Current time:0:00Total duration:18:53

# Stoichiometry example problem 2

## Video transcript

We're told that glucose reacts
with oxygen to give carbon dioxide and water. What mass of oxygen, in grams,
is required to complete the reaction of 25 grams
of glucose? And they also want to know what
masses of carbon dioxide and water are formed? Well let's first just
write the reaction. So they're saying glucose reacts
with oxygen to give carbon dioxide and water. So glucose, C6H12O6, reacts with
oxygen in its molecular form-- there's two
atoms there. The products are carbon
dioxide and water. And you might already
recognize this. This is a stoichiometry
problem. They're saying, hey, we have
25 grams of glucose. How much oxygen is required
to completely react with that glucose? And then how much carbon dioxide
and how much water is going to be produced in grams? That's what stoichiometry
problems are all about. And if you remember from the
last video, the first thing you should always do
is make sure that your equation is balanced. So let's make sure
it's balanced. On the left hand side-- and you
always want to do the most complicated molecules first,
and then do the simplest molecules last, because those
are the easiest ones to balance out. So on the left hand side, here,
I have a 6 carbons. On the entire left hand side. On the right hand side,
I only have 1 carbon. So let me multiply this
over here by 6. And now I have 6 carbons on both
sides of this equation. Let's move to the hydrogens. I have 12 hydrogens
on the left hand side of this equation. The 12 are all sitting right
there in the glucose. How many do I have on
the right hand side? Well I only have 2 hydrogens. So let me multiply
that times 6. And I didn't mess with the
carbons at all, so that shouldn't change anything-- so
now I have 12 hydrogens on both sides of this equation. 12 here, 12 there. And now I can just balance
out the oxygen. I saved that for last because
I just have the oxygen molecule here. That's the easiest
one to balance. So how may oxygens do I have
on the right hand side? I have 6 times 2 here. I have 12 oxygens there. And then I have another
6 oxygens over here. So plus 6 oxygens. I have 18 oxygens on the right
hand side of my equation. So I need to have 18
oxygens on the left hand side of my equation. How many do I have right now? I have 6 oxygens over here. So I'm going to need
12 over here. Right? This is the last thing
I want to mess with. I don't want to put a
coefficient out here. That'll change everything
else. I just want to put a
coefficient here. That'll make everything
balance out. I have 18 on the right
hand side. I already have 6 here. I want to have 12 right here. So let me multiply
this times 6. And now everything
should work out. I have 6 carbons
on both sides. I have 12 hydrogens
on both sides. And I have 18 oxygens. Six here, 12 here. 12 here, 6 here. I have 18 oxygens on both
sides of this equation. Now, the next thing we're going
to want to do is figure out how many moles of the
reactants that they're telling us about that we have. So
they're telling us that we have 25.0 grams of glucose. So let's figure out how many
moles per gram, or how many grams per mole, there are
of a glucose molecule. And since everything here is
dealing with carbons, and oxygens, and hydrogens, let's
look up the atomic weights of all of them. So carbons, hydrogens, and
oxygens are pretty common. So at some point you might
want to memorize their atomic weights. And I want to give proper credit
to the person whose periodic table I'm using. Le Van Han Cedric. I got this off of Wikimedia. It's a creative commons
attribution license. So I want to make sure I
attribute the person who made the periodic table. But we have oxygen. It has an atomic weight
of 15.9999. Usually it's given as 16. But I'll just write
it like that. So oxygen is 15.999. And we're going to
have to figure out carbon, and hydrogen. Let's go back to the
periodic table. We have carbon, has an atomic
weight of 12.011. And hydrogen is 1.0079. So let's do the carbon,
12.011. So the carbon is 12.011. And then the hydrogen-- remember
atomic weights are all the weighted average
of all of the isotopes on earth-- is 1.0079. 1.00-- what was it? Was it one zero or two zeroes? It was two zeroes. 1.0079. So what's the atomic weight of
