Q.An organic compound contains 69% carbon and 4.8% hydrogen, the remainder being oxygen. Calculate the masses of carbon dioxide and water produced when 0.20 g of this compound is subjected to complete combustion.

The key to this problem lies in recognizing the industrial chemistry of sulfuric acid production and understanding what "S–O–S bonds" means structurally. Iron pyrites is $\mathrm{FeS_2}$, and when roasted in air it produces sulfur dioxide. The phrase "catalytic oxidation" points directly to the Contact process, where $\mathrm{SO_2}$ is converted to $\mathrm{SO_3}$ over a vanadium pentoxide catalyst. The real insight comes from identifying which sulfur compound contains an S–O–S linkage — that's the peroxodisulfate or pyrosulfate family.

Let me trace the chemistry step by step:

1. Roasting iron pyrites produces sulfur dioxide When $\mathrm{FeS_2}$ is heated in air:

This is the first gas mentioned in the problem.

2. Catalytic oxidation converts $\mathrm{SO_2}$ to $\mathrm{SO_3}$ In the Contact process, sulfur dioxide is oxidized over a $\mathrm{V_2O_5}$ catalyst at around 450°C:

So compound X is $\mathrm{SO_3}$.

3. Dissolving $\mathrm{SO_3}$ in concentrated $\mathrm{H_2SO_4}$ forms oleum Direct dissolution of $\mathrm{SO_3}$ in water is violently exothermic and produces a corrosive mist. Industrially, $\mathrm{SO_3}$ is absorbed in concentrated sulfuric acid to form oleum (fuming sulfuric acid):

So compound Y is $\mathrm{H_2SO_4}$.

4. Identifying compound Z with S–O–S bonds …

Concept & Intuition

Mass percentage (or weight/weight percentage) tells you how many grams of solute are present in every 100 grams of solution. The key is to remember that the solution mass = solute mass + solvent mass. A common mistake is to divide by the solvent mass alone — that would give a different ratio (like parts per hundred of solvent, not solution). Here, water’s density is 1 g/mL, so 225 mL of water has a mass of 225 g. That makes the total mass easy to find.

Step-by-step

1. Identify the solute and solvent masses.

   Solute: NaCl = 25 g.

   Solvent: water = 225 mL. Since the density of water is 1 g/mL, the mass of water is $225\ \text{g}$.

2. Find the total mass of the solution.

3. Apply the mass percentage formula. $$$$

The key to this problem is understanding what "118% oleum" actually means. Oleum is a solution of sulfur trioxide ($SO_3$) in pure sulfuric acid ($H_2SO_4$). The percentage given — here 118% — is a mass percentage that refers to the mass of $H_2SO_4$ that would be obtained if all the $SO_3$ in the oleum were converted to $H_2SO_4$ by adding water.

So 100 g of 118% oleum, when diluted with enough water, yields 118 g of $H_2SO_4$. That extra 18 g comes from the reaction of $SO_3$ with water: $SO_3 + H_2O \rightarrow H_2SO_4$. This means the oleum contains some free $SO_3$ that is not already combined as $H_2SO_4$.

Let’s work through the neutralisation step by step.

1. Find the mass of free $SO_3$ in 100 g of oleum.

   If the oleum were pure $H_2SO_4$, 100 g would give 100 g of $H_2SO_4$ on dilution. But here we get 118 g — an extra 18 g of $H_2SO_4$. That extra $H_2SO_4$ comes from the reaction of free $SO_3$ with water.

   The molar mass of $SO_3$ is 80 g/mol, and of $H_2SO_4$ is 98 g/mol.

   From the reaction $SO_3 + H_2O \rightarrow H_2SO_4$, 80 g of $SO_3$ produces 98 g of $H_2SO_4$.

   Let $x$ be the mass of free $SO_3$ in 100 g oleum. Then the extra $H_2SO_4$ produced is $\frac{98}{80}x = 1.225x$ grams.

   This extra mass equals 18 g (since 118 − 100 = 18).

   So $1.225x = 18 \implies x = \frac{18}{1.225} = 14.69$ g (approximately).

   More precisely, $x = \frac{18 \times 80}{98} = \frac{1440}{98} = \frac{720}{49}$ g.

2. Find the moles of free $SO_3$.

   Moles of $SO_3 = \frac{\text{mass}}{\text{molar mass}} = \frac{720/49}{80} = \frac{720}{49 \times 80} = \frac{9}{49}$ mol.

   That’s about 0.1837 mol.

3. Find the mass of $H_2SO_4$ already present in the oleum.

   Total mass of oleum = mass of $H_2SO_4$ + mass of $SO_3$ = 100 g.

   So mass of $H_2SO_4$ = $100 - \frac{720}{49} = \frac{4900 - 720}{49} = \frac{4180}{49}$ g.

   Moles of $H_2SO_4 = \frac{4180/49}{98} = \frac{4180}{49 \times 98} = \frac{4180}{4802} = \frac{2090}{2401}$ mol.

   That’s about 0.870 mol.

4. Write the neutralisation reactions with NaOH.

   • $H_2SO_4 + 2NaOH \rightarrow Na_2SO_4 + 2H_2O$ So 1 mol $H_2SO_4$ requires 2 mol NaOH.
   • $SO_3 + 2NaOH \rightarrow Na_2SO_4 + H_2O$ (Note: $SO_3$ first reacts with water to give $H_2SO_4$, which then neutralises; net effect is 1 mol $SO_3$ also requires 2 mol NaOH.) So 1 mol $SO_3$ requires 2 mol NaOH.

5. Calculate total moles of NaOH required.

   Moles of NaOH from $H_2SO_4$ = $2 \times \frac{2090}{2401} = \frac{4180}{2401}$ mol.

