Empirical and molecular formulae 

​​In a nutshell

​Molecular and empirical formulae show how many of each type of atom a compound contains. The molecular formula represents the whole number of each atom type in a compound, whereas the empirical formula is the simplest whole number ratio of each type of atom in a compound. 

Empirical formula 

The empirical formula is the simplest whole number ratio of atoms of each type of element in a compound. 

Empirical formula from experimental data 

The empirical formula of a compound can be calculated using the masses from experimental data. 

procedure 

Example

A hydrocarbon undergoes complete combustion in oxygen to produce $$8.8 \, g$$ of carbon dioxide and $$4.5 \, g$$ of water. What is the empirical formula for this hydrocarbon?

Find the number of moles of carbon dioxide ($$CO_2$$) and water $$(H_2O)$$ using the masses in the question:

$$n \, (CO_2) \, = \, \frac{8.8}{(12 \, + \, (16 \, \times \, 2))} \, = \, \frac{8.8}{44} \, = \, 0.2 \, mol \\ \ \\ n \, (H_2O) \, = \, \frac{4.5}{((1 \, \times \, 2) \, + \, 16)} \, = \, \frac{4.5}{18} \, = \, 0.25 \, mol$$​​

​​

Work out the number of moles of carbon atoms in the hydrocarbon starting material first:

$$1 \, mole$$ of $$CO_2$$ contains $$1 \, mole$$ of carbon atoms
$$0.2 \, moles$$ of carbon atoms 

Now work out the number of moles of hydrogen atoms in the starting material:

$$1 \, mole$$ of $$H_2O$$ contains $$2 \, moles$$ of hydrogen atoms 

$$0.25 \, \times \, 2 \, = \, 0.5 \, moles$$​ of hydrogen atoms 

Write out the mole ratio of atoms in the hydrocarbon:

$$\,C \, \, : \, H \, \,\newline 0.2 : 0.5$$​​

Divide both numbers by the smallest number of moles ($$0.2 \, moles$$):

$$C:H \newline \, 1 \,: \, \frac{5}{2}$$

Multiply by two to make a whole number ratio of atoms:

$$C:H \newline \, 2 \,: \, 5$$

Based on the original masses of carbon dioxide and water, the empirical formula of this hydrocarbon is $$\underline{C_2H_5}$$.

Empirical formula from percentage composition

The empirical formula of a compound can be calculated using the percentage composition of each type of atom. 

procedure

Example 

A compound has the following percentage composition for each type of atom: $$70.5\%$$ carbon, $$13.8\%$$ hydrogen and $$15.7\%$$ oxygen. What is the empirical formula for this compound?

First, calculate the number of moles for each type of atom in the compound:   

$$n \, (C)\, = \, \frac{70.5}{12} \, = 5.875 \, mol \\ \ \\ n \, (H)\, = \, \frac{13.8}{1} \, = 13.8 \, mol \\ \ \\ n \, (O)\, = \, \frac{15.7}{16} \, = 0.98125 \, mol$$

Next, divide each number of moles by the smallest mole value ($$0.98125 \, mol$$): 

$$C \, = \, \frac{5.875}{0.98125} \, = \, 5.99 \\ \ \\ H \, = \, \frac{13.8}{0.98125} \, = \, 14.06 \\ \ \\ O \, = \, \frac{0.98125}{0.98125} \, = \, 1.00$$​​

​

Write this as a ratio of each type of atom:

$$\, \, \, \, \, \, C \, \, \, \, \, : \, \, \, \, \, H \, \, \, \, : \, \, \, \, O \newline \, 5.99 \,: 14.06\, : 1.00$$​​

Using the ratios above, the empirical formula for the compound is $$\underline{C_6H_{14}O}$$. 

Molecular formula

The molecular formula represents the whole number of each type of atom that there is in a compound. 

Molecular formula from the empirical formula 

The molecular formula of a compound can be calculated using the empirical formula and the molar mass of the compound. 

procedure 

Example

The empirical formula of an alcohol is $$CH_3O$$. Given that the molar mass of the alcohol is $$62 \, g \, mol^{-1}$$, find the molecular formula.

Find the molar mass of the empirical formula:

$$M \, = \, (12 \, \times \, 1) \,+\, (1 \, \times \, 3) \, + \, (16 \, \times \, 1)= \, 31 \, g \, mol^{-1}$$

Divide the molar mass ($$62 \, g \, mol^{-1}$$) by the molar mass of the empirical formula:

$$\frac{62}{31} \, = \, 2$$

Multiply the number of each type of atom in the empirical formula by two to find the molecular formula:

$$C_{(1\, \times \, 2)}H_{(3 \, \times \, 2)}O_{(1 \, \times \, 2)} \, = \, C_2H_6O_2$$

The molecular formula for the compound is therefore, $$\underline{C_2H_6O_2}$$. 

Molecular formula from experimental data 

The molecular formula can be calculated from the empirical formula via the use of experimental data again. 

procedure 

Example

A hydrocarbon undergoes complete combustion with oxygen to produce $$4.4 \, g$$ of carbon dioxide and $$1.8 \, g$$ of water. Given the molar mass of the compound is $$112 \, g \, mol^{-1}$$, what is the molecular formula? 

Find the number of moles of carbon dioxide ($$CO_2$$) and water $$(H_2O)$$ using the masses in the question:

$$n \, (CO_2) \, = \, \frac{4.4}{(12 \, + \, (16 \, \times \, 2))} \, = \, \frac{4.4}{44} \, = \, 0.1 \, mol \newline n \, (H_2O) \, = \, \frac{1.8}{((1 \, \times \, 2) \, + \, 16)} \, = \, \frac{1.8}{18} \, = \, 0.1 \, mol$$​​

​​

Work out the number of moles of carbon atoms in the hydrocarbon starting material first:

$$1 \, mole$$ of $$CO_2$$ contains $$1 \, mole$$ of carbon atoms
$$0.1 \, moles$$ of carbon atoms 

Now work out the number of moles of hydrogen atoms in the starting material:

$$1 \, mole$$ of $$H_2O$$ contains $$2 \, moles$$ of hydrogen atoms 

$$0.1 \, \times \, 2 \, = \, 0.2 \, moles$$​ of hydrogen atoms 

Write out the mole ratio of atoms in the hydrocarbon:

$$\,C \, \, : \, H \, \,\newline 0.1 : 0.2$$​​

Divide both numbers by the smallest number of moles ($$0.1 \, moles$$):

$$C:H \newline \, 1 \,: \, 2$$

Use the ratio of each type of atom to find the empirical formula:

$$CH_2$$​​

Find the molar mass of the empirical formula:  

$$M \, (CH_2) \, = \, (12 \, \times \, 1) \,+\, (1 \, \times \, 2) \, = \, 14 \, g \, mol^{-1}$$

Divide the molar mass of the compound ($$112 \, g \, mol^{-1}$$) by the molar mass of the empirical formula:

$$\frac{112}{14} \, = \, 8$$

Multiply the number of each type of atom in the empirical formula by eight to find the molecular formula:

$$C_{(1\, \times \, 8)}H_{(2 \, \times \, 8)} \, = \, C_8H_{16}$$​​

The molecular formula for the compound is therefore, $$\underline{C_8H_{16}}$$.