What is oleum?
Oleum is a solution of "SO"_3 in "H"_2"SO"_4.
Addition of water to oleum converts the free "SO"_3 into "H"_2"SO"_4, and the resulting solution will contain only "H"_2"SO"_4.
"SO"_3 + "H"_2"O" → "H"_2"SO"_4
What is your sample?
Your sample contains 50% "free "SO"_3" and 50 % "combined "SO"_3"" as "H"_2"SO"_4.
How do we express "SO"_3 as % oleum?
% Oleum is the mass of water required to react with the free "SO"_3 in 100 g of oleum.
The reaction is
M_text(r): 80.06color(white)(m)18.02color(white)(mml) 98.08
color(white)(mmll)"SO"_3 + "H"_2"O" → "H"_2"SO"_4
If we use 100 g of your oleum, it will contain 50 g of "SO"_3
"Mass of H"_2"O" = 50 color(red)(cancel(color(black)("g SO"_3))) × ("18.02 g H"_2"O")/(80.06 color(red)(cancel(color(black)("g SO"_3)))) = "11.25 g H"_2"O"
So, the sample would be classed as 11.25 % oleum or 111.25 % "H"_2"SO"_4.
