Calculate the `%` of free `SO_(3)` in oleum ( a solution of `SO_(3)` in `H_(2)SO_(4)`) that is labelled `109% H_(2)SO_(4)` by weight.

Answer

To calculate the percentage of free SO₃ in oleum that is labeled as 109% H₂SO₄ by weight, we can follow these steps:

### Step 1: Understand the meaning of 109% H₂SO₄
The label "109% H₂SO₄" means that in 100 grams of oleum, there are 109 grams of H₂SO₄. This indicates that the solution contains more H₂SO₄ than what would be present if it were a pure solution, implying the presence of free SO₃.

### Step 2: Calculate the total mass of the oleum
Since the oleum is labeled as 109% H₂SO₄, we can say that in 100 grams of oleum, there are 109 grams of H₂SO₄. However, this is not possible since the total mass cannot exceed 100 grams. Therefore, we need to interpret this correctly: the 109 grams of H₂SO₄ includes the mass contributed by free SO₃.

### Step 3: Determine the mass of free SO₃
To find the mass of free SO₃ in 100 grams of oleum, we need to calculate how much SO₃ is present. We know that when SO₃ combines with water (H₂O), it forms H₂SO₄. The reaction is as follows:
\[ \text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 \]

From the stoichiometry of the reaction:
- 1 mole of SO₃ (80 grams) combines with 1 mole of H₂O (18 grams) to form 1 mole of H₂SO₄ (98 grams).

### Step 4: Calculate the amount of water needed
Since we have 109 grams of H₂SO₄, we need to find out how much SO₃ is present. The excess of H₂SO₄ can be attributed to the free SO₃.

To find the mass of SO₃ that would combine with water to form H₂SO₄:
- If we have 100 grams of oleum, we can assume that the excess H₂SO₄ is due to the presence of free SO₃.
- The amount of H₂SO₄ formed from SO₃ can be calculated as follows:
- For every 80 grams of SO₃, we get 98 grams of H₂SO₄.

Let \( x \) be the mass of SO₃ in grams. The mass of H₂SO₄ formed from \( x \) grams of SO₃ can be calculated using the ratio:
\[ \frac{98 \text{ g H}_2\text{SO}_4}{80 \text{ g SO}_3} = \frac{y \text{ g H}_2\text{SO}_4}{x \text{ g SO}_3} \]
This gives us:
\[ y = \frac{98}{80} \times x = 1.225 \times x \]

### Step 5: Set up the equation
Since the total mass of H₂SO₄ in the oleum is 109 grams, we can set up the equation:
\[ y + (100 - y) = 109 \]
Substituting for \( y \):
\[ 1.225x + (100 - 1.225x) = 109 \]
Solving for \( x \):
\[ 100 - 0.225x = 109 \\ -0.225x = 109 - 100 \\ -0.225x = 9 \\ x = \frac{9}{0.225} \approx 40 \text{ grams of SO}_3 \]

### Step 6: Calculate the percentage of free SO₃
Now, we can calculate the percentage of free SO₃ in 100 grams of oleum:
\[ \text{Percentage of free SO}_3 = \left(\frac{40 \text{ g SO}_3}{100 \text{ g oleum}}\right) \times 100 = 40\% \]

### Final Answer:
The percentage of free SO₃ in oleum is **40%**.