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.

thanks for all help i was in a tight spot and this helped me a bunch

Ms mulabaas sucks

Oh my god, thank you.

I believe that would mean you are already at the most simplified ratio i.e. the empirical formula.

Should be C10H12O2.

watch the whole video..

thanks coach!

thank you so much

if it's 2 oxygen, why did you write down 6?

its not 12.14 g of c its 12.14 g of hydrogen

your witting on this is better than mine on paper

wait isn't oxygen a diatomic??

Studying for the exam tomorrow… xD This is some nice revision

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.

thank you I had to retake this test and cramming it all back in to pass.

You put this in real easy to understand terms. Thank you so much. Keep up the good work!

so much sense thou

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?

Cool

Thank you!

you saved my life, thank you sooo much for this! i completely understand everything now 🙂

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

this method is awesome

why is there 6 oxygen its supposed to be 2

Smooth

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

Can anyone explain to me why he put 6 moles of oxygen instead of 2?

thanks dude, really helped

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!

These videos are basically what's allowing me to pass my high school math classes

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

i love you ! you are the only reason am getting through school

you are the best of the best.

These videos are great, but having the "forumlas" typo twice makes me cringe.

My name is KHAAAAAANANNNNANNNNNNNNN

The joy you feel when you understand something that's been troubling you is indescribable.

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

WHY THE MOLECULAR FORMULA WAS ANSWERED FIRST BEFORE EMPIRICAL ?

wouldnt it be 32.0

thank you sooo much.

thumbs down, I hate chemistry

Thank You.. I feel like I am ready for my test now.

thank you very much appreciated your hard work…..👌👍

Good lesson but you messed up the oxygen so I'm not liking it but thanks

damm! the screen truned green i thought! wtf went wrong with my display ..man dont use scary colours

keeps making mistakes..

thumbs up if you are just looking at this right before a test.

tomorrow be exams heat and forget

Isn't the molar mass of carbon 12.01?

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

shouldnt O be 2 not 6 ?

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.

Thank you.

thank

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.

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!

I FREAKING HATE MOLES!!!

Im choked ,spent an hour figuring out why O is 6

What software do you use for this? I need something like this for my school. Thanks!

I'm sorry for asking. but if the percentage of composition more than 100% . can I use this method ?

yo it's C10H12O2

There is a WAY easier way to do this!

Even at 2.0 x speed, this is still slow…:/

how did he get the 12g/mol? 1g/mol? and 16g/mol?

this is good you only need to work on ur mistakes

helpful

two oxygens… and two oxygens

writes a sixNvm he fixed it

I'm so lost for this test that I'm actually coming here to watch Khan Academy….

Why is 2 mols of oxygen O6

Nice 2 buddy

You wrote it as a 6

8.59 its 2 not six sir

thanks a lot god bless you

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