Percentage composition and combustion analysis

​​In a nutshell

The empirical formula is the simplest whole number ratio of each element in a compound. The molecular formula gives the actual number of atoms of each element in a compound. The molecular formula can be found using the empirical formula. Knowing the molecular formula can help you identify the structure of an unknown compound.

Percentage composition

The empirical formula is the simplest whole number ratio of atoms of each element in a compound. The molecular formula can be found from the empirical formula, given the percentage composition.

procedure

Example 

A compound has a molecular mass of $$204$$​. It has the following percentage composition for each type of atom: $$70.5\%$$ carbon, $$13.8\%$$ hydrogen and $$15.7\%$$ oxygen. Calculate the empirical and molecular formula for this compound.

Firstly, calculate the number of moles for each element in the compound:   

$$n \, (C)\, = \, \dfrac{70.5}{12} \, = 5.875 \, mol$$

$$n \, (H)\, = \, \dfrac{13.8}{1} \, = 13.8 \, mol$$​​

$$n \, (O)\, = \, \dfrac{15.7}{16} \, = 0.98125 \, mol$$​​

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

$$C \, = \, \dfrac{5.875}{0.98125} \, = \, 5.99$$​​

$$H \, = \, \dfrac{13.8}{0.98125} \, = \, 14.06$$​​

$$O \, = \, \dfrac{0.98125}{0.98125} \, = \, 1.00$$​​

Write this as a ratio for each element:

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

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

Find the molar mass of the empirical formula:

$$M_r (C_6H_{14}O) = (6 \times 12) + (14 \times 1) + (1 \times 16) = 102 \, gmol^{-1}$$​​

The molecular mass is double the molar mass of the empirical formula. This is known by dividing the molecular mass given by the molar mass of the empirical formula:

$$\dfrac{204}{102} = 2$$​​

To find the molecular formula, multiply the empirical formula by two:

$$C_{(2\, \times \, 6)}H_{(2 \, \times \, 14)} \, O_{(2 \times 1)} = \, C_{12}H_{28}O_2$$​​

Therefore the molecular formula of the compound is $$\underline {C_{12}H_{28}O_2}$$​.

Combustion analysis

Organic compounds which undergo complete combustion form carbon dioxide and water as products. All the carbon and hydrogen atoms in carbon dioxide and water respectively, come from the organic compound. 

Calculating molecular formula using mass

The molecular formula can be calculated from the empirical formula using mass.

procedure 

Example

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

Firstly, find the number of moles of $$CO_2$$ and $$H_2O$$ using the masses provided in the question:

$$n \, (CO_2) \, = \, \dfrac{4.4}{12 \, + \, (16 \, \times \, 2)} \, = \, \dfrac{4.4}{44} \, = \, 0.1 \, mol$$​​

$$n \, (H_2O) \, = \, \dfrac{1.8}{(1 \, \times \, 2) \, + \, 16} \, = \, \dfrac{1.8}{18} \, = \, 0.1 \, mol$$​​

​​

Work out the number of moles of carbon atoms in the hydrocarbon:

$$1 \, mole$$ of $$CO_2$$ contains $$1 \, mole$$ of $$C$$ atoms

Therefore, there are $$0.1 \ moles$$ of $$C$$ atoms

Now work out the number of moles of hydrogen atoms in the hydrocarbon:

$$1 \, mole$$ of $$H_2O$$ contains $$2 \, moles$$ of $$H$$​ atoms 

$$0.1 \, \times \, 2 \, = \, 0.2 \, moles$$​ of $$H$$​ 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 the elements to find the empirical formula:

$$CH_2$$​​

Find the molar mass of the empirical formula:  

$$M_r \, (CH_2) \, = \, (12 \, \times \, 1) \,+\, (1 \, \times \, 2) \, = \, 14 \, gmol^{-1}$$

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

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

Multiply the number of atoms for each element in the empirical formula by eight to find the molecular formula:

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

Therefore, the molecular formula of the compound is $$\underline {C_8H_{16}}$$​.

Calculating molecular formula using volume

Combustion can occur in the gaseous state, therefore volume can be used to calculate combustion equations. This is because gases have the same molar volume when at the same temperature and pressure. 

Example

What is the molecular formula of $$X$$​, given that $$60cm^3$$ of $$X$$ combusts completely with $$480cm^3$$ of oxygen? $$300cm^3$$ of carbon dioxide is produced.

​

Firstly, write out the equation using the volumes provided:

$$60X \, + \, 480 \, O_2 \rightarrow 300 \,CO_2 \,+ \, n \,H_2O$$

​​

Next, simplify the equation. For this example you can divide by 60:

$$X \, + \, 8 \, O_2 \rightarrow 5 \,CO_2 \,+ \, nH_2O$$

​​

Any oxygen atoms that are not in carbon dioxide must be in water. This enables you to find the number of water molecules present:

$$n \, = (8 \times\,2) - (5 \times\,2) = 6$$

​​

This implies that there are six molecules of water. So the balanced equation is:

$$X \, + \, 8 \, O_2 \rightarrow 5 \,CO_2 \,+ \, 6 \,H_2O$$

​​

The equation can be used to identify $$X$$​ as all the carbon atoms will be from molecule $$X$$ to give carbon dioxide and the hydrogen atoms from $$X$$ is found in water. This tells us that there are $$5$$​ carbon atoms and $$12$$ hydrogen atoms in $$X$$. 

Therefore, the molecular formula of $$\underline{X \ is \ C_5H_{12}}$$​.