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.
