Leah here from leah4sci.com and in this MCAT video we're going to look at Empirical versus
Molecular Formula.
You can find this entire video series along with the reactions and stoichiometry practice
quiz by visiting my website leah4sci.com/reactions.
The molecular formula as the name implies gives us the formula or the equation for a
specific molecules.
We get the type of atoms bound to the molecule along with the number that tell us how many
of that atom is found in the molecule.
For example, the molecule water has a molecular formula of H2O.
This tells us that water has hydrogen with a subscript of 2 meaning 2 hydrogen atoms
and Oxygen with an invisible of 1 telling us 1 oxygen.
Anytime you see an atom without a subscript following that atom, assume it's an invisible
and understood 1 so we only have 1 of that atom.
For the molecule ethanol, we have 2 carbon atoms so we write the molecular formula as
carbon for the atom, 2 telling us how many.
We have 1,2,3,4,5,6 hydrogen atoms so we write H with a sub 6 telling us there are 6 hydrogens
and we only have 1 Oxygen so we write the letter O with no number following.
The molecular formula for ethanol is C2H6O.
And this is how it goes for every molecule.
No matter where you are, no matter the conditions, the molecular formula will not change.
Day or Night, Earth or the moon, winter or summer, the molecular formula is like the
social security for that molecule and it doesn't change no matter what.
Unless the molecule itself changes but then it becomes a completely different molecule.
The Empirical Formula is less specific to the individual molecule.
According to google, empirical is an adjective which is based on concern with verifiable
by observation or experience rather than theory or pure logic, ouch!
In the Chemistry setting, the empirical formula is more about the ratio of the atoms in the
molecule.
Let's look at our good friend glucose which is C6H12O6.
Glucose has 6 carbons, 12 hydrogens, and 6 oxygens.
This is the molecular formula because it tells us exactly how many atoms are in glucose.
But to look for the ratio of atoms, we go back to middle school algebra where we look
for the greatest common factor.
The biggest number that I can use to divide everything and get a smaller ratio.
6, 12, and 6 are all divisible by 6.
So when we divide each one by 6 we get a new formula.
C6 divided by 6 is C1 or just C. H12 divided by 6, 12 divided by 6 is 2 gives me H2, O6
divided by 6 is just 1 for CHO.
This is the empirical formula or smallest whole number ratio for the molecule glucose.
Does this mean that anytime you see the empirical formula CH2O it has to be glucose?
Absolutely not!
Let's look at the molecule methanal which you're recognize as formaldehyde, This is
used in a lot of different things like embalming fluid and resins.
If we draw the molecular formula, we have 1 carbon, 2 hydrogens, and 1 oxygen.
The molecular formula for formaldehyde is CH2O.
If we look at the numbers and try to divide by the greatest common factor, will carbon
and oxygen each have just 1, that means the only number we can divide by is 1.
In this case, the molecular formula happens to be the same exact ratio as the empirical
formula.
If you can't break it down anymore, it can be the same.
As you can see, glucose and formaldehyde have the same empirical formula but they're completely
different.
1, sadly makes food taste really really good and the other one's a toxic carcinogen.
So if empirical formula is so ambiguous why would we bother with it especially on the
MCAT?
Even though it's annoying and tedious it makes for a very good analytical method.
On the MCAT you might come across research where you're testing for an unknown substance
and the experiment helps you find the ratio of atoms of the percent composition of the
different atoms and then asks you to use these information to identify the structure or the
molecular formula or something else about this compound.
Let's go with something very simple.
Say you have an unknown amino acid and you're able to identify the different atoms but you
don't know specifically which atom it is.
And in this experiment it turns out that you have the following ratios; 1 sulfur, 1 nitrogen,
2 oxygen, 5 carbon, and 11 hydrogen.
Using this information, first of all, the fact that we have sulfur immediately rules
out ninety percent of the amino acids because only two of them have sulfur atom.
Cysteine and Methionine, so which is it?
Here we have the basic backbone of an amino acid with the R-group attached.
We can't use sulfur because both cysteine and methionine have one sulfur, can't tell
them apart.
Both of them have 1 nitrogen in the amino group.
Both have two oxygens in the carboxyl group, and both have a lot of carbon and hydrogen.
We know there are 2 carbon atoms in the backbone for the alpha carbon and the carboxylic acids
which leaves us the side chain to figure out the difference between them.
This is something you should have memorized, and if you haven't, make sure you study the
amino acid cheat sheet linked below.
The cysteine side chain is just a CH2 bound to an SH.
This gives me a total of 3 carbons telling me this is not cysteine.
The methionine side chain has a CH2 bound to a CH2 bound to Sulfur bound to a CH3.
