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Chapter 7.5c Combustion Analysis Example | Michelle Bunagan | YouTubeToText
YouTube Transcript: Chapter 7.5c Combustion Analysis Example
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This content demonstrates a step-by-step combustion analysis to determine the molecular formula of an organic compound containing carbon, hydrogen, and oxygen, given its mass and the masses of water and carbon dioxide produced upon combustion.
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hi everyone in this last video I just
wanted to do one more quick example of
combustion analysis in this example we
have a compound that contains carbon
hydrogen and also oxygen so when 4.18
grams of this compound is burned we're
able to collect 2.5 grams of water and
6.13 grams of carbon dioxide we're told
the molar mass of the substance we want
to figure out what the molecular formula
is of the compound so just as we did in
the previous video we want to use the
mass of the water and carbon dioxide
collected to figure out how many moles
of hydrogen and carbon were in our
original compound
so that's where we're going to start to
find the moles of carbon
we use the mass of carbon dioxide that
was collected in the combustion analysis
as we did previously we'll want to use
the molar mass of carbon dioxide in
order to convert this to moles the molar
mass is 44.01 grams for every one mole
of carbon dioxide
so grams of carbon dioxide cancel and we
have moles and then we write the
stoichiometric factor based upon the
molecular formula of CO2 to figure out
how many moles of carbon this is we know
that for every one mole of CO2 we have
only one mole of carbon
so this will give us the moles of carbon
that were collected in the carbon
dioxide this is 0.139 moles
we can do the same thing with the water
to figure out moles of hydrogen
so starting with the mass of water which
is 2.50 grams we're able to figure out
how many moles of Water by dividing this
by the molar mass of water which is
18.02 grams for every one mole
of H2O
grams of water cancel and then we can
use the stoichiometric factor based on
the molecular formula of water to know
that this for every one mole of water we
have two moles of hydrogen
doing this calculation we find that we have
have
0.277 moles of hydrogen
now when we got to this point in the
last problem we were able to take the
molar ratio to determine the empirical
formula right the numbers of carbon and
hydrogen that should be in our empirical
formula however if we know that we have
that is made up of carbon hydrogen and
oxygen we can subtract the mass of the
carbon and the mass of the hydrogen and
what should be left would be the mass of
the oxygen
so let's use these numbers of moles to
figure out the mass of hydrogen and the
mass of carbon the mass of carbon we can
and using the molar mass of carbon to
figure out the mass so here we would say
for every one mole of carbon right the
mass would be
12.011 grams
so this gives me a mass of 1.67 grams of
carbon we can do the same thing for the
moles of hydrogen from
0.277 moles of hydrogen I can use the
molar mass of hydrogen where every one
mole of hydrogen is
1.0079 grams
so this tells me that the mass of
hydrogen is
all right so now if we want to figure
out the mass of oxygen we take our
we subtract the mass of carbon which we
found is 1.67 grams and we subtract the
mass of hydrogen which is
0.279 grams
so the mass of oxygen should be
2.231 grams
and just as we did before for carbon and
hydrogen we'll need to know how many
moles this is so the moles of oxygen
would be
and we use the molar mass of oxygen
where one mole of oxygen is
16.00 grams
grams cancel and this tells me that the
and as we learned previously if we want
to figure out what our empirical formula
is we can write our empirical formula
with the number of moles that we've calculated
and then we can divide all of these
numbers by the smallest one so the
and doing these calculations we find
that the empirical formula should be c h
two o now our problem told us that the
molar mass of the substance is
180.156 and they want to know the
molecular formula we just found the
empirical formula in order to determine
the molecular formula we would want to
and divide it by the mass of our
empirical formula so this is 30. 03
grams per mole
so this is the mass of our empirical
formula ch2o when I do this calculation
it's going to tell me the number by
which I need to multiply all of my
subscripts and so this calculation comes
out to 5 which means that to arrive at
my molecular formula for the compound I
would need to multiply all my subscripts
by 5. so this becomes C5 h10
h10 o5
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