What is the %SO3 in Oleum sample that is labelled as 104.5% H2SO4 ?.

SOLUTION: Given % labelling (X) =104.5%, Find %( Free) SO₃=?

Two sample of Oleum are labelled as 109% and 115%,what is the difference between weight of free SO3 in these samples ?.

Given % labelling (X)=109%  and 115% ,finddifference between weight of free SO₃ in these samples ?. 

 Difference between weight of free SO₃ in these samples are= 66.66-44= 26.67 gm

If the percentage free SO3 in an Oleum sample is 20% then label the sample of Oleum in term of percentage H2SO4.?

Percentage labelling of oleum Calculate as:

Calculate the % of free SO3 in an Oleum sample that is labelled as 118 % ?

We can calculate % labelling by Stoichiometric calculation as:

What is OLEUM and it's percentage labelling?

(1) Oleum can be represented by the formula ySO₃.H₂Owhere y is the total molar sulphur trioxide content .the value of y can be varied to different Oleum sample.

(2) Oleum also be expressed  as H₂SO₄.xSO3 where x is molar free suphur trioxide.
(3) Oleum is the solution of of sulphur trioxide in sulphuric acid , it is also known as  fuming sulphuric acid orPyrosulphuric acid (H₂S₂O₇=H₂SO₄+SO₃).
(4) Oleum sample contain two type of SO₃.

(i) Free SO3:- It is that SO₃ which combined with water water to give H₂SO₄ .

(ii) Combined SO3:- The SO₃ which is present in H₂SO₄ itself is known as combined SO₃ and it does not react with water.