Calculating Empirical Formulas - a More Complex Example

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) = 43.19% 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.

Suddenly, I have a problem. If I write out the formula as I have before, I would get C_1.5H₃O. That can't be right.

If, however, I multiply by 2, I'll get a whole number ratio. So, in the end, my empirical formula is C₃H₆O₂.

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 have only whole numbers, you will have 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)