glucose, given all of that? So what's the atomic
weight of glucose? Let me scroll down
a little bit. So the atomic weight of glucose,
C6H12O6, it's going to be equal to 12.011 times 6. 6 times 12.011. Plus 12 times hydrogen. Just to keep things simple,
actually, let me just make this into 16. let me make this into 12, and
let me make this into 1. That's going to make our
math a lot easier. So let me just do
it like that. Let me clear this. So the atomic weight of glucose
is 6 times the atomic weight of carbon. Six times 12. Six times 12. Plus 12 times the atomic
weight of hydrogen. Plus one times 12, or maybe I
have to write 12 times one, just to be consistent. 12 times one. Plus 6 times the atomic
weight of oxygen. Plus 6 times 16. So what is this? This is equal to 72 plus 12. Plus, 6 times 16
is 60 plus 36. It's 96. And that is-- I'll get the
calculator out-- 72. Plus 12, plus 96,
is equal to 180. So the atomic weight of glucose
is equal to 180. Which tells us, that let me
start doing the problem. We have 25.0 grams of glucose. So I'll just write
grams of C6H12O6. We want to write this in terms
of moles of glucose. So we want to cancel out the
grams. So we want the grams in the denominator. Right? Here it's in the numerator. If we divide by that unit, we're
going to cancel it out. So we want the grams of C6H12O6,
or the grams of glucose in the denominator. And we want the moles of glucose
in the numerator. Because then when we perform
this calculation, that will cancel with that, and we will
be left with moles. So how many grams per mole? Well why don't we just
figure it out? The atomic weight is a 180. So it's 180. If we have Avagadro's number of
these molecules, it's going to have a mass, I should say,
of 180 grams per mole. That's the information we got
by figuring out glucose's atomic weight. So let's just perform
this calculation. These units cancel out
with those units. And so we just take 25 times
one, divided by 180. So this is equal
to 25 over 180. And all were left
with in units is moles C6H12O6, or glucose. And what is 25 five
divided by 180? I should say 25.0 just so we
know that we have three significant digits here. And I don't want to
be too particular about significant digits. Sometimes I'm a little bit
loosey goosey about it. And I was already loosey goosey
about the topic weight. But we'll try to be someplace
in the ballpark. So 25 divided by 180
is equal to 0.139. So we have 0.139 moles of-- I'm
tired of writing, I'm just going to write-- moles
of glucose. Now, we want to figure out how
many moles of oxygen-- or, well, the first step we want
to figure out is for every mole of glucose, how many
moles on oxygen? And we know that. It's 1 to 6. So let's write that down. And that's going to tell us how
many moles of oxygen we're going to need. And, remember, we want to get
rid of these moles of glucose. So we want to write that
in the denominator. So moles of glucose required. For every 1 mole of glucose
required, we figured out when we balanced out the equation,
for every mole of glucose, we need 6 moles of oxygen. We need 6 moles of
O2 required. And then what do we get here? Well we have the moles of
glucose canceling out with the moles of glucose. And then we multiply
the 6 times 0.139. So let's just multiply this
times 6 is equal to 0.833. So we need 0-- that's
too fat-- 0.833 moles of oxygen required. Now, we're almost there, we know
how many moles of oxygen are required, molecular
oxygen. Now we just have to figure out
how many grams that is. So let me just write
it over here. We have 0.833 moles of molecular
oxygen are required. We're going to multiply that. Remember, we want to get
this into grams now. So let's try the moles of
O2 in the denomenator. And we're going to put grams
of O2 in the numerator. So how many grams of O2
are there per mole? Well, we know the atomic
weight oxygen is 16. So the atomic weight of
molecular oxygen, where you have two of these, is equal
to 16 times two. Which is equal to 32. So it's 32 grams per mole. So when we perform the
calculation, the moles of O2 in the numerator cancel out
with moles of O2 in the denominator. And we're just going to have
0.833 times 32 is equal to 26-- I'll just say, seven. So this is equal to 26.7
grams of O2 required. Grams of O2, and I'll
write required in that same exact color. So we finished one part
