Oleum is considered as a solution of $S{O}_3$ in $H_{2}S{O}_4,$ which is obtained by passing $S{O}_3$ in solution of $H_{2}S{O}_4.$ When 100g sample of oleum is diluted with desired mass of $H_{2}O$ then the total mass of $H_{2}S{O}_4$ obtained after dilution is known as % labeling in oleum. For example, a oleum bottle labelled as 

 $\text{'}109\% H_{2}S{O}_4\text{'}$ means the 109 g total mass of pure $H_{2}S{O}_4$ will be formed when 100g of oleum is diluted by 9g of $H_{2}O$ which combines with all the free $S{O}_3$ present in oleum to form $H_{2}S{O}_4$ as $S{O}_3+H_{2}O\stackrel{ }{\to }{H}_{2}S{O}_4.$