Oleum 109 % in sulfuric acid means that in 100 g of oleum there is enough SO_3 to make 109 grams of sulfuric acid, once you add the same amount of water in moles, following the reaction:
SO_3 + H_2O -> H_2SO_4
If x is the unknown mass percentage of SO_3, the moles in it must correspond to 9 + x grams of sulfuric acid.
Given the ratio between molar masses of SO_3 and H_2SO_4, that is f = 80.066/98.079 = 0.81634 we have the following mass equation:
x/f = x + 9g
That means: the grams of SO_3 in 100 grams of oleum, when transformed in H_2SO_4 (dividing by f) must become x + 9 grams of sulfuric acid.
This is to minimize algebra and maximize logical and qualitative thinking.
Rearranging the equation to find x we get:
x = 9g*f/(1 - f) = 9g*0.81634/(1-0.81634) = 40.00 g
In fact, if you divide the resulting 40 grams of SO_3 by the factor f = 0.81634 you get 49.00 grams of sulfuric acid as the corresponding amount in moles.
