Question

What is the amount of free $\text{S}{{\text{O}}_{3}}$ in an oleum sample that is labelled as 118%?
A. 40%
B. 50%
C. 70%
D. 80%

Answer 1

Hint: Labelling of oleum will tell us about the amount of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ that will be formed if water is added to it. Assume the mass of $\text{S}{{\text{O}}_{3}}$ and solve the question according to the reaction $\text{S}{{\text{O}}_{3}}+{{\text{H}}_{2}}\text{O}\to {{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$, by using moles and masses. Find the percentage of free $\text{S}{{\text{O}}_{3}}$ by the formula; $\dfrac{\text{mass of S}{{\text{O}}_{3}}}{\text{mass of sample}}\times 100$.

Complete answer:
Let us solve the entire question step by step,
Step (1)- Let the mass of $\text{S}{{\text{O}}_{3}}$ be x grams and mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ be (100-x) grams in 100 grams of oleum. The reaction of $\text{S}{{\text{O}}_{3}}$ with water to form sulphuric acid is $\text{S}{{\text{O}}_{3}}+{{\text{H}}_{2}}\text{O}\to {{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$. It is clear that one mole of $\text{S}{{\text{O}}_{3}}$ reacts with one mole of ${{\text{H}}_{2}}\text{O}$ to form one mole of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$. The moles of $\text{S}{{\text{O}}_{3}}$ are $\dfrac{\text{x}}{80}$ as we know that molecular mass of $\text{S}{{\text{O}}_{3}}$ is 80 grams (sum of atomic mass of three oxygen atoms, $3\times 16$ or 48 grams and one atom of sulphur, 32 grams). So, moles of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ formed is $\dfrac{\text{x}}{80}$.
Step (2)- The weight of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ that would be formed in the reaction will be $\dfrac{\text{x}\times \text{98}}{80}$ . As, the molecular mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ is 98 grams. Sum of mass of two atoms of hydrogen $\left( 1\times 2 \right)$ or 2 grams, mass of sulphur 32 grams and mass of four oxygen atoms $\left( 4\times 16 \right)$ or 64 grams is 98 grams. The formula used here is $\text{given weight= moles}\times \text{molar mass}$.
Step (3)- The total mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ that would be formed in the process when water is added to 118% oleum is 118 grams. The equation to solve this will be weight of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ formed + mass of ${{\text{H}}_{2}}\text{S}{{\text{O}}_{4}}$ already present in oleum gives total mass of oleum. So, the expression will be $\dfrac{98\text{x}}{80}+\left( 100-\text{x} \right)=118$, the value of x will be 80 grams.
Step (4)- Percentage of free $\text{S}{{\text{O}}_{3}}$ will be $\dfrac{\text{mass of S}{{\text{O}}_{3}}}{\text{mass of sample}}\times 100$, mass of $\text{S}{{\text{O}}_{3}}$ is 80 grams. Mass of the sample is 100 grams (assumed). So, the percentage will be $\dfrac{80}{100}\times 100$ or $80%$.
$80%$ is the amount of free $\text{S}{{\text{O}}_{3}}$ in an oleum sample that is labelled as 118%.

The correct option is option ‘d’.

Note:
Oleum is used in sulphonation reactions. Benzene sulphonic acid is produced when oleum is treated with benzene. In that reaction also, $\text{S}{{\text{O}}_{3}}$ is formed as a neutral electrophile which later attacks on the benzene ring to give the product $\left( {{\text{C}}_{6}}{{\text{H}}_{5}}-\text{S}{{\text{O}}_{3}}\text{H} \right)$.