WEBVTT

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We're told that glucose reacts
with oxygen to give carbon

00:00:05.160 --> 00:00:07.240
dioxide and water.

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What mass of oxygen, in grams,
is required to complete the

00:00:11.190 --> 00:00:15.950
reaction of 25 grams
of glucose?

00:00:15.950 --> 00:00:18.990
And they also want to know what
masses of carbon dioxide

00:00:18.990 --> 00:00:20.405
and water are formed?

00:00:20.405 --> 00:00:23.160
Well let's first just
write the reaction.

00:00:23.160 --> 00:00:27.540
So they're saying glucose reacts
with oxygen to give

00:00:27.540 --> 00:00:30.030
carbon dioxide and water.

00:00:30.030 --> 00:00:41.560
So glucose, C6H12O6, reacts with
oxygen in its molecular

00:00:41.560 --> 00:00:43.595
form-- there's two
atoms there.

00:00:43.595 --> 00:00:46.630


00:00:46.630 --> 00:00:51.135
The products are carbon
dioxide and water.

00:00:51.135 --> 00:00:54.200


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And you might already
recognize this.

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This is a stoichiometry
problem.

00:00:58.250 --> 00:01:01.260
They're saying, hey, we have
25 grams of glucose.

00:01:01.260 --> 00:01:03.370
How much oxygen is required
to completely

00:01:03.370 --> 00:01:04.780
react with that glucose?

00:01:04.780 --> 00:01:07.630
And then how much carbon dioxide
and how much water is

00:01:07.630 --> 00:01:09.850
going to be produced in grams?

00:01:09.850 --> 00:01:12.720
That's what stoichiometry
problems are all about.

00:01:12.720 --> 00:01:16.250
And if you remember from the
last video, the first thing

00:01:16.250 --> 00:01:18.210
you should always do
is make sure that

00:01:18.210 --> 00:01:19.550
your equation is balanced.

00:01:19.550 --> 00:01:21.600
So let's make sure
it's balanced.

00:01:21.600 --> 00:01:24.960
On the left hand side-- and you
always want to do the most

00:01:24.960 --> 00:01:27.830
complicated molecules first,
and then do the simplest

00:01:27.830 --> 00:01:30.550
molecules last, because those
are the easiest ones to

00:01:30.550 --> 00:01:32.010
balance out.

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So on the left hand side, here,
I have a 6 carbons.

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On the entire left hand side.

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On the right hand side,
I only have 1 carbon.

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So let me multiply this
over here by 6.

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And now I have 6 carbons on both
sides of this equation.

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Let's move to the hydrogens.

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I have 12 hydrogens
on the left hand

00:01:52.880 --> 00:01:54.060
side of this equation.

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The 12 are all sitting right
there in the glucose.

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How many do I have on
the right hand side?

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Well I only have 2 hydrogens.

00:02:00.910 --> 00:02:04.120
So let me multiply
that times 6.

00:02:04.120 --> 00:02:06.210
And I didn't mess with the
carbons at all, so that

00:02:06.210 --> 00:02:09.650
shouldn't change anything-- so
now I have 12 hydrogens on

00:02:09.650 --> 00:02:10.759
both sides of this equation.

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12 here, 12 there.

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And now I can just balance
out the oxygen.

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I saved that for last because
I just have the oxygen

00:02:17.030 --> 00:02:17.650
molecule here.

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That's the easiest
one to balance.

00:02:19.450 --> 00:02:21.470
So how may oxygens do I have
on the right hand side?

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I have 6 times 2 here.

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I have 12 oxygens there.

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And then I have another
6 oxygens over here.

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So plus 6 oxygens.

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I have 18 oxygens on the right
hand side of my equation.

00:02:35.310 --> 00:02:37.400
So I need to have 18
oxygens on the left

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hand side of my equation.

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How many do I have right now?

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I have 6 oxygens over here.

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So I'm going to need
12 over here.

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Right?

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This is the last thing
I want to mess with.

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I don't want to put a
coefficient out here.

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That'll change everything
else.

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I just want to put a
coefficient here.

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That'll make everything
balance out.

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I have 18 on the right
hand side.

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I already have 6 here.

00:03:00.440 --> 00:03:02.580
I want to have 12 right here.

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So let me multiply
this times 6.

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And now everything
should work out.

00:03:06.470 --> 00:03:08.580
I have 6 carbons
on both sides.

00:03:08.580 --> 00:03:11.640
I have 12 hydrogens
on both sides.

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And I have 18 oxygens.

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Six here, 12 here.

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12 here, 6 here.

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I have 18 oxygens on both
sides of this equation.

