Oleum and its % labelling - Practice Questions & MCQ

Oleum and its % labelling

Oleum is a mixture of SO₃ dissolved in 100% H₂SO₄.  The strength of the Oleum sample is expressed in terms of  % labelling and it is defined as the grams of pure H₂SO₄ that can be obtained from 100 g of the Oleum sample upon dilution with water.

For example, if an Oleum sample is labelled as 109%, it means that upon addition of 9g water to 100 g of Oleum sample, the amount of H₂SO₄ obtained is 109 g.

We can calculate the % of free SO₃ in the sample by simple stoichiometry as follows:

Weight of H₂O added = 9 g

Moles of H₂O added = 0.5

 Moles of SO₃ present = 0.5 (from reaction stoichiometry)

 Weight of SO₃ in the 100g Oleum sample = 0.5  80 = 40 g

 % of free SO₃ in Oleum = 40 %

Let us also discuss a general case 

Let us suppose that the % labelling of Oleum sample is y %. This means that (y-100) g of water is added to 100g Oleum sample

In this manner, the % of free SO₃ in Oleum sample can be calculated.