What is the % of free SO₃ in an oleum that is labelled as${ }^{\text{'}}104.5\mathrm{\%}{\mathrm{H}}_2{{\mathrm{SO}}_4}^{\mathrm{\prime }}$ ?

${\mathrm{H}}_2\mathrm{O}+{\mathrm{SO}}_3⟶{\mathrm{H}}_2{\mathrm{SO}}_4$
$18\mathrm{gr}\text{ of }{\mathrm{H}}_2\mathrm{O}=80\mathrm{gr} {\mathrm{SO}}_3$
$4.5\mathrm{gr}\text{ of }{\mathrm{H}}_2\mathrm{O}=20\mathrm{gr}{\mathrm{SO}}_3$
$\therefore 100\mathrm{gr}\text{ of oleum continues }20\mathrm{gr}{\mathrm{SO}}_3$
$\text{ (or) }20\mathrm{\%}\text{ free }{\mathrm{SO}}_3$