That's a total of 3 in the R-group to an apparent 5 carbons telling me that this is methionine
and all I had to do to figure this out was know the empirical formula or the ratio of
atoms in this molecule.
In this case the ratio was enough to ID the molecule.
Let's try more detailed experiment.
Remember Math will be minimal on the MCAT but it's important to know all the steps in
case you have to calculate one of these pieces.
You don't want this to cost you any points.
Say you're told that the analysis of the unknown diol tells us we have 38.7% carbon, 9.7% hydrogen,
and 51.6% oxygen by mass.
Find the empirical and molecular formula for this molecule.
If you're struggling with percentile math, there's another video for that in this series.
For this problem we're looking for 2 things.
This is a two step problem.
We wanna find the empirical and the molecular formula.
All we are given is percent by mass of each atom so let's analyze and simplify.
We have 38.7% carbon which we'll round up to 39% or maybe even 40, remember on the MCAT
close enough is good enough especially because it's multiple choice.
9.7% hydrogen let's round that to 10%.
And 51.6% Oxygen we can round that to 52% or even 50% again, close enough.
The next thing we wanna do is figure out about how many moles we have.
As you learn when calculating percent by mass, we set a hundred percent to a hundred grams
which means we have about 39 grams of carbon, 10 grams of hydrogen, and 52 grams of oxygen.
Let's convert this into moles because that's the only way to calculate molecular and empirical
formula.
You can find these numbers on the periodic table but I highly recommend memorizing them
for these simple atoms.
1 mole of carbon has a mass of 12 grams which gives us 39 over 12.
Remember to do this without a calculator.
39 over 12 is very difficult but 39 over 13 might be something you recognize as 3.
If that doesn't work, think of what number is close to 39 that divides by 12.
12 times 3 is 36, 39 is slightly greater than 36 so if we round down our answer will be
slightly greater than 3.
Hydrogen is very simple.
1 mole of hydrogen is 1 gram which means that we have 10 moles of hydrogen, and finally
oxygen is 16 grams per 1 mole.
Once again we find ourselves with tricky math, 52 divided by 16.
If this is too hard, see if you can pull something out.
Both 52 and 16 divided by 2 so that would give you 26 because half of 50 is 25 then
2 divided by 2 is 1 giving me 26, divided by 8.
This might be easy to wrap your mind around because 8 times 3 is 24, this is 26 so our
answer is slightly greater than 3.
Our ratio so far is 3 moles of carbon, remember we're estimating.
10 moles of hydrogen and 3 moles of oxygen.
From here, you will divide by the lowest number to find your ratio.
At this point, you will divide by the smallest number to get the lowest possible ratio.
Now, 3 and 10 don't have any common numbers to divide but remember we did some rounding
where our answer is slightly greater than 3 and we wanna keep that in mind.
Slightly greater than 3 divided by slightly greater than 3 gives us 1 carbon.
10 divided by slightly greater than 3 would be 3 because 10 divided by 3 is 3.33 so it's
close enough.
And same thing for oxygen.
Slightly greater than 3 divided by the same number gives us 1 oxygen and this a ratio
of CH3O which is our empirical formula.
And maybe you're thinking, how was I supposed to come up with this without a calculator
on the MCAT?
Wouldn't I just have stopped here?
The reason we know we didn't stopped here is because we're told it is a diol.
Diol tells us that we have 2 alcohols so that means we can't have 3 oxygens, we have to
find a way to reduce this down and we're trying for something close enough.
Another thing you could've done is simply take the whole number.
3 divided by 3 is 1, 10 divided by 3 is something greater than 3, and kept it at that because
on the MCAT you'll have your multiple choice giving you clues about where to go even if
you're slightly off.
We have the empirical formula but now let's find the molecular formula.
To find the molecular formula we need something other than the empirical formula.
We need to be given mass the number of moles or something that tells us how many times
to divide this knowing that the empirical formula times a certain number is going to
be equal the molecular formula.
So how many CH3O go into the molecular formula or in this molecule?
We're told we have a diol.
Diol tells us 2 oxygens.
If the empirical formula only has 1 oxygen, we know we need to double it and therefore
we need to double everything which tells us the empirical formula of CH3O times 2 from
the diol clue is equal to the molecular formula.
What do we have?
CHO. 2 times 1 carbon is 2, 2 times 3 hydrogen is 6, 2 times 1 oxygen is 2 giving me a molecular
formula of C2H6O2.
And if you remember your protecting groups from Organic Chemistry you might recognize
this as ethylene glycol which is used to protect carbonyl groups in the advance reactions.For
more practice like this, try the reactions and stoichiometry practice quiz on my website.
First give this video a thumbs up, then visit my website leah4sc.com/reactions for the entire
reactions and stoichiometry series and the practice quiz.
Again, that's leah4sci.com/reactions.