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Now, the next thing we're going
to want to do is figure

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out how many moles of the
reactants that they're telling

00:03:26.290 --> 00:03:29.150
us about that we have. So
they're telling us that we

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have 25.0 grams of glucose.

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So let's figure out how many
moles per gram, or how many

00:03:40.780 --> 00:03:43.690
grams per mole, there are
of a glucose molecule.

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And since everything here is
dealing with carbons, and

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oxygens, and hydrogens, let's
look up the atomic weights of

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all of them.

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So carbons, hydrogens, and
oxygens are pretty common.

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So at some point you might
want to memorize

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their atomic weights.

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And I want to give proper credit
to the person whose

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periodic table I'm using.

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Le Van Han Cedric.

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I got this off of Wikimedia.

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It's a creative commons
attribution license.

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So I want to make sure I
attribute the person who made

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the periodic table.

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But we have oxygen.

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It has an atomic weight
of 15.9999.

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Usually it's given as 16.

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But I'll just write
it like that.

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So oxygen is 15.999.

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And we're going to
have to figure

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out carbon, and hydrogen.

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Let's go back to the
periodic table.

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We have carbon, has an atomic
weight of 12.011.

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And hydrogen is 1.0079.

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So let's do the carbon,
12.011.

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So the carbon is 12.011.

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And then the hydrogen-- remember
atomic weights are

00:05:02.900 --> 00:05:05.380
all the weighted average
of all of the isotopes

00:05:05.380 --> 00:05:10.050
on earth-- is 1.0079.

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1.00-- what was it?

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Was it one zero or two zeroes?

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It was two zeroes.

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1.0079.

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So what's the atomic weight of
glucose, given all of that?

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So what's the atomic
weight of glucose?

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Let me scroll down
a little bit.

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So the atomic weight of glucose,
C6H12O6, it's going

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to be equal to 12.011 times 6.

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6 times 12.011.

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Plus 12 times hydrogen.

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Just to keep things simple,
actually, let me just make

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this into 16.

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let me make this into 12, and
let me make this into 1.

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That's going to make our
math a lot easier.

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So let me just do
it like that.

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Let me clear this.

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So the atomic weight of glucose
is 6 times the atomic

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weight of carbon.

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Six times 12.

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Six times 12.

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Plus 12 times the atomic
weight of hydrogen.

00:06:17.720 --> 00:06:22.380
Plus one times 12, or maybe I
have to write 12 times one,

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just to be consistent.

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12 times one.

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Plus 6 times the atomic
weight of oxygen.

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Plus 6 times 16.

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So what is this?

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This is equal to 72 plus 12.

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Plus, 6 times 16
is 60 plus 36.

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It's 96.

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And that is-- I'll get the
calculator out-- 72.

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Plus 12, plus 96,
is equal to 180.

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So the atomic weight of glucose
is equal to 180.

00:07:03.870 --> 00:07:08.940
Which tells us, that let me
start doing the problem.

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We have 25.0 grams of glucose.

00:07:13.790 --> 00:07:19.780
So I'll just write
grams of C6H12O6.

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We want to write this in terms
of moles of glucose.

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So we want to cancel out the
grams. So we want the grams in

00:07:28.410 --> 00:07:29.640
the denominator.

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Right?

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Here it's in the numerator.

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If we divide by that unit, we're
going to cancel it out.

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So we want the grams of C6H12O6,
or the grams of

00:07:42.390 --> 00:07:43.730
glucose in the denomenator.

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And we want the moles of glucose
in the numerator.

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Because then when we perform
this calculation, that will

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cancel with that, and we will
be left with moles.

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So how many grams per mole?

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Well why don't we just
figure it out?

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The atomic weight is a 180.

00:08:02.800 --> 00:08:03.970
So it's 180.

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If we have Avagadro's number of
these molecules, it's going

00:08:08.870 --> 00:08:15.140
to have a mass, I should say,
of 180 grams per mole.

00:08:15.140 --> 00:08:19.150
That's the information we got
by figuring out glucose's

00:08:19.150 --> 00:08:19.900
atomic weight.

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So let's just perform
this calculation.

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These units cancel out
with those units.

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And so we just take 25 times
one, divided by 180.

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So this is equal
to 25 over 180.

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And all were left
with in units is

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moles C6H12O6, or glucose.

00:08:42.806 --> 00:08:46.330
And what is 25 five
divided by 180?

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I should say 25.0 just so we
know that we have three

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significant digits here.

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And I don't want to
be too particular

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about significant digits.

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Sometimes I'm a little bit
loosey goosey about it.

