# Question #6547e

Jun 11, 2017

I get 111.25%.

#### Explanation:

What is oleum?

Oleum is a solution of ${\text{SO}}_{3}$ in ${\text{H"_2"SO}}_{4}$.

Addition of water to oleum converts the free ${\text{SO}}_{3}$ into ${\text{H"_2"SO}}_{4}$, and the resulting solution will contain only ${\text{H"_2"SO}}_{4}$.

${\text{SO"_3 + "H"_2"O" → "H"_2"SO}}_{4}$

Your sample contains 50% "free ${\text{SO}}_{3}$" and 50 % "combined $\text{SO"_3}$" as ${\text{H"_2"SO}}_{4}$.

How do we express ${\text{SO}}_{3}$ as % oleum?

% Oleum is the mass of water required to react with the free ${\text{SO}}_{3}$ in 100 g of oleum.

The reaction is

${M}_{\textrm{r}} : 80.06 \textcolor{w h i t e}{m} 18.02 \textcolor{w h i t e}{m m l} 98.08$
$\textcolor{w h i t e}{m m l l} {\text{SO"_3 + "H"_2"O" → "H"_2"SO}}_{4}$

If we use 100 g of your oleum, it will contain 50 g of ${\text{SO}}_{3}$

$\text{Mass of H"_2"O" = 50 color(red)(cancel(color(black)("g SO"_3))) × ("18.02 g H"_2"O")/(80.06 color(red)(cancel(color(black)("g SO"_3)))) = "11.25 g H"_2"O}$

So, the sample would be classed as 11.25 % oleum or 111.25 % ${\text{H"_2"SO}}_{4}$.