
same thing for the hydrogen and the oxygen. Hydrogen has, whether you take
a single isotope or you take the weighted average, its
atomic weight or mass, depending which 1 you use, is
pretty darn close to 1. And so hydrogen’s molar mass
is 1 gram per mole. There’s some decimals here if
you look it up on a periodic table, but I just remember
carbon as 12, hydrogen is 1, and then finally,
oxygen is 16. So oxygen’s is approximately
16 atomic mass units. I guess, if you take a specific
isotope, atomic mass. If you take its weighted
average, atomic weight. And that means that its molar
mass is 16 grams per mole. Now we have the number
of grams of oxygen, hydrogen, and oxygen. So now we should be able to use
this information down here to figure out how many
moles this is. So let’s do this. So it’s 12 grams per mole, but
we want to divide by grams because we want this gram to
disappear in the numerator. So what we could do is, if it’s
12 grams per mole, that’s the same thing as 1 mole for
every 12 grams. All I did is, I took the reciprocal
of this right here. If we have 12 grams for every
mole, that’s also true that we have 1 mole for every
12 grams. And the grams cancel out. And we are left with 120.1
divided by 12, which is pretty close to 10. So this is equal to 10
moles of carbon. So we know how many moles
of carbon we have in 1 mole of eugenol. Now let’s do the hydrogen. So we do the multiplication
here. And so here we have 1 mole of
hydrogen– if we take the inverse of this, or the
reciprocal of this– has a mass of 1 gram. So really, this is just
going to multiply out. So we literally have, we could
say 12.1 moles of hydrogen. But because we’re looking for
formulas, that’s pretty darn close to 12. So we should just say 12
moles of hydrogen. And then finally, we
have our oxygen. And we have 1 mole of oxygen for
every 16 grams of oxygen. 32 divided by– the grams cancel
out, of course– 32 divided by 16 is 2. So this is equal to
2 moles of oxygen. So if we have 1 mole of eugenol,
we have 10 moles of carbon, 12 moles of hydrogen,
and 2 moles of oxygen. So we now have the ratio. Or you could even think of, if
you had 1 molecule of eugenol, we would have 10 carbons,
we have 12 hydrogens, and we have 2 oxygens. And we’re done. This is the molecular formula
we have for every mole. We could think of it in moles
which is a huge number. Or you could think of, for every
molecule of eugenol, we have 10 molecules of carbon,
12 molecules of hydrogen, 6 molecules of oxygen. This right here is molecular
formula. And the empirical formula is
essentially taking this ratio to the simplest terms. And
all of these have a common factor of 2. So the empirical formula, we
essentially just put this ratio in simplest terms. So we
can divide all of them by 2. So let’s divide all
of them by 2. And you have 5 carbons for
every 6 hydrogens. This isn’t O,6. This is 2 oxygens. So let me make that clear. I’ll have to erase a lot
of stuff to get there. There you go. That was 2 oxygens. Don’t want to make
that blunder. We saw we have 2 moles of
oxygen, so this is the molecular formula. And the empirical formula,
at least the 2 is still divisible by 2. So you divide all
of these by 2. You get the simplest ratio. For every 5 carbons, you
have 6 hydrogens, and you have 1 oxygen. Or you can put a 1 right
there if you like. And we’re done.

Only registered users can comment.

  1. thank you i was sick from hs for 3 days in a row and this video saved my life for the homework! the textbook SUCKSSS and 1 mole is 6.02×10^23 for anyone who was wondering.

  2. Great Job! Is this strategy only used for percent composition? If it was already given in grams would it be whatever is given times 1 mole over amu?

  3. Is a mole actually just a ratio between atomic mass units and grams? I know it is probably not that complicated, I am just too tired to wrap my head around it right now

  4. Who can help me with this : I have an alloy of AgCu 72% + wt. 28% , what is the molar mass of this alloy ??

  5. So what if you're only given the percent composition and the elements in the compound? How would you find the empirical formula given only that information? Thanks!

  6. its easier if you just assume 100g and find the empirical formula after that divide the molar mass of the eugenol by the molar mass of the empirical formula and multiply the empirical formula subscripts by that number

  7. Thank you so much. This makes total sense now! I don't know why my Chemistry teacher didn't say this.

  8. you better be my chem teacher 😩 shes annoying, shes given me low marks from her low qualty of teaching argh. thanksss khan!!!

  9. thank you so much! I forgot my notes at school, this helped A LOT. Saved me from the awkwardness of asking someone in my class for an explanation.

  10. i feel like the mistakes like that o6 that ment to be o2 is ment to be on purpose so we sit there and scream at the screen at his mistake wondering wtf he did haha. help us know the rules i assume.

  11. I memorized the steps to solving these problems back when I was taking gen chem. Now as I study for the MCAT, I actually understand what I'm doing and why I'm doing it! Thanks Khan Academy!

  12. God people, just watch the whole video right till the end and you'll see that he does correct himself every time he makes a mistake omfg

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