1. The Determination of an Empirical Formula
Mole Theory Unit
The Determination of an Empirical Formula

2. First – A Review
In previous lessons, we have worked on the ability to take a chemical formula and calculate the percentage of each element in that formula.
The key to this calculation was remembering that a percentage of any part is the amount of that part divided by the total of all of the parts (and then multiplied by 100).

3. An Example
Suppose that we were asked to calculate the percentage of each element in the compound magnesium nitrate.
Step 1 would be to write the chemical formula for the compound.
Magnesium is simply “Mg”
Nitrate is a polyatomic with the formula (NO3)

4. Continuing…
Using the periodic chart, we see that magnesium has an oxidation state of +2 .
From the list of polyatomic ions, we have that Nitrate has an oxidation state of -1 .
Writing all of this and then criss-crossing will yield the formula:
Mg(NO3)2

5. Step 2: Determine the formula weight
Remember that the formula weight is the sum of all of the weights of the atoms in the formula.
Working with this compound, we have:
Mg(NO3)2
The inventory and atomic weights from the periodic chart are:
Mg = 1 x 24
N = 2 x 14
O = 6 x 16
The total will be 148.

6. Step 3: Now for the Percentages
Looking at how we determined the formula weight, we can use those numbers for the percentage of each element.
Mg = 1 x 24
N = 2 x 14
O = 6 x 16
The total will be 148.
% Mg = (24 / 148) x 100
= %
% N = (28 / 148) x 100
= %
% O = (96 / 148) x 100
= %
You should punch through these calculations for practice.

7. Now for the New Topic
The example we just worked through asked us to go from a chemical formula to a set of percentages that describe how nature builds that compound.
In our new task, we will be asked to essentially “work backwards” – to go from a set of percentages back to the chemical formula that would give those percentages if we calculated it that way.

8. Example:
We are told that we have an “unknown” compound.
We do not know its formula, but we are told that it has the following percent composition:
Na = %
S = %
O = %
So we see that the compound contains the elements Na, S, and O. Our task is to determine its proper chemical formula.
This will take several steps.

9. Step 1: Assume that we have 100 grams of the unknown.
Here are the percentages that were given in the previous slide.
Na = %
S = %
O = %
Assuming that we have 100 grams, allows us to simply convert the percentages to grams (since the percentage system is based on the number “100”).

10. So…. Na = 32.39 % S = 22.54 % O = 45.07 % Na = 32.39 grams
The percentages:
Na = %
S = %
O = %
Become grams:
Na = grams
S = grams
O = grams

11. Step 2: Convert the Grams into Moles by dividing each mass by the atomic weight from the Periodic Chart.
Na = grams / 23 = moles Na
S = grams / 32 = moles S
O = grams / 16 = moles O
Each of the numbers that we are using to divide by came from the periodic chart (take a moment and check that out). Note too that we are carrying LOTS of decimal places here – important.

12. Step 3: A Math Technique called “Normalization”.
We will “normalize” the answers to the divisions by dividing all of the answers of moles by the smallest answer in the set. (read that again carefully) Then look at how this is done below.
moles Na / = 2
moles S / = 1
moles O / = 4
We divided all of them by the because it is the smallest number of moles that we got when we did Step 2. If you run the last line through your calculator, you will see that the answer is actually – its OK to round that one to “4”.

13. What Does This Mean?
If the numbers that we end up with when we “normalize” are whole numbers (integers), then we have the numbers for the chemical formula and we are done!
moles Na / = 2
moles S / = 1
moles O / = 4
Since we have integers, the chemical formula of the unknown is
Na2SO4

14. But what if we don’t get all whole numbers when we Normalize?
Good Question – sometimes, nature builds compounds where the numbers are not all divisible by each other.
When that happens, one of the numbers will include an easy fraction.
All that we will have to do is multiply all of the numbers after normalization by the denominator of the fraction.

15. Consider this example:
An “unknown” compound has the following percentage composition:
C = %
H = %
The regular “Step 1” will convert the percentages to grams.
82.76 % C  grams C
17.24 % H  grams H

16. Now for Steps 2 and 3: 82.76 grams C / 12 = 6.897 moles C / 6.897 = 1
17.24 grams H / 1 = moles H / = 2.5
That “2.5” is actually 2 ½ - there is the fraction that was described in the earlier slide. Since the denominator of the fraction ½ is “2” , we will multiply all of the numbers after normalization by “2” .
So, the Carbon will be 1 x “2” = 2
and the Hydrogen will be 2.5 x “2” =
Since we now have integers, we are done! The chemical formula of the unknown is C2H5 .