Hang tight while we fetch the video data and transcripts. This only takes a moment.
Connecting to YouTube player…
Fetching transcript data…
We’ll display the transcript, summary, and all view options as soon as everything loads.
Next steps
Loading transcript tools…
Ch 8.3 - Stoich of Gases, Mixtures, and Reactions | General Chemistry | YouTubeToText
YouTube Transcript: Ch 8.3 - Stoich of Gases, Mixtures, and Reactions
Skip watching entire videos - get the full transcript, search for keywords, and copy with one click.
Share:
Video Transcript
Video Summary
Summary
Core Theme
This content explains how to apply the ideal gas law to calculate gas densities and molar masses, and introduces concepts for handling gas mixtures, including Dalton's Law of Partial Pressures and reactions involving gases.
Mind Map
Click to expand
Click to explore the full interactive mind map • Zoom, pan, and navigate
the topic of this video is stoichiometry
of gaseous substances
mixtures and reactions the learning
objectives are on the screen so go ahead
and pause the video now and write those down
down
um so to address the first learning
objective which is to use the ideal gas
law to compute gas densities and molar masses
masses
i'm going to start there and then we'll
do a quick practice problem and then
proceed so recall that the ideal gas law
is pv equals nrt
kind of just rolls off the tongue um and so
so
what we can do here is also contemplate
what is the
definition of density which is mass over
volume right grams per milliliter is a common
common
form of density that we've encountered
for liquids before
um and so it's mass over volume and so
what this
indicates is that we can actually um use
the ideal gas law to
express the density of the gas using
things like
pressure and the ideal gas constant
temperature and mole amount for example
and i left that volume because volume is
the the key variable that links these two
two
equations so what that means is if we
solve the ideal gas law
for volume volume equals nr t
over p what we can do is then we can also
also
we can substitute that volume in for the volume
volume
for density for a gas and what we end up
with is something that looks like this
density of a gas is equal to the mass of
the gas lower case m multiplied by the
pressure divided by the mole amount of
the gas multiplied by the ideal gas
constant multiplied by temperature in kelvin
kelvin
so what we can do is we can also
separate out the mass
over the mole amount something like
grams per mole
if we think about that as its own term
multiplied by pressure over
rt p over rt this m over n
it's really just grams per mole those
are the units here
and those are the units of molar mass
right so this is also defined as
uh molar mass so quick you can see how
if we if we express
if we merge density with the ideal gas law
law
we can get molar mass density is
specific to a gas okay so
this is where now we're modifying the
ideal gas law making it
turning it into expressions that are
specific to gases the ideal gas law
itself is not specific to any one gas as
long as
it's applicable to any gas that is an
ideal gas
so we can call this capital m uh uh
kind of italicized i'm trying to
distinguish that from
non-italicized capital m which we use
for molarity
so then um the key equations here are
density equals
um the molar mass that's what i'm
talking about here
multiplied by you might want to uh you
know mark that in your notes this is the
molar mass that we're talking about here
not molarity
and we can also do a rearrangement if
you wanted to but
i'm just going to skip ahead to show you
that we can also isolate molar mass
maybe i'll just draw the arrow
over here we can isolate molar mass and express
express
this as a function of the mass of a
sample of gas times the ideal gas
constant r
multiplied by t all over p
v so these are some useful expressions
when thinking about
calculating using density or molar mass
we're trying to
so now that we have the ability to use
molar mass and density
in conjunction with the ideal gas law we
can start to solve
some problems that are really
interesting right it's actually quite
powerful to have these relationships
so here's a practice problem cycle
propane a gas once used with oxygen as a
general anesthetic
that's kind of scary is composed of 85.7
carbon and 14.3 hydrogen uh by mass
find the empirical formula okay so if
1.56 grams of cyclo
propane occupies a volume of 1.00 liters
at 0.984 atmospheres and 50 degrees celsius
celsius
what is the molecular formula this is a
lot of information
but um we can we can use some of the
some of our
skills from prior chapters as well as
what we just derived
to tackle this the first is just an
empirical formula problem we've seen
that before so you assume 100 grams
total and that allows you to quickly
convert from 85.7 percent to 85.7 grams
of carbon
and then we can use the molar mass of carbon
