Here is a typical molecular formula: CH3COOH. It is indeed the molecular formula of acetic acid, which is the major component of vinegar. I will use it as an example to show all information you can learn from this simple formula. Keep in mind that sometimes we combine the same atoms and use a subscript to indicate its quantity. For example, we can write this formula as C2H4O2. The reason why we write the formula as CH3COOH is because it shows another piece of information, i.e. the functional group -COOH to indicate it is a carboxylic acid.
(1) It tells you what types of atoms are in a compound. This compound has C, H, and O atoms. Nothing else. Another example is NaNO3, which indicates that this compound has Na, N, and O atoms.
(2) It shows the number of each atom in a compound. In each acetic acid molecule, you will have 2 carbon atoms, 4 hydrogen atoms, and 2 oxygen atoms. If you add all atoms together, that will give you the total number of atoms in an acetic acid molecule. In this case, there are 8 atoms in total.
(3) It tells you the moles of each atom in one mole of molecules.
You can get this information from the molecular formula directly. For example, if you have one mole of C2H4O2 molecules, then you will have two moles of C atoms, four moles of H atoms and two moles of O atoms. This information can be used for some related calculations, such as if you have 0.5 moles of C2H4O2, then you will have 1 mole C atoms, 2 mol H atoms, and 1.0 mol O atoms.
0.5 mol C2H4O2 × (2 mol C)/( 1 mol C2H4O2)=1.0 mol C
0.5 mol C2H4O2 × (4 mol H)/( 1 mol C2H4O2) = 2.0 mol H
0.5 mol C2H4O2 × (2 mol O)/( 1 mol C2H4O2) = 1.0 mol O
(4) You can determine the molar mass from it.
The molar mass is the mass of one mole of molecules (ions, atoms), which can be calculated from the molecular formula. It is the sum of atomic mass of all atoms. Its unit is g/mol. For C2H4O2, if you add the atomic mass of two C, four H, and two O together, it will give you its molar mass as 60.052 g/mol.
2 × 12.011 + 4 × 1.008 + 2 × 15.999 = 60.052 g/mol
You can find the atomic mass for each element on the Periodic Table. I would suggest you memorize those for atoms from 1 to 20. The molar mass is very useful in chemistry. We frequently use it to convert grams to moles or moles to grams. For example, if you have 20.0 g C2H4O2, it corresponds to 0.333 mol C2H4O2.
20.0 g × (1 mol)/(60.052 g) = 0.333 mol C2H4O2
If you have 0.150 mol C2H4O2, that means you have 9.00 g C2H4O2.
0.150 mol × (60.052 g)/(1 mol) = 9.00 g C2H4O2
Sometimes, you will need to determine the molar mass of an ion. For example, what is the molar mass of CH3COO-? Since ions are formed by losing electrons. Because electrons are very light, so the molar mass of an ion is the same as its neutral form, so when you do the calculation, just ignore the charge and calculate the molar mass from the fomula you have. Fro CH3COO-, molar mass of CH3COO- will be 59.284 g/mol.
2 × 12.011 + 3 × 1.008 + 2 × 15.999 = 59.284 g/mol
(5) (It tells you the molar ratio of atoms in a molecule. For every one mole of C2H4O2 molecules, there are two moles of C atoms, four moles of H atoms and two moles of O atoms, therefore, the molar ratio of C:H:O = 2:4:2 = 1:2:1 (simplest ratio). Therefore, 1:2:1 is the molar ratio of C, H, and O atoms in CH3COOH molecule. This ratio will be the same in any acetic acid molecule no matter where and how you get it. This in fact, is the Proust Law of Constant Proportions. If someone claimed a pure sample is an acetic acid but you found the molar ratio of C:H:O is not 1:2:1, then that sample will not be acetic acid.
(6) Mass percent of each element in a molecule.
Molecular formula can be used to determine the mass percent of each atom in the compound. For C2H4O2, its molar mass is 60.052 g/mol, which means if you have 60.052 g acetic acid (i.e. one mole of C2H4O2, you will have 24.022 g C, 4.032 g H and 31.998 g O atoms:
2 mol C × (12.011 g C)/(1mol C) = 24.022g C
4 mol H × (1.008g H)/(1 mol H) = 4.032 g H
2 mol O × (15.999 g O)/(1 mol O) = 31.998 g O
Therefore, the mass percent for each element will be
C% = (24.022 g C)/(60.052 g C2H4O2) × 100 = 40.0%
H% = (4.032 g H)/(60.052 g C2H4O2) × 100 = 6.71%
O% = (31.998 gO)/(60.052 g C2H4O2) × 100 = 53.4%
(7) Empirical formula of a molecule.
Since the molar ratio of C, H, and O in acetic acid is 1:2:1 (the simplest ratio), so you can express this relationship as CH2O. We call this the empirical formula. It is different from its molecular formula (C2H4O2)). However, they could be the same for some molecules, for example, methane has a molecular formula CH4, the molar ratio of C:H = 1:4 (the simplest ratio), so its empirical formula is also CH4. An empirical formula does not tell you how many C, H, and O atoms in a molecule, molecular formula does. The empirical formula only tells you the molar ratio of atoms.
In practice, we can determine the mass or mass percent of each element in a sample, and we can use that to determine the empirical formula. For example, if you have a sample and from elemental analysis, you found it had 40.0% C, 6.67% H, and 53.3% O, this implies that for every 100 g of sample, there will be 40.0 g C, 6.67 g H and 53.3 g O. You can use this information to determine the molar ratio:
(40.0 g C)/(12.01 g/mol):(6.67 g H)/(1.008 g/mol):(53.3 g O)/(15.999 g/mol) = 3.33:6.61:3.33 = 1:2:1
Therefore, the empirical formula is C1H2O1, i.e. CH2O. You notice the empirical formula for the sample in this example is the same as that for acetic acid, so is this sample the acetic acid? Not really. Since the empirical formula represents a group of compounds that share the same molar ratio of elements, you need other information to identify the exact nature of this sample. Let’s see if you also determine the molar mass of this sample is 60.052 g/mol and it had a COOH group, then most likely this sample will be acetic acid because its empirical formula matches what you found and its molar mass of 60.052 g/mol is two times of formula weight of CH2O (its formula weight is 30.026 g/mol).