80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :

Q. 80 mL of a hydrocarbon on mixing with 264 mL of oxygen in a closed U-tube undergoes complete combustion. The residual gases after cooling to 273 K occupy 224 mL. When the system is treated with KOH solution, the volume decreases to 64 mL. The formula of the hydrocarbon is :

(A) $C_2H_2$

(B) $C_2H_4$

(C) $C_4H_{10}$

(D) $C_2H_6$

Correct Answer: A

Option (A) is correct

Explanation

Let the hydrocarbon be $C_xH_y$. The balanced combustion reaction is: $$C_xH_y(g) + \left(x + \frac{y}{4}\right)O_2(g) \rightarrow xCO_2(g) + \frac{y}{2}H_2O(l)$$ Note: At 273 K (after cooling), $H_2O$ is in liquid state, so its volume is negligible.

Given: Volume of $C_xH_y$ = 80 mL Volume of $O_2$ initially = 264 mL Total Residual Volume (after cooling) = 224 mL (contains $CO_2$ and unreacted $O_2$) Volume after KOH treatment = 64 mL (this is unreacted $O_2$, as KOH absorbs $CO_2$)

Therefore: Volume of $CO_2$ produced = Total Residual - Unreacted $O_2$ = 224 - 64 = 160 mL From the reaction, 80 mL of $C_xH_y$ produces $80x$ mL of $CO_2$. $$80x = 160 \implies x = 2$$

Volume of $O_2$ reacted = Initial $O_2$ - Unreacted $O_2$ = 264 - 64 = 200 mL From the reaction, 80 mL of $C_xH_y$ reacts with $80(x + y/4)$ mL of $O_2$. $$80\left(2 + \frac{y}{4}\right) = 200$$ $$2 + \frac{y}{4} = \frac{200}{80} = 2.5$$ $$\frac{y}{4} = 0.5 \implies y = 2$$

The formula is $C_2H_2$.

Related Theory

1. Introduction to Eudiometry

Eudiometry is a laboratory technique used for the volumetric analysis of gases, particularly for determining the molecular formula of gaseous hydrocarbons or analyzing gas mixtures. The core principle relies on Gay-Lussac's Law of Gaseous Volumes, which states that when gases react, they do so in volumes that bear a simple whole-number ratio to each other and to the volumes of the gaseous products, provided temperature and pressure are constant.

2. General Combustion Equation

For any hydrocarbon containing carbon (C) and hydrogen (H), the general balanced equation for complete combustion is: $$C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$$ If the hydrocarbon also contains oxygen ($C_xH_yO_z$), the formula becomes: $$C_xH_yO_z + \left(x + \frac{y}{4} - \frac{z}{2}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$$

3. Handling Water ($H_2O$) in Eudiometry

A critical step in eudiometry problems is observing the temperature at which the final volume is measured:

If T > 373 K: Water is in the gaseous state (steam). Its volume must be included in the "residual gas" calculation.
If T < 373 K (e.g., Room Temp or 273 K): Water condenses into liquid. Since the volume of liquid water is nearly 1/1600th of its gaseous volume, it is treated as zero (negligible) in gas volume calculations.

In this question, the cooling to 273 K ensures water is liquid, simplifying the residual gas to just $CO_2$ and excess $O_2$.

4. Gas Absorption Reagents

The identification of gases in a mixture is done by passing the mixture through specific reagents that absorb certain gases:

Reagent	Gas Absorbed
KOH / NaOH solution	$CO_2$, $SO_2$, $Cl_2$
Alkaline Pyrogallol	Oxygen ($O_2$)
Ammoniacal $Cu_2Cl_2$	Carbon Monoxide ($CO$)
Turpentine Oil	Ozone ($O_3$)
Anhydrous $CaCl_2$ / $H_2SO_4$	Water Vapor ($H_2O$)

5. Solving Strategy for Hydrocarbons

To find 'x' and 'y' efficiently:

Contraction in Volume: Often, questions mention "contraction on combustion." This is (Initial Volume of Reactants - Final Volume of Gaseous Products). Contraction = $V_{hydrocarbon} + V_{O2(reacted)} - V_{CO2}$ (assuming water is liquid).
Carbon Balance: The volume of $CO_2$ produced is always equal to 'x' times the volume of the hydrocarbon.
Hydrogen Balance: Use the oxygen consumption data to solve for 'y'.