carbon
to calculate how many moles here 7.136 moles
moles
we can do the same thing for hydrogen um
and what we will find
see sorry grams down here i'm trying to
get ahead of myself
grams per 1 mole and that gives us 14.158
14.158
if you normalize these two moles
dividing by both by 7.136 what you'll
one carbon and two hydrogens
for the empirical formula so the
empirical formula is ch2
if you take the molar mass of
of ch2 as the empirical unit
this is 14.03
grams per empirical unit
okay so i'm referring to this in
parentheses as the empirical unit
so that's the first part of the problem
we've done that before that's why i kind of
of
blasted through it now we can use the
expression that we just derived which is the
the
molar mass is equal to the mass
times r times t over pv
so what we can do is we can actually
calculate the molar mass here
and compare it to the empirical mass the the
the
14.03 value if they're the same the
empirical mass is the the
uh uh actual formula
from that we need to to that's the final
answer the empirical formula is the
molecular formula
if it's if the molar mass is greater
than we need to adjust our formula so that
that
from the empirical to the molecular formula
formula
so um everything is given to us here we
have 1.56 grams is the mass
we can use the ideal gas constant we do
have to convert temperature
and we have to um i think that's the
only thing like to convert is temperature
temperature
so i'm going to rewrite all i'm going to
write this all out so the problem stated
1.56 grams
i'm using the uh the gas constant 0.0821
this is
rounded from the one that we used before
liter atmospheres
and also the temperature is going to be
50 degrees celsius plus 273
gives us 323 kelvin we can we cannot
leave it in degree celsius
the atmosphere pressure is 0.984
atmospheres and the volume was 1.00 liters
liters
you'll notice here uh that the
atmospheres cancel
out the volume in liters cancels out the
temperature in kelvin cancels out
but grams and moles are left and
specifically it's grams per mole
so that is that's good because that's
the unit of molar mass and we calculate 42.0
42.0
grams per mole so clearly that is larger
than the empirical
mass unit so what you can do is take 42.0
42.0
grams per mole divide that by 14.03
grams per mole
and what you'll get is three so what we
have to do is multiply our
empirical formula by 3
to get the molecular formula c3h6
so you see this is a powerful analysis method
method
the next thing that we're going to talk
about is actually gas mixtures and specifically
specifically
dalton's law of
partial pressures
and the the the main concept of the law
is that um if you have a gas mixture this
this
the total pressure is the sum of all the individual
individual
pressures exerted by each individual gas
so you imagine if we had
these three gases shown here in their
own individual containers at those
pressures if we
put them all into one container the new
total pressure is the sum of all of
those individual pressures
so uh mathematically we express this by
saying that the total pressure p
total is equal to the partial pressure so
so
we we denote partial pressure as a
subscript with whatever that gas is in
this case
a just as a placeholder plus the partial
pressure of b plus the partial pressure
of c plus the partial pressure
of whatever is oops this is supposed to
be dot
um and essentially what this with the
the the generic form is the
summation of all partial pressures in
the system so p
i okay so again these are the pressures exerted
exerted
by each individual gas okay
also worth noting is so how do we um if
we have a mixture how do we denote the
partial pressure
if we only know the total pressure we
accomplish that with something
uh called the mole fraction so it's chi sub
sub
a multiplied by p total so this
is sort of like a it is the fraction the
mole fraction
of the gas over the the um
uh which represents the the amount of the
the
the gas in moles over the total moles of
gases in the
in the mixture so what does that look
like chi
sub a is actually equal to the moles of a
a
over the total number of moles okay just
expressed as a fraction so all of the
uh mole fractions of all the
gases in a mixture should add up to one
okay so let's maybe
look at a quick problem that deals with partial
partial
partial pressure in the interest of time
i'm only going to do part a of this
problem i'll leave part b to you
a gas mixture is used for anesthesia
used for anesthesia contains 2.83 moles
of oxygen
and 8.41 moles of nitrous oxide the
total pressure of the mixture is 192 kilopascals
kilopascals
so we're going to solve for a what what
are the mole fractions of o2
and n2o and then i'll leave you to do
the um
uh actually sorry i'm not going to do
this part a i'm going to do the
calculations for
oxygen so a and b but just for oxygen
and then you do
nitrous oxide following that logic so