of the question. We have two more parts left. If we go back to the problem, it
asks us, we already figured out, what mass of oxygen
is required for the complete reaction? We did that part, so we can
write a little check there. We did that first part. But they also ask, what masses
of carbon dioxide and water are formed? So now we need to figure out
carbon dioxide and water. And so we go back here. We know we have 0.139
moles of glucose. We figured that out
in the first part. So 0.139 moles of glucose. And let's do the carbon dioxide
first. We know for every mole of glucose, we're
going to produce 6 moles of carbon dioxide. We see that right there. There's a one out
here implicitly. So for every mole of glucose,
you're going to produce 6 moles of carbon dioxide. So let's write that. So for every one mole of
glucose, you're going to have 6 moles of CO2 produced. We just got that directly
from the equation. And I wrote it this way, instead
of one over the 6, because I wanted it to cancel
out with the moles of glucose. Moles of glucose in the
numerator, moles of glucose in the denominator. So that canceled
out with that. And this is going to be
equal to 6 times 0.139 moles of CO2 produced. And we know what that is. We already multiplied 6
times 0.139 before. This is going to be equal to
0.833 moles of CO2 produced. We've already done
that calculation. And actually we could even--
well, I won't skip too many steps-- but we see over here,
for every one mole of glucose, 6 moles of this, 6 moles of
this, and 6 moles of that. So that's why we had the
0.833 moles of oxygen. And we're also getting the 0.833
moles of carbon dioxide. And, if we did the exact same
thing, we've performed the exact same calculation. For every mole of this
used, you have a mole of that produced. Or every 6 of this, you have
6 of that, 6 of that. So you're also going to have,
by the very same logic, I mean, you could do this
again with the water. You're also going to have 0.833
moles of water produced. Right? For every mole of carbon
dioxide, you have a mole of water. Here it's 6 for 6, but
it's a 1:1 ratio. You could think about
it like that. Now let's figure out how
many grams this is. We have the same number of moles
of carbon dioxide and the same number of
moles of water. But they're going to have
different number of grams. So the actual number of molecules
are the same, but the actual mass of those molecules are
going to be different. So what's carbon dioxide's
atomic weight? Just to remind ourselves. Carbon has atomic weight
of 12, oxygen of 16. So it's going to be 12 plus 16
times two, which is equal to 12 plus 32, which
is equal to 44. So if we are starting off with
0.833 moles of carbon dioxide that are produced, we want to
figure out how many grams per mole-- we're going to put the
moles of carbon dioxide in the denominator because we want
it to cancel with that. And then we have the grams
of carbon dioxide. We know carbon dioxide's
atomic weight is 44. So that means there are 44
grams. If we have a mole of carbon dioxide, if we have
6.022 times 10 to the 23 carbon dioxide molecules. So this is going to
be 44 times 0.833 is-- what do we get? 0.833 times 44 is equal to 36
point-- let's just say, 7. 36.7 grams of CO2 are going
to be produced. And then we're almost done. We just have to figure out how
many grams of water are going to be produced. We know it's the same number
of moles, but how many grams is that? So water's atomic weight is
going to be 2 times 1, right? Because we have to
2 hydrogens. Plus 16. Which is equal to 18. So if we're starting off with
0.833 moles, or if we're producing 0.833 moles of water,
we just multiply that times how many grams
per mole of water? So 1 mole of H2O-- once again
we want this in the denominator so it cancels out
with this in the numerator. 1 mole of H20 has a mass of 18
grams. Or 18 grams of H20 for every mole of H20. So this cancels with that and
we get-- get the calculator out-- 18 times 0.833
is equal to 14.994. So we can just round
that to 15.0. So we have 15.0 grams
of water produced. And we're done! We are done. We figured out that if we start
off with that 25 grams of glucose, like they told us
at the beginning of the problem, we're going to require
26.7 grams of oxygen to react with it, we're going
to produce 36.7 grams of carbon dioxide, and
15 grams of water.