this video to go over solution stoichiometry
so we'll talk about how we use
these dilution units that we learn about in the solutions chapter
in calculations so first of all as a review I just want to go over real quickly
what molarity is, molarity denoted by a
uppercase M is a term that we use to describe the number of moles
of our solute in one liter of solution
alright so molarity is equal to moles of solute
over liter of solution this is molarity and this is what you'll come into contact
with
during solution stoichiometry problems so let's look at a quick example
here we have a 5.5 molar
hydrochloric acid solution so what does that mean
well if it says its 5.5 molar
then that means we have 5.5 moles
of HCl
in one liter
of solution and we can also write this
in a different way we can say that there are
5.5 moles in
one liter of solution now you notice that this
is an equality, an equality is something that we use in conversions
remember an equality we can get two conversion factors from
every equality conversion factor is simply writing an equality in the form of a
fraction
so our two
conversion factors from this equality would be
5.5 moles over one liter
or one liter
over 5.5 moles
now we use these conversion factors to answer a question similar to this
if we're asked to convert 10 liters of a 5.5 molar
HCl solution to moles let's say I wanted to make
10 liters of a hydrochloric acid solution that had a concentration a 5.5
molar how many moles of HCl do I need in order to do this
in order to set up this conversion we find out
look at the problem and
determine what it gives us so it tells us
that we have 10 liters
of solution
this is what we're wanting to make or this is what we're wanting to convert
from we're going to multiply that 10 liters
by a conversion factor that will allow us to convert
to moles
of HCl now since we know that our solution is 5.5 molar
we're gonna pick one of these two conversion factors
in order to complete this conversion so the conversion factor
that we're gonna pick is
this first one, it's going to go right here now why did we pick that conversion
factor
we picked it so that our units cancel out we've got solution
liters of solution here liters of solution here those
variables will cancel out the only unit that we're left with is moles
of HCl and that's what we're trying to convert to over here
so you take you 10 multiply by 5.5
and then divide by one we find that we need
55
moles of HCl in order to make ten liters of our 5.5 molar
HCl solution so this is just the
introduction to using molarity
in the stoichiometric problems let's look at a little bit more difficult
problem too solve
so we have 0.025 liters
of a potassium chloride solution that has a molarity of 0.500 molar
and we're going to add the solution to a silver nitrate
solution according to this reaction
and we want to know what mass of silver chloride
would we form by adding this potassium chloride solution
so what you'll notice here from our balanced equation
is that we have potassium chloride an aqueous soluble salt
were adding that to silver nitrate also an aqueous
soluble salt to form
potassium nitrate, aqueous soluble salt
and silver chloride but silver chloride is insoluble
so it forms an insoluble salt that will precipitate
out a solution what I want to know is if I add this 0.025 liters
of potassium chloride to my silver nitrate
how much silver chloride can I form from this reaction
so the first step is we need to determine a pathway
to answering this question
we're given
a volume of potassium chloride in our question
it tells us that we're adding
0.025 liters
of this potassium chloride solution
it also gives us the molarity of the potassium chloride solution
so we can convert from volume
volume of potassium chloride to moles of potassium chloride
in order to do this conversion the conversion factor that we're going to
use
is similar to the one that we used here it's the conversion factor from that
equality which tells us how many moles of solute
are in a liter of solution so in order to
answer this conversion we use the molarity
of the potassium chloride that it gives us in the problem
however this does not answer the problem
we have to do yet another conversion so from moles of potassium chloride
we can then convert
to moles of silver nitrate
I'm sorry, silver chloride this is what were eventually wanting to get to
if we know the number of moles of potassium chloride we can convert
to the number of moles of silver chloride using our mole to mole ratio
like I said in order to do this conversion we're going to use
the mole to mole ratio which comes from our balanced equation
this gives us the moles of silver chloride that we can produce from this
reaction
but yet once again that does not solve our problem
because the problem ask what mass of silver chloride
we're going to need or we're going to form in this reaction
so we have a third conversion
we can go from moles of silver chloride
to mass
mass of silver chloride anytime we're going from moles of a compound
to mass of that same compound we're gonna use the molar mass
of that compound for this conversion
the molar mass comes directly from the periodic table
in order to determine the molar mass of silver chloride
you're going to get the mass, the molar mass or the atomic mass of silver from the periodic
table
and add that to the atomic mass of chlorine also from the periodic table
so let's start this conversion here
we've got 0.025
liters of our
potassium chloride solution
our first conversion factor
we're going to use molarity of that potassium chloride solution
in order to solve for the number of moles
KCL
so the molarity of the potassium chloride from our original problem
is 0.500 molar
that also means it's 0.500 moles of potassium chloride
per liter
moles per liter you'll see that our liter here
cancels out with liter and the value that we're left with
is moles of potassium chloride from this reaction we find
that a half molar
solution of potassium chloride with a volume of 0.025 liters
gives us
0.0125 moles of potassium chloride
that's our first conversion our second conversion
will start with the answer to our last conversion
and for this conversion
we need to convert from moles of potassium chloride
to moles of silver chloride we've already said in order to do this we're
going to use the mole to mole ratio
coming from our balanced chemical equation here's our balanced chemical
equation
coefficients that go in front of
each reactant and product
is what tells us our mole to mole ratio you'll see here
that one mole
of potassium chloride will produce
one mole of silver chloride so that is our mole to mole ratio
of potassium chloride to silver chloride is a one to one ratio
this is gonna be our second conversion factor here
one mole of potassium chloride
can form one mole
of silver chloride
so in order to solve for this equation we take our 0.0125 moles of
potassium chloride
we multiply it by one and then divide by one
and our units that cancel out are going to be our moles
of potassium chloride so this time we are solving for moles
of silver chloride
when we solve this conversion we find it's equal to
0.0125 moles of silver chloride and then finally our last
conversion let's start with the answer to our previous
0.0125
moles of silver chloride
and
we need to convert moles of silver chloride to
grams of silver chloride
like I said you're going to use the molar mass
for this conversion to go from moles of a compound to mass
of the same compound the molar mass
of silver chloride is 143.4 grams
of silver chloride per mole of silver chloride
this is gonna make up our third conversion factor
143.4 grams
silver chloride per
one mole look at our units that cancel out once again
moles of silver chloride cancel out with moles of silver chloride
the unit we're left with is the unit that we're looking for which is our mass
or our gram value of silver chloride we take our 0.0125 moles
multiply that by 143.4
then divide by one and we find
that we can make 1.79
grams of silver chloride
when we add 0.025 liters of potassium chloride
that has a molarity of 0.500 molar
hopefully this helps with your calculation questions
coming from the solutions chapter dealing with stoichiometry of solutions