What is oleum?
We can consider oleum as 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 109 % oleum?
109 % oleum means that we must add 9 g of water to 100 g of the oleum to convert all the "SO"_3 to "H"_2"SO"_4.
How much "SO"_3 is in 109 % oleum?
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
"Mass of SO"_3 = 9 color(red)(cancel(color(black)("g H"_2"O"))) × ("80.06 g SO"_3)/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "40 g SO"_3
So, 100 g of 109 % oleum contains 40 g "SO"_3 and 60 g "H"_2"SO"_4.
Calculate the mass percent of oxygen in 109 % oleum
In "SO"_3,
"Mass of O" = 40 color(red)(cancel(color(black)("g SO"_3))) × "48.00 g O"/(80.06 color(red)(cancel(color(black)("g SO"_3)))) = "24 g O"
In "H"_2"SO"_4,
"Mass of O" = 60 color(red)(cancel(color(black)("g H"_2"SO"_4))) × "64.00 g O"/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "39 g O"
"Total mass of O = 24 g+ 39 g = 63 g O"
"% O" = (63 color(red)(cancel(color(black)("g"))))/(100 color(red)(cancel(color(black)("g")))) × 100 % = 63 %
