The formula of a gaseous hydrocarbon, which requires 6 times of its own volume of O2 for complete ox

Q: Why do we ignore the volume of $H_2O$ in these calculations?

In standard eudiometry or combustion problems at room temperature, the water produced condenses into a liquid. The volume of a liquid is extremely small (negligible) compared to the volume of gases, so it doesn't contribute to the final gaseous volume.

Q: Can we use moles instead of volumes for this problem?

Yes! According to Avogadro's Law, at constant temperature and pressure, the volume of a gas is directly proportional to its number of moles ($V \propto n$). So, a volume ratio of 1:6:4 is exactly the same as a molar ratio of 1:6:4.

Q: How do we memorize the general combustion equation?

You don't need to memorize it! Just balance it logically: $C_xH_y \rightarrow xCO_2 + (y/2)H_2O$. Now count the oxygen atoms on the right: $2x$ from $CO_2$ and $y/2$ from $H_2O$. Total oxygen atoms = $2x + y/2$. Since oxygen gas is diatomic ($O_2$), divide by 2 to get the coefficient: $x + y/4$.

Q: What if the hydrocarbon contained oxygen, like $C_xH_yO_z$?

The balancing logic remains the same, but you must account for the oxygen already present in the molecule. The coefficient for $O_2$ would become $(x + y/4 - z/2)$.

Q: Why did we assume the hydrocarbon is a gas?

The problem explicitly states "gaseous hydrocarbon". Even if it didn't, combustion analysis (eudiometry) relies on measuring gas volumes, so the reactants and products (except water) are treated as ideal gases.

Question

The formula of a gaseous hydrocarbon, which requires 6 times of its own volume of $O_2$ for complete oxidation and produces 4 times its own volume of $CO_2$ is $C_xH_y$. The value of $y$ is ……… .

Sigma Percentile · Accepted Answer

### The Mystery of the Unknown Hydrocarbon

Imagine you are a chemical detective in a laboratory, and you are handed a sealed flask containing an unknown gaseous hydrocarbon, $C_xH_y$. Your mission is to find its exact molecular formula.

To do this, you perform a classic combustion analysis. You spark the gas with oxygen and measure the volumes of the gases before and after the reaction. The problem gives us two crucial clues: the hydrocarbon requires exactly **6 times its own volume** of oxygen ($O_2$) to burn completely, and it produces exactly **4 times its own volume** of carbon dioxide ($CO_2$).

How do we translate these physical volume measurements into the molecular subscripts $x$ and $y$? Let's dive into the chemistry!

### The Master Equation

Every time you face a combustion problem involving a hydrocarbon, your first instinct should be to write down the **general balanced chemical equation**. This equation is the master key that unlocks the stoichiometry of the reaction.

For any hydrocarbon $C_xH_y$, the balanced combustion equation is:

$$C_xH_y(g) + \left(x + \frac{y}{4}ight)O_2(g) \longrightarrow xCO_2(g) + \frac{y}{2}H_2O(l)$$

Notice the state symbols. At standard room temperature conditions, the water produced condenses into a liquid ($l$). Because the volume of a liquid is negligible compared to the volume of gases, we completely ignore the water when dealing with gaseous volume ratios!

### Decoding the Volumes with Gay-Lussac's Law

Now, we need a bridge to connect the moles in our equation to the volumes given in the problem. Enter **Gay-Lussac's Law of Combining Volumes**.

This beautiful law states that when gases react at the same temperature and pressure, the ratio of their volumes is exactly equal to the ratio of their stoichiometric coefficients.

Looking at our master equation, we can write the volume ratio as:

$$1 	ext{ vol of } C_xH_y : \left(x + \frac{y}{4}ight) 	ext{ vol of } O_2 \longrightarrow x 	ext{ vol of } CO_2$$

This means if we start with $V$ liters of our hydrocarbon, we will need $V \left(x + \frac{y}{4}ight)$ liters of oxygen, and we will produce $V(x)$ liters of carbon dioxide.

### The Final Calculation

Let's use our first clue. The volume of $CO_2$ produced is 4 times the volume of the hydrocarbon.

Comparing this to our theoretical ratio, the coefficient of $CO_2$ is simply $x$. Therefore, we can instantly conclude:

$$x = 4$$

We have found the number of carbon atoms! Our hydrocarbon is a $C_4$ molecule.

Now for the second clue. The volume of $O_2$ required is 6 times the volume of the hydrocarbon.

According to our master equation, the coefficient for oxygen is $\left(x + \frac{y}{4}ight)$. Setting this equal to 6 gives us:

$$x + \frac{y}{4} = 6$$

We already know that $x = 4$. Let's substitute this value into the equation:

$$4 + \frac{y}{4} = 6$$

Subtracting 4 from both sides:

$$\frac{y}{4} = 2$$

Multiplying by 4, we get our final answer:

$$y = 8$$

The mystery is solved! The unknown hydrocarbon is $C_4H_8$, which is butene. The value of $y$ is exactly **8**.

Animated Solution for Chemistry - Basic Concepts in Chemistry: The formula of a gaseous hydrocarbon, which requires 6 times of its own volume of $O_{2}$ for complete oxidation and produces 4 times its own volume of $C O_{2}$ is $C_{x} H_{y}$ . The value of $y$ is ……… .

Enter Numerical Value:

Visualized Solution

\text{Combustion Setup}

\text{General Combustion Equation}

\text{Gay-Lussac's Law of Volumes}

\text{Calculating } x

\text{Calculating } y

\text{Solving for } y

\text{Extensions}

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Mystery of the Unknown Hydrocarbon

The Master Equation

Decoding the Volumes with Gay-Lussac's Law

The Final Calculation

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