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This content explains the principles of gravimetric analysis, a quantitative chemical analysis technique that uses mass measurements of separated components to determine the composition of a sample. It details two specific methods: precipitation analysis and combustion analysis, illustrating each with stoichiometric calculations.
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in this video we're continuing to learn
about quantitative chemical analysis
here we're going to describe the
fundamental aspects of gravimetric
analysis and we're going to perform
stoichiometric calculations using
typical gravimetric data so graphometric
analysis is the type of analysis where
we have a sample that is our analyte we
subject it so some sort of treatment
that causes a change or reaction and we
can then separate it from other
components of the sample if we make Mass
measurements of this
separated component then we can find out
information about the original sample
right how much of it we have or what its
concentration would be so as one example
we could precipitate something from a
solution get the mass of that
precipitate and then be able to use
stoichiometric calculations to figure
out how many moles of a given component
were precipitated
and then we can use that information to
figure out for example the concentration
of the original unknown so in this
example we have a solid mixture that
contains magnesium sulfate that solid
mixture is then dissolved in water and
treated with an excess of barium nitrate
and this causes a precipitation of
barium sulfate so we can see that our
original component that's of interest is
magnesium sulfate when it reacts with
the barium nitrate the barium will
precipitate out with the sulfate of our
original component
we have our balanced chemical equation
where we see the reaction of the
magnesium sulfate with the barium
nitrate to form a our precipitate which
we can see here is the solid and our
spectator ions magnesium and nitrate so
the question is what is the
concentration or mass percent of the
magnesium sulfate that is in the mixture
so in gravimetric analysis we're going
to use the mass of the product from our
reaction right this is the mass of the
barium sulfate precipitate that was
formed from the reaction so that's where
we're going to start from that we'll be
able to use the molar mass to calculate
the moles of barium sulfate
if we know the male the moles of barium
sulfate we can use the stoichiometric
factor from our balanced chemical
equation to calculate the moles of
magnesium sulfates that were reacted
from that we can use the molar mass to
calculate the mass of magnesium sulfate
that we started with and then finally
using the sample Mass right the samp the
mass of the solid mixture we can
calculate what percentage of it was
magnesium sulfate so our flow diagram
tells us to start with the mass of the
barium sulfate so this was
we have a massive barium sulfate but we
would want to know the number of moles
so that we can use our stoichiometric
Factor so we're going to use the molar
mass of barium sulfate to calculate
moles of barium sulfate and this is 233
0.43 grams
for every one mole
taking this product grams will cancel
and we'll have moles of barium sulfate
so now we can use our stoichiometric
factor from our chemical equation we
should see that for every one mole of
barium sulfate that is produced
all right so now we know how many moles
of magnesium sulfate we reacted we want
to find the mass percent of magnesium
sulfate so we'll want to know what Mass
this corresponds to
so we'll use the molar mass of magnesium
sulfate this is one
120.37 grams of magnesium sulfate for
doing this calculation we find that we have
0.3181 grams of magnesium sulfate that
were reacted
now our question is asking what is the
mass percent of magnesium sulfate in our
original mixture so there could have
been other ionic compounds present in
the solid mixture we're not exactly sure
what the other components were but we
know that only a portion of it is
magnesium sulfate so to find our mass
percent we're going to want to take the
and compare this to the sample it's
the mass of the mixture
and to calculate a mass percent you'll
want to multiply by a hundred
so here let's plug in the numbers that
we have the mass of magnesium sulfate
that we found is
and our denominator the mass of the
entire mixture is
0.4550 grams
multiplying this by a hundred percent to
get our percentage we find that the mass
percent of magnesium sulfate is
69.91 another type of gravimetric
analysis is combustion analysis so in
this case the elemental composition
right how much of each element you have
for hydrocarbons and other types of
compounds may be determined using
combustion analysis in this case a wage
sample of the compound is heated to a
high temperature under stream of oxygen
right we know that oxygen is required in
order to combust or burn a sample so the
combustion results in the formation of
products we know for hydrocarbons which
contain carbon and hydrogen our products
are always going to be carbon dioxide
and water as shown in the chemical
equation on the slide in the diagram
that's shown here right our combustion
takes place in the furnace but then the
products of our reaction are swept
through into separate pre-weighed
collection devices so the first device
that you see here is a water absorber
meaning that it's going to absorb the
water that is the product of our
combustion reaction
other substances would continue to flow
through to the next pre-wave chamber
where we have a carbon dioxide absorber
and so any carbon dioxide that is
produced in our combustion reaction
would be absorbed now from these
pre-weighed collection devices we're
able to determine how much carbon
dioxide and how much water is produced
by our combustion reaction and with this
information we can calculate how many
moles of hydrogen for example are in the
water that's collected or how many moles
of carbon are in the carbon dioxide that
has been collected and once we know how
many moles of carbon or hydrogen we have
we can then use that information to
determine the empirical formula of your
unknown compound so let's look at an
example of a combustion analysis in this
question polyethylene a hydrocarbon
polymer is used to produce food storage
bags and many other flexible plastic
items a combustion analysis of a mass of
this sample yields
.00394 grams of carbon dioxide and
.00161 grams of water so using this
information we want to determine the
empirical formula of polyethylene
now we have a flow diagram to show us
what our approach should be we can see
that we're going to start with the mass
of carbon dioxide and the mass of water
that result from our combustion analysis
we're going to then use the molar masses
of these two products to determine the
moles of each of them so we'll determine
the moles of carbon dioxide and the
moles of water that have been produced
using the molecular formula of carbon
dioxide we can then calculate how many
moles of carbon have been collected in
the combustion analysis and how many
moles of hydrogen have also been
collected once we have moles of these we
can then use the molar ratio to
determine the empirical formula we're
first going to figure out how many moles
of carbon we have collected in our
combustion analysis they collected
according to our flow diagram we're
going to use the molar mass of carbon
dioxide to calculate how many moles of
CO2 that is
the molar mass of carbon dioxide is
grams of CO2 are going to cancel so
we'll have moles of CO2 and our flow
diagram says we're going to use the
stoichiometric factor in our molecular
formula for CO2 to convert this to moles
of CO2 now I can see from the formula
for every one mole of CO2 I only have
doing this calculation I calculate that
I have
8.95 times 10 to the minus 5th moles of carbon
carbon
now let's do the same thing for hydrogen
we can figure out the moles of hydrogen
by using the mass of water that was
collected in the combustion analysis
this is
next I'll use the molar mass of water to
convert this to moles of water the molar
mass of water is 18.02
grams of water are going to cancel and
my next step is to use the molecular
formula of water to write a
stoichiometric factor relating moles of
H2O to moles of hydrogen now based upon
the formula H2O I can see that I have
two moles of hydrogen for every one mole
of water doing this calculation allows
me to find that I have
1.79 times 10 to the minus fourth moles
of hydrogen collected from my combustion analysis
analysis
now the last last step to find my
empirical formula is to take the ratio
of the number of moles we learned
previously that when we do this we want
to divide by the smallest number of
moles which we have found for moles of
carbon so I'm going to set my ratio as
1.79 times 10 to the minus fourth moles
of hydrogen
divided by the smaller number which is
8.95 times 10 to the minus fifth moles
of carbon
doing this calculation I find that the
ratio is actually 2. so this means that
I have two moles of hydrogen for every
one mole of carbon in my empirical
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