# Question #1cf45

Aug 14, 2017

40,0%

#### Explanation:

Oleum 109 % in sulfuric acid means that in 100 g of oleum there is enough $S {O}_{3}$ to make 109 grams of sulfuric acid, once you add the same amount of water in moles, following the reaction:

$S {O}_{3} + {H}_{2} O \to {H}_{2} S {O}_{4}$

If $x$ is the unknown mass percentage of $S {O}_{3}$, the moles in it must correspond to 9 + $x$ grams of sulfuric acid.

Given the ratio between molar masses of $S {O}_{3}$ and ${H}_{2} S {O}_{4}$, that is $f = \frac{80.066}{98.079} = 0.81634$ we have the following mass equation:

$\frac{x}{f} = x + 9 g$

That means: the grams of $S {O}_{3}$ in 100 grams of oleum, when transformed in ${H}_{2} S {O}_{4}$ (dividing by $f$) must become $x + 9$ grams of sulfuric acid.

This is to minimize algebra and maximize logical and qualitative thinking.

Rearranging the equation to find $x$ we get:

$x = 9 g \cdot \frac{f}{1 - f} = 9 g \cdot \frac{0.81634}{1 - 0.81634} = 40.00 g$

In fact, if you divide the resulting 40 grams of $S {O}_{3}$ by the factor $f = 0.81634$ you get 49.00 grams of sulfuric acid as the corresponding amount in moles.