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Stoichiometry
Solving Problems Involving Mass/Mole Relationships
Finding Empirical and Molecular Formulas Page [1 of 2]
Suppose you’re an analytical chemist and you get called out to a site where there’s been a horrible chemical spill, and
your job is to figure out what the chemical is. Well, you know, you could eat some, or something like that. Put some
on your skin, and that might get you some information. But let’s look at a piece of information that you might be able
to get that might be a little less dangerous. And that piece of information is the analysis by mass of the various
elements that make up the molecule.
For instance, here’s a problem. The analysis of a compound—and let’s not worry about the analytical technique at
this point. We find, though, that it consists of 39.9 % carbon, 6.7% hydrogen, and 53.4% oxygen by mass. What can
we learn from this information about the unknown compound in the chemical spill? And these kinds of problems are
what are known as empirical formula questions. I’ll define what “empirical formula” is in a moment, but let’s go
through the mechanics of taking these pieces of information and getting something that looks a little more familiar.
Now the trick to working empirical formula problems is to assume some certain mass of material. And 100 grams is
very convenient, so let’s assume that our sample consists of 100 grams. Why is that useful? Well if we assume that
our sample consists of 100 grams, since these are all percentages by mass, that 100 grams consists of 39.9 grams of
carbon, 6.70 grams of hydrogen, and 53.4 grams of oxygen. And these add up to, at least pretty close to, 100 grams.
So all of our mass is accounted for.
These things in grams, mass, it’s okay, but it’s not really useful. Remember that if we’re trying to understand about
the composition of a molecule, how many atoms of a particular kind of molecule, what we need to do is we need to get
this into moles. And the way to get into moles is to take the masses and divide by the molar mass, or multiply by 1
over the molar mass, and that looks like this. So I worked this out ahead of time. We take the mass of carbon, 39.9
grams times 1 mole over 12.0 grams. That’s the molar mass of carbon. And that gives us 3.33 moles of carbon.
Similarly for hydrogen, the molar mass of hydrogen is 1.01 grams. So this represents 6.63 moles of hydrogen. Again,
it’s that the mass of hydrogen we had in our 100 gram sample, 6.63 moles of it was hydrogen, and finally, 3.34 moles
of it is oxygen. Okay, what does that get us? Well it tells us something. It tells us that the relative ratio of atoms in
our molecule is 1 to 2 to 1. It’s exactly the same as the ratio of the number of moles of atoms in our sample. So the
way we do this is, if we imagine that our sample consists of 3.33 atoms of carbon to 6.63 atoms of hydrogen to 3.34
atoms of oxygen—And we can’t talk about fractional oxygens and fractional carbons, but if we divide through by 3.33,
we can see that this becomes CH2O. And again, this represents the ratio of atoms in the molecule. Now this is not
the molecule. Why is it not the molecule? Because it turns out that any product of these numbers and a small integer
is also going to give a molecular formula that’s the same as this. This is what is known as an empirical formula. And
if you didn’t understand that last statement, don’t worry because I’m going to come back to it.
The molecular formula is the exact atomic composition of the molecule. So here’s a molecule of benzene, and it has
exactly the composition, C6H6, that means six atoms of carbon, six atoms of hydrogen. Here is ethanol. That’s the
molecular formula for ethanol, ozone, chlorine, and formaldehyde.
The empirical formula, which is what we derived in our calculation on the previous couple of pages, is the formula of
the molecule written with smallest integer subscripts. And why is it written with smallest integer subscripts? Because
it turns out we are limited in the amount of information that we can glean from the analysis by mass. In other words,
this just happens to be the best we can do.
And what we saw on the previous page was that the empirical formula for our unknown compound is CH2O. That
means, for every carbon atom in the molecule, there are two hydrogen atoms and one oxygen atom. Note that this
empirical formula also happens to be exactly the molecular formula for formaldehyde. So it might be that our
compound is formaldehyde, and formaldehyde does happen to be a colorless liquid. But note that CH2O is also the
empirical formula for a bunch of other molecules. So here’s the molecular formula for acetic acid, and it has exactly
the same ratio of atoms. And remember, if we did an analysis of this by mass, it would have the same composition by
mass as the unknown problem. In other words, this has exactly the same empirical formula as CH2O.
