# Question c06a7

Mar 24, 2016

Here's what's going on here.

#### Explanation:

Oleum is simply a saturated solution of sulfur trioxide, ${\text{SO}}_{3}$, dissolved in sulfuric acid, ${\text{H"_2"SO}}_{4}$.

You'll sometimes see this oleum referred to as disulfuric acid, ${\text{H"_2"S"_2"O}}_{7}$, or fuming sulfuric acid.

Oleum will always have a percentage of sulfuric acid that is greater than 100%. This happens because the percentage of sulfuric acid includes the mass of water needed to react with the dissolved sulfur trioxide to form sulfuric acid

${\text{SO"_text(3(aq]) + "H"_2"O"_text((l]) -> "H"_2"SO}}_{\textrm{4 \left(a q\right]}}$

In essence, the concentration of oleum tells you how much water must combine with the free sulfur trioxide present in $\text{100 g}$ of oleum in order to produce sulfuric acid.

Let's take, for example, a 10color(red)(5)% oleum solution. The percent concentration tells you that you need $\textcolor{red}{5} \textcolor{w h i t e}{a} \text{g}$ of water in order to react with all the free sulfur trioxide present in $\text{100 g}$ of this oleum sample to form $\text{105 g}$ of sulfuric acid.

Take a look at the above equation. Notice that one mole of sulfur trioxide will react with one mole of water to form one mole of sulfuric acid.

Use the molar masses of the three compounds to determine the ratio that exists between their masses.

${\text{For SO"_3: " "" "M_M ~~ "80 g mol}}^{- 1}$

${\text{For H"_2"O": " "" "M_M ~~ "18 g mol}}^{- 1}$

${\text{For H"_2"SO"_4:" " M_M ~~ "98 g mol}}^{- 1}$

So, this tells you that $\text{80 g}$ of sulfur trioxide reacts with $\text{18 g}$ of water to produce $\text{40 g}$ of sulfuric acid.

In this example, $\textcolor{red}{5}$ grams of water will react with

5 color(red)(cancel(color(black)("g water"))) * "80 g SO"_3/(18color(red)(cancel(color(black)("g water")))) = "22.22 g SO"_3

This means that $\text{100 g}$ of 105% oleum will contain $\text{22.22 g}$ of sulfur trixoide.

The percent of free sulfur trioxide will thus be $\text{22.2%}$.

Implicitly the percent of sulfuric acid in this oleum will be

"% H"_2"SO"_4 = 100 - 22.22 = 77.78%

Notice that this oleum solution will produce $\text{105 g}$ of sulfuric acid. If $\text{5 g}$ of water react with $\text{22.22 g}$ of sulfur trioxide, you will get

5color(red)(cancel(color(black)("g water"))) * ("98 g H"_2"SO"_4)/(18color(red)(cancel(color(black)("g water")))) = "27.22 g H"_2"SO"_4#

Since the $\text{100-g}$ sample of oleum contains $\text{77.78 g}$ of sulfuric acid, you will end up with

${\text{77.78 g " + " 27.22 g" = "105 g H"_2"SO}}_{4}$

So, as a conclusion, the percentage of free sulfur trioxide tells you much sulfur trioxide you get per $\text{100 g}$ of a given oleum solution.