Estimation of Carbon and Hydrogen

A known mass of the organic compound is burnt in excess oxygen in the presence of copper(II) oxide (to ensure complete combustion). All the carbon in the compound is converted to $\ce{CO2}$ and all the hydrogen to $\ce{H2O}$:

$\ce{C_xH_y + O2 ->[CuO] x CO2 + \frac{y}{2} H2O}$

The products are passed through two absorption chambers:

1. Anhydrous calcium chloride ( $\ce{CaCl2}$ ) — absorbs $\ce{H2O}$
2. Potassium hydroxide ( $\ce{KOH}$ ) — absorbs $\ce{CO2}$

The increase in mass of each absorber gives the mass of $\ce{H2O}$ ( $m_1$ g) and $\ce{CO2}$ ( $m_2$ g) produced from a known mass of compound ( $m$ g).

Since all the carbon in the compound ends up in $\ce{CO2}$:

• Molar mass of $\ce{CO2}$ = 44 g/mol; contains 12 g of C per mole
• So 44 g of $\ce{CO2}$ contains 12 g of C

$\text{Percentage of C} = \frac{12}{44} \times \frac{m_2}{m} \times 100$

Since all hydrogen ends up in $\ce{H2O}$:

• Molar mass of $\ce{H2O}$ = 18 g/mol; contains 2 g of H per mole
• So 18 g of $\ce{H2O}$ contains 2 g of H

$\text{Percentage of H} = \frac{2}{18} \times \frac{m_1}{m} \times 100$

Where $m$ = mass of organic compound, $m_1$ = mass of $\ce{H2O}$ formed, $m_2$ = mass of $\ce{CO2}$ formed.