How do these differ? Well basically these numbers are these numbers, 1, 2, 1, multiplied by a small integer. Multiply
it by two and you get C2H4O2. Multiply 1, 2, 1 by three and you get 3, 6, 3. That happens to be the molecular formula
for glycerin. It’s not necessarily the case that the unknown is glycerin, because there are lots of molecules that
probably have this molecular formula. I am just saying glycerin is a possibility. And glycerin also happens to be a
Stoichiometry
Solving Problems Involving Mass/Mole Relationships
Finding Empirical and Molecular Formulas Page [2 of 2]
colorless liquid. And finally, you can take this on to infinity. In other words, any molecule that has a molecular formula
CN H2N ON is going to give exactly the same empirical formula.
So we know more, but we still don’t know what our unknown is. In order to get a better idea of what our unknown is,
and we’re still, at the end of all of this, not going to know exactly what our unknown is, we need another piece of
information. And another piece of information that we could get might be the molar mass of the unknown. We can
use mass spectrometry, for instance. And suppose we fi nd out that the molar mass of our unknown is 60 grams per
mole. How are we going to use that piece of information together with the information that we already have to get an
idea of what the molecular formula is, or to get the molecular formula? And the answer is to recognize that the
empirical formula weight, or how much the molar mass of the empirical formula divided into the molar mass of the
molecular formula, or the molecular formula weight, is a small integer. And that small integer tells us how many
empirical formula units there are in the molecular formula.
Let’s see if I can make that clearer. The empirical formula weight is the molar mass of this thing. And the molar mass
of CH2O is 30 grams, because we have one mole of atoms of carbon, two moles of atoms of hydrogen, and one mole
of atoms of oxygen. And we add 12 grams, which is the molar mass of carbon, 2 grams, which is the molar mass of
two moles of hydrogen, and 16 grams, the molar mass of oxygen. And that gives us 30 grams per mole for the
empirical formula weight. The molecular formula weight is exactly how much each of these guys weighs. How much
a mole of each of these guys weighs. And you can see that if you divide the mass of a mole of this stuff by the mass
of a mole of this stuff, it tells you how these guys are related to these guys. In other words, this is exactly twice that.
This is exactly three times that. This is exactly four times that. And that idea is expressed in the idea that N, which is
the ratio between the molecular formula weight and the empirical formula weight, is going to tell us what our molecular
formula is.
So we plug in molecular formula weight, that’s 60 grams per mole, and the empirical formula weight. We calculated
that. That’s the mass of a mole of CH2O, and that’s 30 grams per mole. And so this is equal to 2. What is this 2
again? This tells us how many empirical formulas there are in the molecular formula. So in other words, the
molecular formula is equal to CH2O times 2, or C2H4 O2. Where this two is not multiplying the whole thing, but it is
multiplying the coefficients here. In other words, C1H2O1, and then each one of these times 2, because we had 2
times the molecular formula. The molecular formula weight was twice what the empirical formula weight was.
So we’ve actually gotten as far as a molecular formula for our unknown compound. And this corresponds to, among
other things, acetic acid, which is the acid in vinegar. So it could be our unknown compound is, in fact, acetic acid.
It’s not necessarily acetic acid, but the point is, we’ve been able to use two pieces of information, the analysis by
mass, the carbon, hydrogen, oxygen composition by mass, and the molar mass of the unknown compound in order to
determine a molecular formula. And ultimately this piece of information can be used to learn more about the
compound. For instance, we might find out that it’s an acid. That would further solidify the idea that it’s acetic acid.
So it’s sort of a mystery of which we’ve gotten part of the way. We can figure out now what the molecular formula is
from the two pieces of information that I showed you.