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Empirical Formulas from Composition
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 moles there are of each element. We take the 4.5 grams of carbon and
multiply by the conversion factor that changes from grams to moles by using the atomic
weight and we get 0.375 moles of carbon. (Notice that I carried an extra digit here rather
than rounding off right away. This is because I'm not finished calculating with that
number yet. If I rounded off now, that would change the number and it would change the
results of the following calculations. So I won't round off yet, I will wait until the
calculations are done and then round off.) 
Determine the empirical formula of methane given that 6.0 g of methane
can be decomposed into 4.5 g of carbon and 1.5 g of hydrogen. 
C 
4.5 g x 
1 mole =
12.0 g 
0.375 moles C 
H 
1.5 g x 
1 mole =
1.0 g 
1.5 moles H 
Mole ratio 

H =
C= 
1.5 /.375 =
0.375/.375= 
4.0 =
1.0= 
4
1 
Empirical formula 
CH_{4} 



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
every one 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 H_{4}C, but the formula is generally written
as CH_{4}. 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 follow through the calculations, you can see that the amount of
iron is changed from grams of iron to moles of iron, 0.155 moles. Then the amount of
oxygen is changed from grams to moles. The sample contains 0.233 moles of oxygen. Next, we
need to get the mole ratio of oxygen to iron. Because there's more oxygen, I will
divide both by iron. 0.233 divided by 0.155 comes out to be 1.50. Notice that in this case it does not
come 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. 
Determine the empirical formula of the compound made when 8.65 g of iron
combines with 3.72 g of oxygen. 
Fe 
8.65 g x 
1 mole =
55.8 g 
0.155 moles 
O 
3.72 g x 
1 mole =
16.0 g 
0.233 moles 
Mole ratio 

O =
Fe= 
0.233/0.155 =
0.155/0.155= 
1.50 x 2=
1.00 x 2= 
3
2 
Empirical formula 
Fe_{2}O_{3} 



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_{2}O_{3}. If you were to write O_{3}Fe_{2},
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 Fe_{2}O_{3}. 
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 3to1
ratio, 3 oxygen's to every one sulfur, so the empirical formula is SO_{3}. 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 
1 mole =
32.1 g 
1.25 moles 
O 
60 g x 
1 mole =
16.0 g 
3.75 moles 
Mole ratio 

O =
S = 
3.75/1.25=
1.25/1.25= 
3
1 
Empirical formula 
SO_{3} 



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
1to2to1 ratio of carbon to hydrogen to oxygen and the empirical formula is CH_{2}O.
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 
1 mole =
12.0 g 
3.33 moles 
H 
6.7 g x 
1 mole =
1.0 g 
6.7 moles 
O 
53.3 g x 
1 mole =
16.0 g 
3.33 moles 
Mole ratio 

C 
3.33 =
3.33 
1.00 
= 1 


H 
6.7 =
3.33 
2.01 
= 2 


O 
3.33 =
3.33 
1.00 
= 1 
Empirical formula 
CH_{2}O 



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 1.26to1.26to1.90. 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 11/2. Now, we have
a 1 to 1 to 11/2 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
onehalf. After doing that we come up with a 2 to 2 to 3 ratio and that gives us the
empirical formula Na_{2}S_{2}O_{3}. In other cases the fractional
mole amount might be 11/3 or 21/4. 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 
1 mole =
23.0 g 
1.26 moles 
S 
40.5 g x 
1 mole =
32.1 g 
1.26 moles 
O 
30.4 g x 
1 mole =
16.0 g 
1.90 moles 
Mole ratio 

Na 
1.26 =
1.26 
1.00 x 2= 
2 


S 
1.26 =
1.26 
1.00 x 2= 
2 


O 
1.90 =
1.26 
1.51 x 2= 
3 
Empirical formula 
Na_{2}S_{2}O_{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.
19alt.).
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 cutoff would be appropriate.
Practice with Determining Empirical Formulas from Composition
Try your hand at working the problems in Example 19. After you've done that, check your
answers below then continue. (If, after completing these, you would like to try some
additional problems of this type, you will find several in a "Practice Problems"
page in the WrapUp for this lesson. Click here to go
there now.)
Answers
The formulas for the compounds in Example 19 are SO_{2}, CH_{3}, C_{2}H_{6}O,
and N_{2}H_{8}Cr_{2}O_{7}.
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Clackamas Community College
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