Stoichiometry
10.0 Percentage Labeling of Oleum
10.0 Percentage Labeling of Oleum
Oleum or fuming sulphuric acid contains $S{O_3}$ gas dissolved in sulphuric acid. When water is added to oleum, $S{O_3}$ reacts with ${H_2}O$ to form ${H_2}S{O_4},$ thus mass of the solution increases.
\[S{O_3} + {\text{ }}{H_2}O \to {H_2}S{O_4}\]
The total mass of ${H_2}S{O_4},$ obtained by diluting $100g$ of sample of oleum with desired amount of water, is equal to the percentage labeling of oleum.
$\therefore $ % labelling of oleum = Total mass of ${H_2}S{O_4}$ present in oleum after dilution.
= mass of ${H_2}S{O_4}$ initially present + mass of ${H_2}S{O_4}$ produced on dilution.
