The proportion of an element in a given compound does not vary.

The proportion of carbon in carbon dioxide will always be \(\frac{12}{44}\)

The proportion of hydrogen in water will always be \(\frac{2}{18} = \frac{1}{9}\)

Example 1:

Suppose a 1.87g sample of an organic compound is completely burned, producing 2.65g of carbon dioxide and 1.63g of water. This information can be used to find the empirical formula of the organic compound.

Answer:

Mass of carbon in organic compound sample = \( 2.65 \times \frac{12}{44} 0.723g\)

Mass of hydrogen in organic compound sample = \(1.63 \times \frac{1}{9} = 0.181g\)

These two masses add up to give 0.904g - we know that the remaining mass in the organic compound must be made up of oxygen.

Therefore, there was 0.966g of oxygen. The usual table can now be made to find the empirical formula.

The table shows that the empirical formula is \(CH_3O\)