00:08:58.080 --> 00:09:00.370
And I was already loosey goosey
about the topic weight.

00:09:00.370 --> 00:09:02.160
But we'll try to be someplace
in the ballpark.

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So 25 divided by 180
is equal to 0.139.

00:09:12.420 --> 00:09:20.060
So we have 0.139 moles of-- I'm
tired of writing, I'm just

00:09:20.060 --> 00:09:23.100
going to write-- moles
of glucose.

00:09:23.100 --> 00:09:28.180
Now, we want to figure out how
many moles of oxygen-- or,

00:09:28.180 --> 00:09:31.160
well, the first step we want
to figure out is for every

00:09:31.160 --> 00:09:34.150
mole of glucose, how many
moles on oxygen?

00:09:34.150 --> 00:09:34.860
And we know that.

00:09:34.860 --> 00:09:36.400
It's 1 to 6.

00:09:36.400 --> 00:09:37.660
So let's write that down.

00:09:37.660 --> 00:09:40.216
And that's going to tell us how
many moles of oxygen we're

00:09:40.216 --> 00:09:41.060
going to need.

00:09:41.060 --> 00:09:43.810
And, remember, we want to get
rid of these moles of glucose.

00:09:43.810 --> 00:09:45.565
So we want to write that
in the denominator.

00:09:45.565 --> 00:09:49.870


00:09:49.870 --> 00:09:56.490
So moles of glucose required.

00:09:56.490 --> 00:10:00.600
For every 1 mole of glucose
required, we figured out when

00:10:00.600 --> 00:10:03.680
we balanced out the equation,
for every mole of glucose, we

00:10:03.680 --> 00:10:06.510
need 6 moles of oxygen.

00:10:06.510 --> 00:10:13.060
We need 6 moles of
O2 required.

00:10:13.060 --> 00:10:16.360
And then what do we get here?

00:10:16.360 --> 00:10:20.190
Well we have the moles of
glucose canceling out with the

00:10:20.190 --> 00:10:21.750
moles of glucose.

00:10:21.750 --> 00:10:26.180
And then we multiply
the 6 times 0.139.

00:10:26.180 --> 00:10:32.850
So let's just multiply this
times 6 is equal to 0.833.

00:10:32.850 --> 00:10:38.770
So we need 0-- that's
too fat-- 0.833

00:10:38.770 --> 00:10:43.330
moles of oxygen required.

00:10:43.330 --> 00:10:47.720
Now, we're almost there, we know
how many moles of oxygen

00:10:47.720 --> 00:10:49.610
are required, molecular
oxygen.

00:10:49.610 --> 00:10:52.130
Now we just have to figure out
how many grams that is.

00:10:52.130 --> 00:10:53.820
So let me just write
it over here.

00:10:53.820 --> 00:11:01.850
We have 0.833 moles of molecular
oxygen are required.

00:11:01.850 --> 00:11:04.010
We're going to multiply that.

00:11:04.010 --> 00:11:07.470
Remember, we want to get
this into grams now.

00:11:07.470 --> 00:11:10.570
So let's try the moles of
O2 in the denomenator.

00:11:10.570 --> 00:11:13.350


00:11:13.350 --> 00:11:17.090
And we're going to put grams
of O2 in the numerator.

00:11:17.090 --> 00:11:21.300
So how many grams of O2
are there per mole?

00:11:21.300 --> 00:11:24.980
Well, we know the atomic
weight oxygen is 16.

00:11:24.980 --> 00:11:27.460
So the atomic weight of
molecular oxygen, where you

00:11:27.460 --> 00:11:33.520
have two of these, is equal
to 16 times two.

00:11:33.520 --> 00:11:36.400
Which is equal to 32.

00:11:36.400 --> 00:11:42.640
So it's 32 grams per mole.

00:11:42.640 --> 00:11:46.400
So when we perform the
calculation, the moles of O2

00:11:46.400 --> 00:11:48.950
in the numerator cancel out
with moles of O2 in the

00:11:48.950 --> 00:11:49.680
denomenator.

00:11:49.680 --> 00:11:59.750
And we're just going to have
0.833 times 32 is equal to

00:11:59.750 --> 00:12:02.940
26-- I'll just say, seven.

00:12:02.940 --> 00:12:09.570
So this is equal to 26.7
grams of O2 required.

00:12:09.570 --> 00:12:13.650
Grams of O2, and I'll
write required in

00:12:13.650 --> 00:12:15.650
that same exact collar.

00:12:15.650 --> 00:12:17.940
So we finished one part
of the question.

