>> Welcome to a screencast on combustion analysis.
This is our example, propanoic acid is a compound
that contains carbon, hydrogen, and oxygen.
And in combustion analysis, when things burn, they require oxygen.
And if they burn completely, combust completely, they produce carbon dioxide and water.
So, we have .1236 grams of propanoic acid.
Burns completely.
Produces .421 grams of carbon dioxide and .172 grams of H2O.
And the question is to find the empirical formula for propanoic acid.
Well, first of all, here's kind of a schematic of how combustion analysis works.
We have a sample of our substance.
In this case, our small mass of propanoic acid.
We run a lot of oxygen past it, so the sample's in a little typically glass cylinder,
open on or where there's an inlet and an outlet.
We heat it so it'll burn.
And then, the products of the combustion are pushed through or out the end
of this glass cylinder and into and through a U-tube here.
Connected to another U-tube there.
And then waste goes out the end.
And we heat it to make sure that the combustion occurs and everything reacts.
And then, the first set of U-tubes contains a material that absorbs carbon dioxide.
And the second set of or second U-tube contains a material that absorbs water.
And so, what ends up happening is by the time the reaction is done,
this U-tube will weigh more than it was, than it weighed when it started.
And its new mass is due to absorbed carbon dioxide.
And then, similarly, this U-tube will weigh more than it did because it has absorbed
or had water stick to the material in here.
And our sample is a carbon hydrogen, in this case, oxygen compound.
And the key thing is where does all of the carbon from the sample that we burned end up?
And of course, the answer is all of the carbon combines with oxygen to make CO2.
So, however many moles of carbon we started with, in this sample, they all arrive here.
And then, a similar question, what happens to all the hydrogen in the sample?
And all of the hydrogen in the sample also combines with oxygen and ends up bonded to it,
forming the molecule or the compound H2O.
And then, that's absorbed and sticks to this U-tube.
And then, the oxygen, we're not actually able to track directly.
But as you'll see in a bit, we can deduce how much oxygen had to be there
from knowing the amounts of CO2 and H2O that are collected during the combustion analysis.
So, let's work our way through it, starting with, in this case, with .421 grams of CO2.
And in this problem, we're going to not show explicitly every calculation.
But so, these are things you should make sure you know how to do.
But hopefully, you can say I can determine the molar mass of carbon dioxide.
And then, use that to convert from grams of CO2 to moles of CO2, which turns out to be .00957.
And oops. We can, therefore, say since one mole of CO2 contains one mole of carbon,
.00957 moles of CO2 contains .00957 moles of carbon.
Because we only care about the carbon, at this point.
We do the same basic idea with the H2O.
Start with .172 grams of H2O.
You should be able to determine molar mass of H2O and then convert from grams to moles.
And this time, we get .00954 moles of H2O.
And then, what we're interested in is the hydrogen.
And since one mole of H2O contains two moles of hydrogen,
then .00954 moles of water contains twice that many moles, .0191 moles of hydrogen.
Okay. Now, we note that since we have the moles of hydrogen and carbon,
and these came from our original sample which we know the mass of.
We can then determine how many grams of carbon this number of moles is.
And simple conversion using the atomic weight of carbon.
Same idea with hydrogen.
Determine the number of grams that corresponds to, using its atomic weight.
And then, we recall that our initial sample was .236 grams.
We just went through the calculation to find that of that,
.115 grams are carbon and .0191 grams are hydrogen.
So, if we subtract these two masses, the masses of carbon
and hydrogen, from the mass of the sample.
We can deduce that that mass .102 grams must be the mass of oxygen in the sample.
We weren't able to get it directly, but we're able to get it via some calculations.
And then, of course, we might be interested in knowing how many moles that is.
So, convert grams of oxygen into moles of oxygen, and you get .00638.
Now, we're almost there and we're back in familiar territory.
Here is a summary.
Moles of carbon, moles of hydrogen, and moles of oxygen that we know.
Remember that this is our ratio.
And if you can't look at this and immediately see what it is as a whole number ratio,
this advice would be divide by the smallest of those numbers
and that will still give us a ratio.
And we do, when we do that division, we have 1.5 moles of carbon to 2.99 moles
of hydrogen to one mole of oxygen.
Rounding these off gives 1.5 to three to one.
And then, noting that we can't have 1.5, let's say carbon atoms in the formula,
we're going to have to double everything to keep the same ratio.
But to have all whole numbers which are going to be legitimate in the formula.
So, three moles carbon to six moles hydrogen
to two moles oxygen gives us an empirical formula of C3H6O2.
Here's the last thing for this chapter.
It's worth noting that we can determine things like atomic masses
and molar masses using a device known as the mass spectrometer.
It's a much more modern method used to determine masses compared to say combustion analysis
which also, by the way, only gives us empirical masses.
And we are often interested in the molecular mass of a substance.
And here's how it works.
We have a sort of glass tube curved like this.
We input our substance.
It's hit with an electron beam which ionizes it and may partially fragment it, as well.
We have an accelerating grid that pulls the particles, the ionized particles
of the ionized molecular fragments in this direction.
It goes through a slit, so it's made into kind of a narrow beam.
And then, that beam is passed by a magnet which will deflect the path
of the charged particles either a little or a lot, depending upon how large the particles are.
These, the larger particles get deflected less.
The smaller particles get deflected more.
And we can adjust the strength of the magnetic field
such that we can make the various particles go through the slit here and arrive at a detector.
And we can determine the mass of the particles based upon the strength of the magnetic field.
And so, what ends up happening is we get something that looks
like this, which is a mass spectrum.
It will give the mass of the fragments and how many of each of those reach the detector.
So, the height of each of these corresponds to the number of particles like that.
And then, what one can do is look at this spectrum
and determine the structure of the molecule.
And in this case, we are going to see
that it has a molar mass pretty close to about 130, it looks like.
Because these would be the particles that are only ionized.
And then, here are pieces.
And it's a little beyond the scope of the class, but we can use this to determine formulas.
And that's it for combustion analysis.