Calculating Empirical Formulas
The percent composition can be used to calculate the empirical formula, but it will require us to covert
between mass and moles, since percent composition is done in grams and formulas are written to
represent mole or atom ratios.
Here is a sample problem:
A compound consists of 80% carbon and 20% hydrogen. What is the empirical formula?
Our first step is to get rid of the percents. A percent alone is unusable, since it is just a ratio. So, for
instance, if 45% of the students taught by your chemistry teacher are girls, I still have no idea how many
girls or boys are being taught because I don’t know the total number of students.
To get past that, we make a simple assumption—that we have 100 g of the substance. That way 80%
carbon turns into 80g of carbon. We could of course choose any amount but 100g makes the math easy.
So, we can now translate that question into:
A 100 g sample of a compound consists of 80g of carbon and 20g of hydrogen. What is the empirical
formula?
Now we’ll convert the grams to moles:
80𝑔 𝐶 𝑥
20𝑔 𝐻 𝑥
1 𝑚𝑜𝑙𝑒 𝐶
= 6.66 𝑚𝑜𝑙𝑠 𝑜𝑓 𝐶
12.0107 𝑔 𝐶
1 𝑚𝑜𝑙𝑒 𝐻
= 19.84 𝑚𝑜𝑙𝑠 𝑜𝑓 𝐻
1.00794 𝑔 𝐻
That means that the compound contains 6.66 mols of C for every 19.84 mols of H. This is the correct
mole ratio, but it isn’t pretty. To make it a simple ratio, we simply divide both mole amounts by the
smallest number (in this case 6.66 moles)
6.66 𝑚𝑜𝑙𝑠 𝐶
=1𝐶
6.66 𝑚𝑜𝑙𝑠
19.84 𝑚𝑜𝑙𝑠 𝐻
= 2.98 𝐻
6.66 𝑚𝑜𝑙𝑠
At this point we get to do a little playing with the significant figures. 2.98 is trying very hard to be 3, so
the mole ratio is 1C:3H. Therefore the empirical formula is:
𝐶𝐻3
Calculating Empirical Formulas continued
Assuming that you’ve already looked at the first example, let’s try a more complex one.
A compound contains 40.00% C, 6.72% H and 53.28% O. What is the empirical formula?
To solve this we need to convert the percentages to grams, convert the grams to moles and then divide
by the smallest mole amount (to find the simplest ratio).
40.00% 𝐶 → 40.00𝑔 𝐶𝑥
1 𝑚𝑜𝑙 𝐶
3.33 𝑚𝑜𝑙 𝐶
= 3.33 𝑚𝑜𝑙𝑠 𝐶 →
=1𝐶
12.0107 𝑔
3.33 𝑚𝑜𝑙
6.72% 𝐻 → 6.72𝑔 𝐻 𝑥
1 𝑚𝑜𝑙 𝐻
6.67 𝑚𝑜𝑙 𝐻
= 6.67 𝑚𝑜𝑙 𝐻 →
~2𝐻
1.00794 𝑔
3.33 𝑚𝑜𝑙
53.28% 𝑂 → 53.28𝑔 𝑂 𝑥
1 𝑚𝑜𝑙 𝑂
3.33 𝑚𝑜𝑙 𝑂
= 3.33 𝑚𝑜𝑙𝑠 𝑂 →
=1𝑂
15.9994 𝑔
3.33 𝑚𝑜𝑙
So the empirical formula for this compound is:
𝐶𝐻2 𝑂
Calculating Empirical Formulas – even more
Now you’ve seen a very simple example and a more complex example. Let’s try one more.
A compound contains 48.65% C and 8.16% H with the rest of the compound being O. What is the
empirical formula?
Here we were not given the percent of O, but we can obtain that easily through subtraction:
100% − (48.65 + 8.16) = 43.20% 𝑂
Now we can set up the problem, just like the last one. I’ve converted to grams in my head to save some
space.
48.65𝑔 𝐶 𝑥
1 𝑚𝑜𝑙 𝐶
4.05 𝑚𝑜𝑙 𝐶
= 4.05 𝑚𝑜𝑙 𝐶 →
= 1.5 𝐶
12.0107 𝑔
2.70 𝑚𝑜𝑙
8.16𝑔 𝐻 𝑥
1 𝑚𝑜𝑙 𝐻
8.10 𝑚𝑜𝑙 𝐻
= 8.10 𝑚𝑜𝑙 𝐻 →
=3𝐻
1.00794 𝑔
2.70 𝑚𝑜𝑙
43.20𝑔 𝑂 𝑥
1 𝑚𝑜𝑙 𝑂
2.70 𝑚𝑜𝑙 𝑂
= 2.70 𝑚𝑜𝑙 𝑂 →
=1𝑂
15.9994 𝑔
2.70 𝑚𝑜𝑙
Suddenly, I have a problem. If I take those answers and write out the formula as I have before I would
end up with:
𝐶1.5 𝐻3 𝑂
That can’t be right…
However, I can get a simple WHOLE-number ratio simply by multiplying everything by 2. That would give
me:
𝐶3 𝐻6 𝑂2
The lesson to be learned:
After dividing by the smallest mole amounts you will often (maybe even usually) have only whole
numbers. In those cases when you don’t, they will be simple fractions.
You should keep an eye out for the following decimals:
X.5 (half)
X.333 (one third) or X.666 (2 thirds)
X.25 (one quarter) or X.75 (three quarters)
Calculating Molecular Formula
Once you have an empirical formula, you can calculate the molecular formula if you know the molar
mass. Here’s a sample problem:
A compound has the empirical formula C6H12O6. If the molar mass is approximately 180 g/mol, what is
the molecular formula?
To solve this we need to remember that the molecular formula must be a multiple of the empirical
formula (since empirical is the simplest ratio). That means that the molar mass must be a multiple (the
same multiple) of the empirical mass. If that’s enough info to figure it out, great. If not, here is a way to
organize yourself:
Calculating Molecular formulas continued
Now that you’ve done one example, let’s try another:
A compound has an empirical formula of C3H6O2. If the molar mass is approximately 175 g/mol what is
the molecular formula?
Here’s the table set up initially
C12H24O8