Empirical Formulas from Composition
Because you can calculate the composition given the
formula, you should be able to work the other way around
and get formulas from their compositions. That's our next
topic, determining the empirical formula of a compound,
starting with some information about its composition.
Here is an overview of the process. First change the weights
or weight percentages to moles. Then calculate the mole
ratio. Express the mole ratio in whole numbers. Next, switch
your thinking from moles to atoms and remember that the
mole ratio is equal to the atom ratio. The last step is to
write the formula using the atom ratio.
This approach will necessarily give you the simplest ratio of
atoms in the formula, and not necessarily the actual number
of atoms in a molecule. Other information must be available
to determine molecular formulas. We will address that issue
later in this lesson.
Weights or Weight Percentages
Moles
Mole Ratio
Atom Ratio
Express as Formula
Five examples follow, each with an additional step or degree of complexity. (These are also shown
in Example 18 in your workbook.) Contact your instructor if these are not sufficient. Then try the
Quick Quiz on Determining Empirical Formulas before moving on.
First Example
In this first example, we are starting with mass data and will get a simple (x:1) atom ratio. We are
given information about how much carbon and how much hydrogen there is in a particular sample
of methane. Specifically 6.0 grams of methane contains 4.5 grams of carbon and 1.5 grams of
hydrogen. What's involved in determining the empirical formula is to eventually get the ratio of
atoms within the compound. As mentioned before, the way we go about doing this is to start with
mass and change it to moles. Once we get the mole ratio, then the mole ratio will be the same as
the atom ratio and from that we can get the empirical formula.
Starting with 4.5 grams of carbon and 1.5 grams
of hydrogen, we need to find out how many
Determine the empirical formula of methane given
moles there are of each element. We take the
that 6.0 g of methane can be decomposed into 4.5 g
4.5 grams of carbon and multiply by the
of carbon and 1.5 g of hydrogen.
conversion factor that changes from grams to
C
4.5 g x
1 mole =
0.375 moles C
moles by using the atomic weight and we get
12.0
g
0.375 moles of carbon. (Notice that I carried an
H
1.5 g x
1 mole =
1.5 moles H
extra digit here rather than rounding off right
1.0
g
away. This is because I'm not finished calculating
with that number yet. If I rounded off now, that Mole ratio
H=
1.5 /.375 = 4.0 = 4
C=
0.375/.375= 1.0= 1
would change the number and it would change
the results of the following calculations. So I
Empirical formula
CH4
won't round off yet, I will wait until the
calculations are done and then round off.)
Next, do the same thing with hydrogen, 1.5 grams times the conversion factor of one mole over 1.0
grams, which is the atomic weight for hydrogen, and we get 1.5 moles of hydrogen. To get the
mole ratio, divide the each mole amount by the smaller amount. In this case, we divide both by
carbon simply because there is more hydrogen (by moles) than carbon. When we divide 1.5 by
0.375, it comes out to be 4.0; and when we divide 0.375 by itself we get 1.0. As a mole ratio that
should be interpreted as 4.0 mol H to 1.0 mol C. Since we want to interpret this as an atom ratio
and since atoms come in whole numbers, the ratio is expressed as a nice even 4 to 1 ratio. That
tells us there are four hydrogen atoms to everyone carbon atom. So the empirical formula is CH 4.
At this point don't worry about what order you put the hydrogen and carbon; we will deal with that
later in the course. For now you could write H4C, but the formula is generally written as CH4.
As an aside let me point out that these rules are not terribly rigid as long as you keep in mind what
you are doing and how you are doing it. For example, what if you set up the mole ratio with the
smaller number on top. You don't get 4:1, you get 0.25 instead. But 0.25 = 1/4 and if you keep
track of the units or the order of the ratio, you still get one mole of carbon for four moles of
hydrogen.
Second Example
A number of complicating factors can arise. The next few examples take these into account one by
one. In this second example, we still start with mass data, but we will not get a simple (x:1) atom
ratio.
Here we're given the weights of iron and oxygen
that combine with one another to form a
particular compound. Specifically, 8.65 grams of
iron combines with 3.72 grams of oxygen. If you Determine the empirical formula of the compound
follow through the calculations, you can see that made when 8.65 g of iron combines with 3.72 g of
oxygen.
the amount of iron is changed from grams of iron
Fe
8.65 g x 1 mole =
0.155 moles
to moles of iron, 0.155 moles. Then the amount
55.8
g
of oxygen is changed from grams to moles. The
3.72 g x 1 mole =
0.233 moles
sample contains 0.233 moles of oxygen. Next, we O
16.0 g
need to get the mole ratio of oxygen to iron.
Because there's more oxygen, I will divide both
Mole ratio
O = 0.233/0.155 = 1.50 x 2= 3
Fe= 0.155/0.155= 1.00 x 2= 2
by iron. 0.233 divided by 0.155 comes out to be
1.50. Notice that in this case it does not come
Empirical formula
Fe2O3
out to be a nice whole number. But to get a
formula, we are going to need a whole number
ratio of oxygen atoms to iron atoms.
We cannot work with 1.50 atoms. We need whole numbers of oxygen and iron. So, here you need to
be able to change from decimal fractions into whole number fractions. You need to realize that
1.50 is the same as 1 1/2; and if you convert that into an improper fraction, you come up with 3/2
or 3 to 2 ratio. If there are 3 moles of oxygen to 2 moles of iron, then there are 3 oxygen atoms to
2 iron atoms. So, the empirical formula for the compound is Fe 2O3. If you were to write O3Fe2, that
is fine for now. We'll get into proper formula writing later on. Again the numbers in the ratio could
have been inverted. If we put 0.155 moles Fe over 0.233 moles O, we would get 0.665 and if you
recognize that as the same as 2/3, then you still get 2 moles Fe to 3 moles O, which gives you the
same formula Fe2O3.
