# Example of finding reactant empiricalÂ formula

## Video transcript

This problem asks us, suppose
you isolate an acid from clover leaves and know that it
contains only the elements carbon, hydrogen, and oxygen. Heating 0.513 grams of the
acid in oxygen produces points-- let me write these
down or let me underline them-- so heating 0.513 grams of
the acid in oxygen produces 0.501 grams of carbon dioxide
and 0.103 grams of water. What is the empirical formula
for the acid? So let's just write
the equation here. And once again, you can't
balance this equation because we don't know how many carbons
or hydrogens or oxygens we have in the acid. But let's just write it
down in general terms. So we have the acid. It has x carbons, y hydrogens,
and z oxygens, and we're going to heat it in the presence of
an abundance of oxygen. And when we do that the problem
tells us that this produces-- let me do it in
another color-- this produces some carbon dioxide
and some water. And they tell us the amounts
of carbon dioxide and water measured in grams, and they
tell us the amount of our original substance. Let me write that here. So we're heating 0.513 grams
of our original substance. And then that produces 0.501
grams of carbon dioxide, and 0.103 grams of water. Now we're going to do this
problem very similar to the way that we've done
the last one. Remember, we just have figure
out the empirical formula. We just have to figure out the
ratio, the simplest whole number ratios of carbon to
hydrogens to oxygens. The only thing that makes this
slightly more difficult than the last one is we have oxygens
here, and the oxygens are kind of mixed in with
the two products. So we can't just look at the
moles of our two products and say oh, we must have this
many moles of oxygen. What we're going to do here is
we're going to figure out the moles of carbon dioxide using
this information there. And that'll tell us the moles of
carbon that must have come out of our mystery acid, because
this was the only source of carbon
that might have generated the carbon dioxide. Then we're going to use the
water to figure out the moles of hydrogen in the product,
which has to be the same as the moles of hydrogen in
our mystery substance. And then we can figure out the
mass of carbon and hydrogen out of this 0.513 grams, we'll
be able to figure out because we'll know how many moles of
carbon and how many moles of hydrogen we started off with. We'll be able to know their
mass, and then we'll be able figure out how many grams of
oxygen we must have had in our original substance. And then we can figure out
the moles from there. And then figure out
the ratios. If that confused you, hopefully
just solving through it will make a lot more sense. So first let's just figure out
the moles of carbon in our product, and the only product
here that had carbon was the carbon dioxide. And that'll tell us how many
moles of carbon we had in our original mystery
acid over here. So we are end up with 0.501
grams of carbon dioxide-- that tells us right there. And we want to convert this into
moles of carbon dioxide. So we want to multiply this
times moles of carbon dioxide per grams of carbon dioxide. And if you care about
dimensional analysis you know that because these grams of
carbon dioxide will cancel with these grams of
carbon dioxide. So how many grams is one
mole of carbon dioxide? And we've done this
multiple times. The molar mass of carbon
dioxide-- you have 1 carbon plus 2 oxygens, so 16 times 2. That is 12 plus 32 or 44. So one mole of carbon dioxide
has 44 grams of carbon dioxide, or 44 grams has 1 mole.
but we wrote it moles per gram because we want to
cancel out with the grams right there. So let's cancel that out, and
you're going to get 0.501 divided by 44 moles
of carbon dioxide. Let's get the calculator out. So we've got 0.501 divided by
44 is equal to 0.-- we have three significant digits here,
and this is just a pure number or pretty close to a pure
number-- so 0.11-- I'll round it up to 4-- let me put it on
the side so I don't forget this-- so this tells us that we
have 0.0114 moles of carbon dioxide in our product. Now that also tells us for
every molecule of carbon dioxide we also have
one atom of carbon. So however many carbon dioxide
molecules we have, we have that same number of carbon. Remember, a mole is
just a number. When I say 0.0114 moles I'm
just telling you a number. So this tells me right there--
I'll stay in blue-- this tells me that we also have 0.0114
moles of carbon. That's what that tells me,
because for every carbon dioxide I have a carbon. And the carbons in the product
all had to come from my mystery substance. So how many grams of this
mystery substance must have been carbon? Well I had this many
moles of carbon in our original reactant. They all end up over here. Let's figure out their grams. So
let's multiply-- so times-- now here we want to do grams per
mole, because we want to end up in grams. We want to
cancel out the moles. We want grams of carbon
per mole of carbon. And the molar mass of
carbon is 12-- it's a good one to memorize. So we have 12 grams of carbon
per mole of carbon-- we've got 12 over 1 if you like. So what does this equal? How many grams of carbon are
we dealing with in this mystery substance right there? Get the calculator back. So if we multiply-- it's
actually this exact same number, let's just multiply it
by 12-- so times 12 is 0.137. So let me write that down. So we have 0.137 grams
of carbon in our original reactant. So of this 0.513 grams
of this mystery acid, 0.137 grams are carbon. So that's what we got so far,
so we have two good take-aways. How many moles of carbon we had
over here, and how many grams of carbon that
translates to. Now let's do the same thing
for the hydrogen. Same drill. Let's do the hydrogen--
we're using that information right there. So we're starting off with 0.103
grams of water, and we want to turn this into
moles of water. So let's multiply it times
moles of water per grams of water. Once again, to cancel out the
grams of water, numerator and denominator. And what is the molar
mass of water? You have two hydrogens, so it's
2 times 1 plus the molar mass of oxygen, which is
16, so it equals 18. So 1 mole of water has 18 grams,
or 18 grams per mole. But we wrote 18 in the
