What is the percentage of free ${\mathrm{SO}}_3$ in an oleum sample that is labelled as $104.5\mathrm{\%}{\mathrm{H}}_2{\mathrm{SO}}_4$?

Oleum is mixture of  ${\mathrm{H}}_2{\mathrm{SO}}_4$ and ${\mathrm{SO}}_3$ gas. When we add water to oleum, ${\mathrm{SO}}_3$ present in it reacts with water to from ${\mathrm{H}}_2{\mathrm{SO}}_4$ according to the following equation :

${\mathrm{SO}}_3+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4$

$104.5\mathrm{\%}$labeled oleum means 100g oleum reacts with ${\mathrm{H}}_2\mathrm{O}$ to form $104.5\mathrm{\%}{\mathrm{H}}_2{\mathrm{SO}}_4$ Hence mass of water is 4.5g.
According to stoichiometry coefficient 

18g water combines with 80g ${\mathrm{SO}}_3$.

$\therefore$4.5g of ${\mathrm{H}}_2\mathrm{O}$ combines with 20g of ${\mathrm{SO}}_3$.

$\therefore$ 100g of oleum contains 20g of ${\mathrm{SO}}_3$

Hence 20% of ${\mathrm{SO}}_3$ was free.