How do we figure this out? We know that:
1) the number of oxygen atoms can be found by using Avogadro’s number, if we
know the moles of oxygen atoms;
2) the number of moles of oxygen atoms can be found if we know the number of
moles of each molecule;
3) the number of moles of each molecule can be found if we know the mass of each
compound
We know from the information given that we have an equal mass of each
compound, but no real numbers to plug in and find moles. So what can we do?
Since we have an equal mass of each compound, if we calculate the percent by
mass of oxygen in each compound, we could just compare their ratios, since we
know that the mass is directly related to number of moles (which is directly related
to the number of atoms).
So let’s start by finding the % by mass of oxygen in H2SO4:
To find % mass, we use the equation:
As we did with the first example we worked in class, we assume we have 1 mole
of the compound, which has a molar mass of 98.086 g/mol (mass of the whole):
For every mole of H2SO4 molecules, there are 4 moles of oxygen atoms, which
has a molar mass of 16.00 g/mol (mass of the part):
Now we just punch in the numbers and find:
Now we can do the same for the other two compounds!
Assuming 1 mole of sugar (mass of the whole: 342.296g/mol), we have 11 moles
of oxygen atoms per mole of sugar molecules:
And for potassium chlorate, assuming 1 mole of the compound, we have 3 moles
of oxygen atoms per mole of KClO3 molecules:
Now that we have the % by mass of each compound, we can just compare the
ratios:
Compound
H2SO4
C12H22O11
KClO3
% by mass oxygen
65.25%
51.42%
39.17%
So we can see that H2SO4 has the greatest percent oxygen by mass, so if we
had an equal mass of each compound, the mass of oxygen in H2SO4 would be
the greatest, meaning that it has the greatest number of oxygen atoms.
***Also note that subscript does NOT indicate which compound has the greater %
by mass of oxygen! Sugar has 11 moles of oxygen atoms per mole of compound,
but has less % mass of oxygen than H2SO4, which has only 4 moles of oxygen
atoms per mole of compound!***
Now, what would happen if we were told the mass of each
compound in the initial problem? Say we have 50.0 grams
of each of the compound in the table above. How many
atoms of oxygen are in each?
So how do we get from grams of a compound to atoms of oxygen? In any problem
where I need to convert one type of unit to another unit, I like to set up my
dimensional analysis with units only so I can figure out where I’m going, kind of like
a road map. Then, once I plot my “route”, I go back and fill in the numbers for my
conversion factors.
In this problem, I’m starting with grams, and I want to get to number of atoms of
oxygen:
I know that to get number of atoms, I will have to use Avogadro’s number at some
point in the calculation, and to be able to use Avogadro’s number, I need to know
moles. But I have grams! So I’ll need to convert my grams to moles…and I can do
that with molar mass (g/mol):
Now I have moles, but since it is moles of the compound, not just oxygen. I’ll need
to convert the moles of the compound to moles of oxygen atoms:
Well, I have moles of oxygen atoms, so I can figure out the number of moles of
oxygen atoms with Avogadro’s number, which is number of “things” per mole:
So now all of my units should cancel out correctly, leaving me with the number of
oxygen atoms!
So all I have to do is plug in the numbers and solve…
Note that the starting amount given only has three significant figures, so your
answer should have the same.
Now we can do the same calculation for the other two compounds:
For sugar (C12H22O11):
For potassium chlorate (KClO3):
So we want to find the formula and name of some iron oxide compound. What do
we do?
Well, we know that the new compound is some “iron oxide” meaning that it must be
composed of X moles of iron and Y moles of oxygen:
FexOy
So if we want to write a formula, we need to figure out the molar ratios of iron and
oxygen in the new compound so we can figure out the subscripts. But how can we
even know how much of each element there is in the new compound?
Remember that we learned about the Law of Conservation of Mass earlier this
semester, which means that whatever amount of matter we start with before the
reaction will be the amount of matter we end up with after the reaction.
Since we start with 12.00 grams of iron, that means that the new compound
contains 12.00 g of iron as well (assuming all of the iron is in the new compound
now). So the moles of iron in the new compound should be:
So we have moles of iron, but what about the oxygen? We know from the problem
that the mass of the new compound is17.16 grams. If 12.00 grams of the compound
is iron, then the remainder must be oxygen:
Now that we have the mass of oxygen in the new compound, we can find moles of
oxygen:
Molecular formulas are always ratios of whole numbers, since they represent the
number of atoms in a molecule (you can’t have half an atom!), so we have to figure
out the whole number ratio by dividing by the smaller of the two numbers:
We still don’t have whole numbers, though, so we’ll need to multiply by some factor
to get them to the closest whole number ratio. I see that if I multiply 1.5 by 2, I will
have 3; so multiplying both numbers by 2 gives me:
So now I see that my molecular formula is:
Fe2O3
So here we have another problem in which we have to find the molecular
formula, but this time we do not know how much we started with, or how much
we ended up with, we are only given the mass of carbon in propane and its molar
mass. So what can’t do what we did on the previous problem. What can we do?
Another way we can figure out ratios of substances in a compound is to find their
percent composition by mass.
First, we’ll start by calculating the % by mass of carbon, since we are given the
amount of carbon (mass of part) and the molar mass (mass of whole):
If propane is 81.71% carbon, then the % hydrogen must be:
In order to find the molecular formula of propane, though, we need to find the
moles of each particle to figure out the ratios. As I talked about in lecture, percent
composition is independent of the amount of a substance. So we can choose any
amount to calculate the moles. And since percents are out of 100%, it’s easiest
to just set our amount to 100 grams.
So if we have 100 grams of propane, then there are 81.71 grams of carbon and
18.29 grams of hydrogen. Using molar masses, we can find moles of each:
Then we find the ratio of the two:
And finally we find the closest whole number ratio:
So we have found the formula:
C 3H 8
But is this the empirical formula or the molecular formula? Well, we know the
molar mass of propane should be 44.094 g/mol, so let’s check what the molar
mass of C3H8 would be:
It checks out! So C3H8 is the molecular formula, but since it is the smallest whole
number ratio, it is also the empirical formula!
This problem is very similar to the one we just did, but this problem was really
nice and gave us two of the three % compositions we need! To find % by mass of
oxygen, we just have to subtract the % by mass of carbon and hydrogen from
100%:
Now that we have all three % compositions, we can find the moles of each
particle as we did before by assuming that we have 100 grams of the compound,
which means we have:
40.00 grams of carbon, 6.71 grams of hydrogen and 53.29 grams of oxygen.
Using molar mass to find moles, we get:
Then we divide by the smallest number to get the ratios:
Well, that was nice, we got whole number ratios! So the molecular formula is
CH2O then? Well, we know our compound’s molar mass is 120 g/mol, so let’s
check:
That doesn’t add up to 120 g/mol at all! So this must be our empirical formula.
How do we come up with the molecular formula then? Well, let’s look at the ratio
of the molar mass of our compound versus the molar mass of the empirical
formula, CH2O:
It looks like the molar mass of the compound is 4 times larger than the molar
mass of the empirical formula. So if we were to multiply all of the molar ratios by
4, we would have a formula of:
C4H8O4
Does that check out?
Yes it does!
So our molecular formula is
C4H8O4.