00:12:17.940 --> 00:12:20.830
We have two more parts left.

00:12:20.830 --> 00:12:25.180
If we go back to the problem, it
asks us, we already figured

00:12:25.180 --> 00:12:27.680
out, what mass of oxygen
is required for

00:12:27.680 --> 00:12:28.470
the complete reaction?

00:12:28.470 --> 00:12:31.500
We did that part, so we can
write a little check there.

00:12:31.500 --> 00:12:32.500
We did that first part.

00:12:32.500 --> 00:12:36.690
But they also ask, what masses
of carbon dioxide and water

00:12:36.690 --> 00:12:37.490
are formed?

00:12:37.490 --> 00:12:41.410
So now we need to figure out
carbon dioxide and water.

00:12:41.410 --> 00:12:43.800
And so we go back here.

00:12:43.800 --> 00:12:47.930
We know we have 0.139
moles of glucose.

00:12:47.930 --> 00:12:50.550
We figured that out
in the first part.

00:12:50.550 --> 00:12:58.350
So 0.139 moles of glucose.

00:12:58.350 --> 00:13:05.140
And let's do the carbon dioxide
first. We know for

00:13:05.140 --> 00:13:09.940
every mole of glucose, we're
going to produce 6 moles of

00:13:09.940 --> 00:13:10.770
carbon dioxide.

00:13:10.770 --> 00:13:11.670
We see that right there.

00:13:11.670 --> 00:13:13.560
There's a one out
here implicitly.

00:13:13.560 --> 00:13:17.380
So for every mole of glucose,
you're going to produce 6

00:13:17.380 --> 00:13:19.290
moles of carbon dioxide.

00:13:19.290 --> 00:13:21.130
So let's write that.

00:13:21.130 --> 00:13:29.040
So for every one mole of
glucose, you're going to have

00:13:29.040 --> 00:13:34.660
6 moles of CO2 produced.

00:13:34.660 --> 00:13:36.780
We just got that directly
from the equation.

00:13:36.780 --> 00:13:39.710
And I wrote it this way, instead
of one over the 6,

00:13:39.710 --> 00:13:41.980
because I wanted it to cancel
out with the moles of glucose.

00:13:41.980 --> 00:13:45.050
Moles of glucose in the
numerator, moles of glucose in

00:13:45.050 --> 00:13:46.920
the denominator.

00:13:46.920 --> 00:13:48.870
So that canceled
out with that.

00:13:48.870 --> 00:13:55.960
And this is going to be
equal to 6 times 0.139

00:13:55.960 --> 00:14:01.840
moles of CO2 produced.

00:14:01.840 --> 00:14:04.810


00:14:04.810 --> 00:14:05.710
And we know what that is.

00:14:05.710 --> 00:14:09.040
We already multiplied 6
times 0.139 before.

00:14:09.040 --> 00:14:17.300
This is going to be equal to
0.833 moles of CO2 produced.

00:14:17.300 --> 00:14:19.790
We've already done
that calculation.

00:14:19.790 --> 00:14:22.960
And actually we could even--
well, I won't skip too many

00:14:22.960 --> 00:14:29.120
steps-- but we see over here,
for every one mole of glucose,

00:14:29.120 --> 00:14:32.890
6 moles of this, 6 moles of
this, and 6 moles of that.

00:14:32.890 --> 00:14:38.850
So that's why we had the
0.833 moles of oxygen.

00:14:38.850 --> 00:14:43.570
And we're also getting the 0.833
moles of carbon dioxide.

00:14:43.570 --> 00:14:46.300
And, if we did the exact same
thing, we've performed the

00:14:46.300 --> 00:14:47.480
exact same calculation.

00:14:47.480 --> 00:14:49.515
For every mole of this
used, you have a

00:14:49.515 --> 00:14:50.580
mole of that produced.

00:14:50.580 --> 00:14:53.180
Or every 6 of this, you have
6 of that, 6 of that.

00:14:53.180 --> 00:14:55.510
So you're also going to have,
by the very same logic, I

00:14:55.510 --> 00:14:57.470
mean, you could do this
again with the water.

00:14:57.470 --> 00:15:06.980
You're also going to have 0.833
moles of water produced.

00:15:06.980 --> 00:15:07.270
Right?

00:15:07.270 --> 00:15:09.330
For every mole of carbon
dioxide, you

00:15:09.330 --> 00:15:10.100
have a mole of water.

00:15:10.100 --> 00:15:12.620
Here it's 6 for 6, but
it's a 1:1 ratio.

00:15:12.620 --> 00:15:14.530
You could think about
it like that.