Third Example
In this third example we start with a percentage
composition instead of weights. Simply assume
you have a 100 gram sample, 40 percent of that
is sulfur, 60 percent is oxygen. Thus you have 40
g of sulfur and 60 g of oxygen. (It doesn't really
matter whether you assume a 100g sample or any
other size sample, because the ratio will remain
the same.) Change those weights into moles by
using atomic weights as is shown, then take the
mole ratio of oxygen to sulfur, 3.75 to 1.25. That
comes out to be a 3-to-1 ratio, 3 oxygen's to
every one sulfur, so the empirical formula is SO3.
That was very much like the first example,
except that in this one we started with
percentages rather than weights.
The composition of a compound is 40% sulfur and
60% oxygen by weight. What is its empirical formula?
S
40 g x
O
60 g x
Mole ratio
O=
S=
Empirical formula
1 mole =
32.1 g
1 mole =
16.0 g
3.75/1.25=
1.25/1.25=
SO3
1.25 moles
3.75 moles
3
1
Fourth Example
The new aspect in this fourth example is that we
are dealing with three elements. The starting
part of the calculation is the same. Since we're
dealing with percents again, assume a 100 gram
sample. Thus we have 40.0 grams of carbon, 6.7
grams of hydrogen, 53.3 grams of oxygen. The
atomic weight of each of those elements is used
to calculate moles. For C, H, and O respectively,
there are 3.33 moles, 6.7 moles, and 3.33 moles.
Now, we're faced with what to do with a ratio of
three elements rather than just two. Luckily, we
can use the same method that we used in the
previous examples. Simply divide each of those
numbers by the smallest number of moles that is
there; and if you do that, you come up with the
simple relationship of 1.00 mole C, 2.01 mole H
and 1.00 mole O. In whole numbers that is a 1to-2-to-1 ratio of carbon to hydrogen to oxygen
and the empirical formula is CH2O. Since these
calculated values are very close to whole
numbers, they can be rounded off. You cannot
always do that, as is shown in the next example.
Pure formaldehyde consists of 40.0% carbon, 6.7%
hydrogen, and 53.3% oxygen. What is its empirical
formula?
C
40.0 g x
H
6.7 g x
O
53.3 g x
Mole ratio
C
H
O
Empirical formula
1 mole =
12.0 g
1 mole =
1.0 g
1 mole =
16.0 g
3.33 =
3.33
6.7 =
3.33
3.33 =
3.33
CH2O
3.33 moles
6.7 moles
3.33 moles
1.00
=1
2.01
=2
1.00
=1
Fifth Example
This example starts with percentage data and
deals with more than two elements, but also has
a mole ratio that doesn't come out nice and
simple. Take a moment to follow through the
calculations until you get down to determining
the mole ratio. Now, look at the mole ratio of
sodium to sulfur to oxygen. It is 1.26-to-1.26-to1.90. Again, we divide by the smallest because
we're dealing with three elements. This gives us
a new ratio of 1.00 to 1.01 to 1.51. The 1.00 is as
close to 1 as you can get using three digits so
that presents no problem. However, the 1.51 is
not close enough to either 1 or 2 to be rounded
off that much. But it is close enough to 1.5 to be
treated as 1-1/2. Now, we have a 1 to 1 to 1-1/2
ratio, we can get that into a whole number ratio
by multiplying each number by 2 in order to clear
the fraction. We use 2 because it is the
denominator in one-half. After doing that we
come up with a 2 to 2 to 3 ratio and that gives us
the empirical formula Na2S2O3. In other cases the
fractional mole amount might be 1-1/3 or 2-1/4.
In any case that you have to deal with a
fractional mole amount multiply everything by
the denominator of the fraction.
Determine the empirical formula of a compound
that is 29.0% sodium, 40.5% sulfur, and 30.4 %
oxygen by weight.
Na
29.0 g x
S
40.5 g x
O
30.4 g x
Mole ratio
Na
S
O
Empirical formula
1 mole =
23.0 g
1 mole =
32.1 g
1 mole =
16.0 g
1.26 =
1.26
1.26 =
1.26
1.90 =
1.26
Na2S2O3
1.26 moles
1.26 moles
1.90 moles
1.00 x 2= 2
1.00 x 2= 2
1.51 x 2= 3
If you have any questions about any of those examples, please check with the instructor. Make sure
any problems and questions that you have are taken care of. If you need or want more practice on
figuring out empirical formulas, you can find more exercises in your text or you can ask the
instructor for some extra practice exercises (Ex. 19-alt.).
Comment on Rounding Off
Students often ask how much they can round off these kind of calculations. How far away from a
whole number can we get and still round it off? That depends on how accurate the data was. If you
want to round 1.05 to 1, you are saying that the data was off by 5%. But if you want to round 1.25
to 1, you are saying that the data was off by 25%. That's not very good data on which to base an
empirical formula. Besides, it might have been off 25% the other way. At this point in your
chemistry studies, I would say that if you have to round off more than 5% to get a whole number,
you should not round off. Under other circumstances a different cut-off would be appropriate.