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

1.

First find the number of moles of each product in the reaction using the equation:

$n \, = \, \frac{m}{M}$

2.

Work out the number of moles for each type of atom in the compound using the number of moles calculated in step 1.

3.

Divide number of moles for each type of atom by the smallest mole value to find the empirical formula.

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

1.

Calculate the number of moles for each type of atom, using the percentage composition as the mass.

$n \, = \, \frac{m}{M}$

2.

Divide each number of moles by the smallest value for the number of moles to find a ratio for the number of each type of atom.

3.

Use the ratio to find the empirical formula for the compound.

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

1.	First find the molar mass of the empirical formula.
2.	Divide the molar mass of the compound by the molar mass of the empirical formula.
3.	Multiply the number of each type of atom in the empirical formula by the answer to step 2 to find the molecular formula.

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

1.	First find the number of moles of each product in the reaction using the equation: $n \, = \, \frac{m}{M}$
2.	Work out the number of moles for each type of atom in the compound using the number of moles calculated in step 1.
3.	Divide number of moles for each type of atom by the smallest mole value to find the empirical formula.
4.	Find the molar mass of the empirical formula.
5.	Divide the molar mass of the compound by the molar mass of the empirical formula.
6.	Multiply the number of each type of atom in the empirical formula by the answer to step 5 to find the molecular formula.

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, $\underline{C_8H_{16}}$ .