# Question 9af40

Jun 2, 2017

The mass percent of oxygen in 109 % oleum is 63 %.

#### Explanation:

What is oleum?

We can consider oleum as 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 ${\text{H"_2"SO}}_{4}$.

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

What is 109 % oleum?

109 % oleum means that we must add 9 g of water to 100 g of the oleum to convert all the ${\text{SO}}_{3}$ to ${\text{H"_2"SO}}_{4}$.

How much ${\text{SO}}_{3}$ is in 109 % oleum?

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

${\text{Mass of SO"_3 = 9 color(red)(cancel(color(black)("g H"_2"O"))) × ("80.06 g SO"_3)/(18.02 color(red)(cancel(color(black)("g H"_2"O")))) = "40 g SO}}_{3}$

So, 100 g of 109 % oleum contains 40 g ${\text{SO}}_{3}$ and 60 g ${\text{H"_2"SO}}_{4}$.

Calculate the mass percent of oxygen in 109 % oleum

In ${\text{SO}}_{3}$,

$\text{Mass of O" = 40 color(red)(cancel(color(black)("g SO"_3))) × "48.00 g O"/(80.06 color(red)(cancel(color(black)("g SO"_3)))) = "24 g O}$

In ${\text{H"_2"SO}}_{4}$,

$\text{Mass of O" = 60 color(red)(cancel(color(black)("g H"_2"SO"_4))) × "64.00 g O"/(98.08 color(red)(cancel(color(black)("g H"_2"SO"_4)))) = "39 g O}$

$\text{Total mass of O = 24 g+ 39 g = 63 g O}$

"% O" = (63 color(red)(cancel(color(black)("g"))))/(100 color(red)(cancel(color(black)("g")))) × 100 % = 63 %#