00:15:14.530 --> 00:15:18.710
Now let's figure out how
many grams this is.

00:15:18.710 --> 00:15:20.920
We have the same number of moles
of carbon dioxide and

00:15:20.920 --> 00:15:23.330
the same number of
moles of water.

00:15:23.330 --> 00:15:26.400
But they're going to have
different number of grams. So

00:15:26.400 --> 00:15:30.340
the actual number of molecules
are the same, but the actual

00:15:30.340 --> 00:15:32.480
mass of those molecules are
going to be different.

00:15:32.480 --> 00:15:39.950
So what's carbon dioxide's
atomic weight?

00:15:39.950 --> 00:15:41.410
Just to remind ourselves.

00:15:41.410 --> 00:15:45.520
Carbon has atomic weight
of 12, oxygen of 16.

00:15:45.520 --> 00:15:50.810
So it's going to be 12 plus 16
times two, which is equal to

00:15:50.810 --> 00:15:55.510
12 plus 32, which
is equal to 44.

00:15:55.510 --> 00:16:06.370
So if we are starting off with
0.833 moles of carbon dioxide

00:16:06.370 --> 00:16:09.550
that are produced, we want to
figure out how many grams per

00:16:09.550 --> 00:16:12.320
mole-- we're going to put the
moles of carbon dioxide in the

00:16:12.320 --> 00:16:15.220
denominator because we want
it to cancel with that.

00:16:15.220 --> 00:16:18.360
And then we have the grams
of carbon dioxide.

00:16:18.360 --> 00:16:21.090
We know carbon dioxide's
atomic weight is 44.

00:16:21.090 --> 00:16:24.510
So that means there are 44
grams. If we have a mole of

00:16:24.510 --> 00:16:27.620
carbon dioxide, if we have
6.022 times 10 to the 23

00:16:27.620 --> 00:16:29.680
carbon dioxide molecules.

00:16:29.680 --> 00:16:34.590
So this is going to
be 44 times 0.833

00:16:34.590 --> 00:16:38.130
is-- what do we get?

00:16:38.130 --> 00:16:46.500
0.833 times 44 is equal to 36
point-- let's just say, 7.

00:16:46.500 --> 00:16:56.220
36.7 grams of CO2 are going
to be produced.

00:16:56.220 --> 00:16:57.530
And then we're almost done.

00:16:57.530 --> 00:17:00.380
We just have to figure out how
many grams of water are going

00:17:00.380 --> 00:17:01.040
to be produced.

00:17:01.040 --> 00:17:02.590
We know it's the same number
of moles, but how

00:17:02.590 --> 00:17:04.329
many grams is that?

00:17:04.329 --> 00:17:11.470
So water's atomic weight is
going to be 2 times 1, right?

00:17:11.470 --> 00:17:13.460
Because we have to
2 hydrogens.

00:17:13.460 --> 00:17:15.839
Plus 16.

00:17:15.839 --> 00:17:18.500
Which is equal to 18.

00:17:18.500 --> 00:17:26.160
So if we're starting off with
0.833 moles, or if we're

00:17:26.160 --> 00:17:33.680
producing 0.833 moles of water,
we just multiply that

00:17:33.680 --> 00:17:36.920
times how many grams
per mole of water?

00:17:36.920 --> 00:17:41.200
So 1 mole of H2O-- once again
we want this in the

00:17:41.200 --> 00:17:45.110
denominator so it cancels out
with this in the numerator.

00:17:45.110 --> 00:17:51.990
1 mole of H20 has a mass of 18
grams. Or 18 grams of H20 for

00:17:51.990 --> 00:17:53.700
every mole of H20.

00:17:53.700 --> 00:17:59.510
So this cancels with that and
we get-- get the calculator

00:17:59.510 --> 00:18:13.180
out-- 18 times 0.833
is equal to 14.994.

00:18:13.180 --> 00:18:16.630
So we can just round
that to 15.0.

00:18:16.630 --> 00:18:25.230
So we have 15.0 grams
of water produced.

00:18:25.230 --> 00:18:27.200
And we're done!

00:18:27.200 --> 00:18:28.520
We are done.

00:18:28.520 --> 00:18:34.910
We figured out that if we start
off with that 25 grams

00:18:34.910 --> 00:18:37.180
of glucose, like they told us
at the beginning of the

00:18:37.180 --> 00:18:42.600
problem, we're going to require
26.7 grams of oxygen

00:18:42.600 --> 00:18:46.740
to react with it, we're going
to produce 36.7 grams of

00:18:46.740 --> 00:18:52.180
carbon dioxide, and
15 grams of water.