Analyzing an Element: Divining the Empirical and Molecular Formula

Written: February 2015
Originally Published: Half Measures: Chemical Reactions: Elemental Analysis.

As a certain wise man wise once said, elemental analysis is not the sexiest thing around. But is a very necessary thing. Learn it, and then you can move onto more interesting concepts. In short, elemental analysis is the determining of the empirical and molecular formulas for some compound. We'll examine each of these in turn.

I. Empirical Formula

First, then, the empirical formula. Simply put, it is the ratio of atoms to each other in one molecule of a certain compound. For example, the empirical formula for water is H20, because for every one oxygen atom in a water molecule, there are two hydrogen atoms (e.g., they are in a 1:2 ratio). Or again, the empirical formula for glucose is CH2O, because for every one carbon and one oxygen atom, there are two hydrogen atoms. (e.g., the number of combined oxygen and carbon atoms is equal to the number of hydrogen atoms in any molecule of glucose).

In order to find a compound's empirical formula from a sample of that compound, one must know (1) the percentage of each element in the sample, and (2) the molecular mass of each element in the compound (keep in mind that a compound is a combination of two or more elements). To give an example, suppose we are presented with a sample of an unknown compound. But we do know that this compound is composed of 29% chlorine, 18.8% sodium, and 52.2% oxygen, and we know the molecular mass of each element. For simplicity's sake, we'll assume the sample is 100 grams. Given this information, we can find the compound's empirical formula.

1. Find the mass of each element in the unknown substance - Because we've assumed that the unknown sample is 100 grams, all we need to do is multiple 100 by the percent of each element. In the present example, the mass of chlorine is 29 grams (e.g., 100 grams x 0.29 = 29 grams), sodium is 18.8 grams, and oxygen is 52.2 grams.
2. Find the molar mass of each element - First, look up the mass of each element in AMUs (this is the element's molecular mass); use the periodic table. Second, change the unit from AMUs to grams (this is the element's molar mass). In this example, the molar mass of chlorine is 35.5 grams, sodium is 23 grams, and the molar mass of oxygen is 16 grams. 
3. Divide the actual mass of each compound by its molar mass - This will tell you how many moles of each element there are in the unknown sample. For example, if we're considering chlorine, divide 29 grams by 35.5 grams per mole to get 0.817 moles of chlorine. In other words, there are about .817 moles of chlorine in the unknown sample. By the same method, we find that there are 0.817 moles of sodium and 3.26 moles of oxygen in the unknown sample. We can put them into a ratio of 0.817:0.817:3.26; for every 0.817 moles of chlorine and 0.817 moles of sodium, there are 3.26 moles of oxygen.
4. Simplify the ratio - Divide each number in the ratio by the smallest member of that ratio. In the present case, divide each number by 0.817. This gives us a ratio of 1:1:4. Therefore, the empirical formula for our unknown compound is NaClO4, also known as sodium perchlorate.

II. Molecular Formula

The molecular formula represents the actual number of each element in one molecule of a compound, not just the ratio of each element to each other; sometimes these things are the same, sometimes they are not. For example, the empirical formula for sodium perchlorate is NaClO4. This is also its molecular formula. But, other compounds are not so simple.

1. C6H12O6 - Glucose
2. C4H8O4 - Erythrose
3. C3H6O3 - Lactic Acid
4. C2H4O2 - Acetic Acid
5. CH2O - Ribose

All the above compounds have the same empirical formula: CH2O. But each compound has different multiples of that formula per molecule. What makes glucose different from erythrose, for example, is that glucose has more atoms of each element per molecule than does erythrose; nevertheless, they both have the same ratio of those elements to each other, and so have the same empirical formula. Initially, it can be difficult to understand the difference between the two formulas. Think of it like this: the empirical formula gives us the relative number of atoms of an element in a compound. It tells us how many atoms of each element their are relative to the other atoms in that compound (as all ratios do). The chemical formula, on the other hand, gives the absolute or actual number of atoms of an element in a compound.

In order to find the molecular formula, you must know (1) the the molecular mass of the sample, and (2) the empirical formula of the sample. Suppose, for example, the molecular mass of an mystery compound is 90 AMU. Now we know the empirical formula of this mystery sample is CH20. To find the molecular mass of CH20, simply find the molecular mass of carbon, hydrogen, and oxygen, and add them together; we find this to be 30 AMUs. In other words, there are three "CH2O"s in this unknown sample. We therefore multiple CH2O by 3 to get C3H6O3. This is the sample's chemical formula. Therefore, the sample under consideration is lactic acid.

Resources

"Chemistry: the Central Science" - In-class textbook.