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.
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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