for o2 the
mole fraction of o2 is going to equal
the moles of o2
over the total moles so this is going to
be equal to 2.83
moles over and i'm going to add 2.83 plus
plus
8.41 that's those are the mole amounts
provided this gives us 0.252
mole fraction for oxygen so pretty much
what that means is that 25.2 percent of
the total pressure is responsible
um uh is due to oxygen
so so with the way that we solve that
mathematically is we say the partial
pressure of o2 is equal to
the mole fraction of o2 times the total pressure
pressure
um and so it should be again 25.2
percent but
so we take 0.252 multiplied by
192 kilopascals and that gives us
48.4 kilopascals as the partial pressure
of oxygen
so you can go ahead and do nitrous oxide
on your own so
in some for some reactions when you want
to isolate a gas um it's generated from
an aqueous solution
so if we're trying to measure the
pressure uh uh
of the gases collected above an aqueous
solution we also have to account for the
the amount of water that's actually also
in the gas phase that exerts some type
of pressure
the the pressure over a liquid is the
vapor pressure
so whenever you do calculations um
collected over water you have to account
for the vapor pressure
we just treat it just like in the
dalton's law of partial pressures
the the vapor pressure will be p sub h2o
it is temperature dependent so there's a
table that you can look up um
these these are values that you will not
have to memorize but you'll see that as
the temperature
if we look at vapor pressure of water as
a function of temperature the vapor pressure
pressure
as the temperature increases so too will
the vapor pressure
increase so you typically have to look
up what the vapor pressure of water is
at your given reaction temperature
to then accurately account for the the
partial pressure of water vapor
in the total pressure of of your
okay and then the final thing that i
want to talk about
is um reaction uh uh of gases
so um here i've just copied and pasted
the the um the problem itself and the
solution i'm just gonna walk through it
the concept here is that this is to deal
with the fact that one mole of gas
if it's an ideal gas at stp
will occupy 22.4 liters of volume
so what that means is that it doesn't
matter what the identity of the gas is
if it's an ideal gas
um the volume is indicative of the the
mole amount right there's there's
there's a direct correlation between
volume and mole amount
so what that means is for stoichiometric
purposes you can just use the
the volume ratio of a given reaction
to um to uh
solve the problem with dimensional
analysis so here it's telling us what volume
volume
of o2 gas measured at 25 degrees celsius
and 760 uh torr
is required to react with 2.7 liters of propane
propane
measured under the same conditions of
temperature and pressure so actually you
notice this
this is not even stp necessarily um it's
just that we know
if it's an ideal gas and and and the uh
it's the same temperature and pressure
then the mole amount is reflected in the volume
volume
so the solution here actually you have
to write out the balanced
equation it's a combustion reaction um and
and
when it's balanced we see here instead
of thinking about uh
just mole amounts you can think um
directly in units of of volume
instead of one mole reacts with with
with five moles of oxygen we can say one volume
volume
of this gas c3h8 propane reacts with five
five
volumes of oxygen to produce three
volumes of carbon dioxide and four
volumes of water vapor
even though it says liquid there but so
what we can say here then is if you
wanted to solve for
the volume of oxygen starting from 2.7
liters of
of propane you do not have to go through
mole amount because
that's kind of a redundant thing the
volume is indicative of the mole amount
and vice versa for ideal gases so what
you do is you start with
the liter quantity for propane
and you use the balanced chemical
reaction to say that for every one liter
of propane that reacts
we it will consume five liters of
uh or five volumes of oxygen so here
we're sticking with units of liters
because that's the starting point
and then that easily allows us to
calculate the volume of o2 we do not
have to go through moles
it's sort of a shortcut process to to compare
compare
how many volumes of one gas will react
with it how many volumes of another gas
to produce how many volumes of
Click on any text or timestamp to jump to that moment in the video
Share:
Most transcripts ready in under 5 seconds
One-Click Copy125+ LanguagesSearch ContentJump to Timestamps
Paste YouTube URL
Enter any YouTube video link to get the full transcript
Transcript Extraction Form
Most transcripts ready in under 5 seconds
Get Our Chrome Extension
Get transcripts instantly without leaving YouTube. Install our Chrome extension for one-click access to any video's transcript directly on the watch page.
