Molecular Formulas
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Molecular Formulas

The last topic for this lesson is determining the molecular formula. The empirical formula represents the simplest ratio of atoms of different elements contained in a particular compound. With molecular compounds we generally want to know more. We want to know the actual number of each kind of atom that is contained in the molecule.

For example hydrogen peroxide has a 1:1 ratio of hydrogen to oxygen. Thus its empirical formula is HO. But molecules of hydrogen peroxide contain four atoms, two each of hydrogen and oxygen. Thus, hydrogen peroxide has a molecular formula of H2O2. How do we know that?

Another example is the compound benzene. It has a molecular formula of C6H6. There are actually 12 atoms in the molecule. But the empirical formula only gives you CH, a 1-to-1 ratio. Another example is glucose, it has an empirical formula of CH2O. That empirical formula doesn't distinguish it from quite a number of other compounds, including other types of sugars and also formaldehyde, that have the same empirical formula. But if you knew that the molecule contains 6 carbon atoms and 12 hydrogen atoms and 6 oxygen atoms, then that is quite a bit more information. How do we know these things?

Elements are not immune from this issue either. Empirical formulas do not show that some elements have molecules in which the atoms are paired up, such as H2 and O2. Again, how do we know that?

The way you can go about determining the molecular formula involves the fact that the molecular weight can be determined independently of the formula. You did (or will do) this in the experiment for this lesson. It can also be done by using Avogadro's Law along with gas densities or number of other ways that we don't have time to go into in this lesson. But the molecular weight can be determined. And if that molecular weight is compared with the formula weight that's obtained from an empirical formula, then you can figure out the molecular formula from that information.

The principle of the process is this. Start with the empirical formula (EF) and use whatever multiple of it gives the correct molecular weight (MW). In the tables shown here you can see that H2O and H2O2 are the molecular formulas (MF) that match the molecular weights.
Water
EF = H2O
MW = 18 g/mole
Possible
MF
Corresponding
MW (g/mole)
H2O
H4O2
H6O3
18
36
54
Hydrogen peroxide
EF = HO
MW = 34 g/mole
Possible
MF
Corresponding
MW (g/mole)
HO
H2O2
H3O3
17
34
51

One example of how to do this is shown here (and in example 20 in your workbook). The empirical formula gives the formula weight of 15.0. The molecular weight is 30.0 g/mole which is just twice the empirical formula weight. You can figure that out by dividing the molecular weight by the empirical formula weight. In this case you get that it's two times heavier. That means that the molecular formula must contain twice as many atoms as the empirical formula but still in the same ratio. That means twice as many of each kind of atom. In this case the ratio is 1-to-3. Twice that many would be 2-to-6, so the molecular formula is C2H6.
What is the molecular formula of a compound if its empirical formula is CH3 and its molecular weight is 30?.
Empirical formula weight of CH3 = 12.0 + 3.0 =15.0
Molecular weight is (30.0 / 15.0) = 2.0 times as heavy as the empirical formula weight, so the molecular formula must contain twice as many atoms as the empirical formula (but in the same ratio).
Molecular formula is C2H6.

Practice with Determining Molecular Formulas

The next thing I want you to do is work on Exercise 21 and determine the molecular formulas for those materials. Check your answers below, then continue with the lesson. (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 Wrap-Up for this lesson.  Click here to go there now.)

Answers

The molecular formulas for the chemicals in Exercise 21 are C4H10, P4, C6H6, and H2O2. Again, if you had any trouble getting any of these molecular formulas, check with the instructor to figure out why.

 

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