denominator, because we want the grams into the
denominator. So these cancel out and
we have 0.103 divided by 18 moles of water. Let me get the calculator
back. So we have 0.103 divided
by 18 moles of water. Let me put this on the side
so I can still see it, very bad memory. So this tells us that
we have 0.00572 moles of H2O, of water. We just converted these grams
into the number of molecules that we actually have. Now if we have this many moles
of H2O, how many moles of hydrogen do we have? For every molecule of
water we have two molecules of hydrogen. So you multiply that times 2--
so let me do that right here-- so 0.00572 moles of H2O times,
we have 2 moles of hydrogen for every one mole, I
should say, of H2O. So we're just going to multiply
this by 2 to get moles of hydrogen. So let me get back the--
let's multiply it by 2. So times 2 is equal to 0.114. So this is equal to-- let
me get that back. I always forget how many
zeros there are, so I remember it now. OK, so it is equal to 0.0114
moles of hydrogen. So so far we figured out the
number of moles of carbon in our product and its mass that
must have been in the original reactant, the mystery
reactant. And we've done the moles
for the hydrogen. We have this many moles
of hydrogen. So we must have started
off with that many moles over here. Let's figure out the
mass of that. So what is the molar-- we want
to figure out this in grams-- so times-- the molar mass
of hydrogen, it is 1 gram per mole. So hydrogen's easy--
times 1 gram of hydrogen per mole of hydrogen. I wanted the moles in the
denominator so it cancels out. Moles cancel with moles. So we also have-- I'll do it in
that same orange color-- we also have 0.0114 grams
of hydrogen. So that's another important
take-away. Now we have moles of carbon,
moles of hydrogen. If we can figure out moles of
oxygen in our original acid then we can figure out the
smallest whole number ratio between them, then we would
be able to find our empirical formula. And to figure out moles of
oxygen we need to first figure out how many grams of
oxygen we have here. And to do that we have a total
of 0.513 grams of this thing. We subtract out how much of it
is carbon, how much of it is hydrogen, and that'll tell
us how much is oxygen. Let me scroll down
a little bit. So if we want to figure out
how much is oxygen, we literally just take 0.5-- let me
write 0.513 grams-- let me just write this-- minus
the grams of carbon. So minus 0.137-- let me do in
that same color so you know where I'm getting these numbers
from-- minus our grams of carbon, 0.137 grams of carbon
minus-- I'll do it in the same orange color--
minus the-- where's our grams of hydrogen? Minus 0.011-- let me, just so
you get the whole idea. We started off with 513 grams
of acid, whatever it is, minus-- let me write it this way
even better-- CxHyOz minus our 0.137 grams of carbon,
minus our 0.0114 grams of hydrogen, will give us our grams
of oxygen that we're dealing with. So let's figure that out. Because everything in that
substance, everything in this mystery acid right here has
to be either carbon, hydrogen, or oxygen. We know how much is carbon
and hydrogen, the leftover has to be oxygen. So we have 0.513 minus 0.137
minus 0.0114 is equal to 0.36-- I'll just round
it-- 0.365. So this is equal to 0.365
grams of oxygen. Let me make sure that
I got that right. Yeah, 0.365 grams of oxygen. So let's figure out how
many moles this is. So we're starting off, or this
mystery acid has 0.365 grams of oxygen, and we want to
convert that to moles. So times-- we want moles of
oxygen in the numerator, grams of oxygen in the denominator,
so they cancel. And we know the molar mass
for oxygen is 16. 16 grams per mole-- right, the
16 has to match up with the grams. So this tells us that
we have 0.365 divided by 16 grams of oxygen. So that's the number
right there. I'll just divide it by 16. So 0.22, I'll say 8
just to round it. Or 0.0228-- let me put it aside,
I want to make sure I don't lose any zeros here. So this tells us that we have
0.0228-- these guys cancel out-- moles of oxygen. We're almost done. We now know, we've now been
able to solve the moles of carbon in our original reactant
because all of this carbon had to come from here. We have our moles of hydrogen. We were able to figure out from
this information because all the hydrogen had
to come from here. And now we know the
moles of oxygen in our original reactant. Our moles, because the leftover
out of this 0.513 grams had to be oxygen, so we
were able to figure out the moles of oxygen. So let's see if we can figure
out the smallest whole number ratio between the three. So the ratio-- and I'll write
the ratio like this-- we have-- let me do it in the same
colors-- we have 0.114 moles of carbon, to--
and then we have 0.0114 moles of hydrogen. These look nice, they're
actually the same numbers. So we're going to have a 1:1
ratio, to-- then we have the oxygen, 0.0228 moles
of oxygen. And this looks like exactly
twice times that. So if we divide everything, if
we divide-- this is a ratio, this is another way of writing
the ratio-- if we divide everything by 0.114-- let
me be careful here. This up here was 0.0114
moles of carbon. So this was a 0 over here. Let me clear it out. I want to be nice
and neat here. So let me clear this. I'm always afraid that I'm going
to lose a zero here. Our moles of carbon
was 0.0114. So let's divide everything by
this amount because that is kind of the smallest number of
moles, and that's exactly that number there. Do that in white. So you divide by 0.0114,
0.0114, 0.0114, and what do you get? What does this turn into? This turns into-- this is just
a 1-- so this is 1 mole of carbon for every-- this is
just a 1 right here-- for every-- I'll do it in the orange
color-- for every 1 mole of hydrogen-- for every--
this is exactly 2, right? 228 divided by 140, that's 2. This is exactly 2-- for every
2 moles of oxygen. So that tells us our empirical
formula for our mystery acid. It must be-- we don't know the
molecular formula, but we know the simplest ratios, whole
number ratios-- it must be 1 carbon for every 1 hydrogen, and
for every 1 carbon and 1 hydrogen we must
have 2 oxygens. So it's CHO2. That is the empirical formula
for our mystery acid. Hopefully you found that fun.