6. Common Mistakes to Avoid

Forgetting Oxygen in Compound: If the compound is an alcohol or ether, don't forget the 'z' term in the oxygen balance.
Units: Ensure all volumes are in the same units (mL or L). Eudiometry relies on ratios, so as long as units are consistent, the math works.
Excess Oxygen: "Residual gas" almost always contains the leftover oxygen. Many students mistake the residual gas for just $CO_2$.

7. Real-world Application and JEE Relevance

In industrial chemistry, Orsat apparatus uses these principles to analyze flue gases from furnaces to ensure complete combustion and environmental compliance. For JEE aspirants, this topic tests the intersection of Stoichiometry and the Gaseous State. It requires a firm grasp of balanced chemical equations and the ability to interpret laboratory observations (like "volume decrease") into mathematical constraints.

8. Degree of Unsaturation (DU)

Once the formula $C_2H_2$ is found, we can determine its structure. $DU = x - y/2 + 1 = 2 - 2/2 + 1 = 2$. This implies the molecule could have two double bonds or one triple bond. Given it's a simple $C_2$ hydrocarbon, it is ethyne (acetylene) with a triple bond.

Frequently Asked Questions (FAQs)

1. What is the role of KOH in this experiment? KOH is an alkaline solution that reacts with acidic gases like $CO_2$. The volume decrease after adding KOH specifically tells us how much $CO_2$ was present in the gas mixture.

2. Why does the volume of oxygen decrease during combustion? Oxygen is a reactant that combines with carbon and hydrogen to form $CO_2$ and $H_2O$. The decrease represents the amount of $O_2$ chemically consumed.

3. What if the temperature was 400 K instead of 273 K? At 400 K, water would be gaseous. The residual volume would include $CO_2$, excess $O_2$, and steam. You would need an extra step (like cooling) to separate the water volume.

4. Can eudiometry identify isomers? No, eudiometry only provides the molecular formula (e.g., $C_3H_6$). It cannot distinguish between propene and cyclopropane without additional chemical tests.

5. What is "Initial Contraction"? It is the difference between the total volume of reactants ($Hydrocarbon + O_2$) and the volume of the product mixture before any absorption.

6. Is nitrogen ever used in these problems? Yes, if a compound contains nitrogen, it is usually evolved as $N_2$ gas, which is not absorbed by KOH or pyrogallol.

7. Why 273 K? 273 K is a standard temperature (0°C). It ensures that water is liquid and allows for easy comparison using molar volume ratios.

8. What is the molecular formula for acetylene? Acetylene is $C_2H_2$. It is the simplest alkyne.

9. How do you find the volume of O2 reacted if it isn't given? By using the formula $V_{reacted} = V_{initial} - V_{leftover}$. The leftover volume is usually the volume remaining after KOH and other absorbers have been used.

10. Can I use moles instead of volumes? Yes! According to Avogadro's Law, volume ratios are the same as mole ratios for gases at constant T and P. Volume in mL can be treated directly as "molar units".

Expert Contribution by: JEE NEET Experts

10+ years of experience in physical chemistry and competitive exam coaching.

Reagent	Gas Absorbed
KOH / NaOH solution	\(CO_2\), \(SO_2\), \(Cl_2\)
Alkaline Pyrogallol	Oxygen (\(O_2\))
Ammoniacal \(Cu_2Cl_2\)	Carbon Monoxide (\(CO\))
Turpentine Oil	Ozone (\(O_3\))
Anhydrous \(CaCl_2\) / \(H_2SO_4\)	Water Vapor (\(H_2O\))