   Moles of NaOH from $SO_3$ = $2 \times \frac{9}{49} = \frac{18}{49}$ mol.

   Total = $\frac{4180}{2401} + \frac{18}{49}$.

   Convert to common denominator: $\frac{18}{49} = \frac{18 \times 49}{49 \times 49} = \frac{882}{2401}$.

   So total = $\frac{4180 + 882}{2401} = \frac{5062}{2401}$ mol.

   Simplify: $5062 \div 2401 \approx 2.108$? That doesn’t match the options — something’s off.

Let’s use the oleum percentage definition directly.

100 g of 118% oleum contains enough $SO_3$ to produce 118 g of $H_2SO_4$ on full dilution. That means the total sulfur content is equivalent to that in 118 g of $H_2SO_4$.

Moles of $H_2SO_4$ in 118 g = $\frac{118}{98} = \frac{59}{49}$ mol.

Each mole of $H_2SO_4$ (or $SO_3$) requires 2 moles of NaOH for complete neutralisation. …

The key insight is that percentage composition by weight tells us what fraction of the solution's mass is solute. When you pour two solutions together, you're simply adding their solutes together and their solvents together. The new percentage is the total solute mass divided by the total solution mass.

Think of it this way: if you have a glass of lemonade that's 20% sugar and another that's 30% sugar, pouring them together doesn't create some mysterious new concentration—you just count up all the sugar and all the liquid.

Step-by-step calculation

1. Find the solute mass in the first solution.

   A 20% solution by weight means 20 g of solute per 100 g of solution. For 200 g:

2. Find the solute mass in the second solution.

   A 30% solution contains 30 g of solute per 100 g of solution. For 300 g:

3. Calculate the total solute in the mixture.

   Simply add the two solute masses:

4. Calculate the total mass of the mixture. …

The concept here is percentage composition by mass. To compare which compound has the least nitrogen, we don’t need to memorize numbers — we just calculate the mass of nitrogen relative to the total mass of one mole of each compound. The lower the ratio, the lower the percentage.

Let’s work through each option step by step.

1. Ammonium sulphate — formula: $(\text{NH}_4)_2\text{SO}_4$

   • Nitrogen atoms: 2 (each N = 14 g/mol) → total N mass = $2 \times 14 = 28$ g/mol
   • Molar mass:
     • N: 28
     • H: $8 \times 1 = 8$
     • S: 32
     • O: $4 \times 16 = 64$
     • Total = $28 + 8 + 32 + 64 = 132$ g/mol
   • % N = $\frac{28}{132} \times 100 \approx 21.21\%$

2. Urea — formula: $\text{CO}(\text{NH}_2)_2$

   • Nitrogen atoms: 2 → total N mass = 28 g/mol
   • Molar mass:
     • C: 12
     • O: 16
     • N: 28
     • H: $4 \times 1 = 4$
     • Total = $12 + 16 + 28 + 4 = 60$ g/mol
   • % N = $\frac{28}{60} \times 100 \approx 46.67\%$

3. Ammonium nitrate — formula: $\text{NH}_4\text{NO}_3$

   • Nitrogen atoms: 2 → total N mass = 28 g/mol
   • Molar mass:
     • N: 28
     • H: $4 \times 1 = 4$
     • O: $3 \times 16 = 48$
     • Total = $28 + 4 + 48 = 80$ g/mol
   • % N = $\frac{28}{80} \times 100 = 35.00\%$

4. Calcium nitrate — formula: $\text{Ca}(\text{NO}_3)_2$

   • Nitrogen atoms: 2 → total N mass = 28 g/mol
   • Molar mass:
     • Ca: 40
     • N: 28
     • O: $6 \times 16 = 96$
     • Total = $40 + 28 + 96 = 164$ g/mol …

Concept & Intuition

Percentage composition is a straightforward idea: it’s the fraction of the total mass contributed by a particular element, multiplied by 100. For nitrogen compounds, we compare how much nitrogen is “packed” into each gram of the compound. A heavier non‑nitrogen part (like calcium or extra oxygen) dilutes the nitrogen content. So we expect compounds with large, heavy counterions to have lower nitrogen percentages.

Step‑by‑Step Calculation

1. Ammonium sulphate – formula: $(\text{NH}_4)_2\text{SO}_4$

   • Nitrogen atoms: 2 × 14 = 28 g/mol
   • Molar mass:
     • N: 2×14 = 28
     • H: 8×1 = 8
     • S: 32
     • O: 4×16 = 64
     • Total = 28 + 8 + 32 + 64 = 132 g/mol
   • %N = $\frac{28}{132} \times 100 \approx 21.21\%$

2. Urea – formula: $\text{CO(NH}_2)_2$

   • Nitrogen atoms: 2 × 14 = 28 g/mol
   • Molar mass:
     • C: 12
     • O: 16
     • N: 28
     • H: 4×1 = 4
     • Total = 12 + 16 + 28 + 4 = 60 g/mol
   • %N = $\frac{28}{60} \times 100 \approx 46.67\%$

3. Ammonium nitrate – formula: $\text{NH}_4\text{NO}_3$

   • Nitrogen atoms: 2 × 14 = 28 g/mol
   • Molar mass:
     • N: 28
     • H: 4×1 = 4
     • O: 3×16 = 48
     • Total = 28 + 4 + 48 = 80 g/mol
   • %N = $\frac{28}{80} \times 100 = 35.00\%$

4. Calcium nitrate – formula: $\text{Ca(NO}_3)_2$

   • Nitrogen atoms: 2 × 14 = 28 g/mol
   • Molar mass:
     • Ca: 40
     • N: 28
     • O: 6×16 = 96
     • Total = 40 + 28 + 96 = 164 g/mol
   • %N = $\frac{28}{164} \times 100 \approx 17.07\%$ …