At 300 K and 1 atm, 15 mL of a gaseous hydrocarbon requires 375 mL air containing 20% O2 by volume f

Q: Why do we completely ignore the volume of water formed in the final calculation?

The problem explicitly states that the water formed is in liquid form. Under standard conditions, the volume occupied by a liquid is infinitesimally small compared to the volume occupied by gases. Therefore, its contribution to the total $330\text{ mL}$ is negligible and mathematically ignored.

Q: What exactly is the 'inert part' of the air, and why does it matter?

Air is a mixture, primarily consisting of Nitrogen (about $78\%$) and Oxygen ($21\%$). During combustion, only the Oxygen reacts with the hydrocarbon. The Nitrogen and other trace gases are 'inert'—they don't react. They remain in the container from start to finish, so their volume must be included when calculating the final total gas volume.

Q: Why did we multiply the stoichiometric coefficients by 15?

According to Gay-Lussac's law of combining volumes, the ratio of the volumes of reacting gases is identical to the ratio of their stoichiometric coefficients in the balanced equation. Since $1$ volume of the hydrocarbon corresponds to $15\text{ mL}$, we must multiply all other gaseous coefficients by $15$ to find their respective volumes in milliliters.

Q: Is $C_2H_{12}$ a real chemical compound?

No, it is absolutely impossible. The maximum number of hydrogen atoms a two-carbon hydrocarbon can have is $6$ (Ethane, $C_2H_6$). The data provided in the question is chemically inconsistent, which is why the calculated answer does not match any of the real-world options provided.

Q: What would change if the water was formed as a gas (steam) instead of a liquid?

If the water were in a gaseous state, it would contribute significantly to the final volume. We would have to include its volume, which is $15 \times \frac{y}{2}\text{ mL}$, in our final equation: $V_{\text{final}} = V_{CO_2} + V_{H_2O(g)} + V_{\text{inert}}$.

Question

At $300	ext{ K}$ and $1	ext{ atm}$, $15	ext{ mL}$ of a gaseous hydrocarbon requires $375	ext{ mL}$ air containing $20\%\ O_2$ by volume for complete combustion. After combustion, the gases occupy $330	ext{ mL}$. Assuming that the water formed is in liquid form and the volumes were measured at the same temperature and pressure, the formula of the hydrocarbon is

Options:

(A) $C_3H_8$

(B) $C_4H_8$

(C) $C_4H_{10}$

(D) $C_3H_6$

Sigma Percentile · Accepted Answer

### The Setup: Decoding the Air Mixture

Imagine you are conducting a combustion experiment in a sealed chamber. You start with $15	ext{ mL}$ of an unknown gaseous hydrocarbon, $C_xH_y$. To burn it completely, you pump in $375	ext{ mL}$ of air.

Here is the first critical trap: air is not pure oxygen! The problem states that air contains $20\%\ O_2$ by volume. This means the actual volume of oxygen available for the reaction is just a fraction of the total air.

Let's calculate the exact volume of oxygen:
$$V_{O_2} = \frac{20}{100} 	imes 375	ext{ mL} = 75	ext{ mL}$$

If $75	ext{ mL}$ is oxygen, what is the rest? The remaining $300	ext{ mL}$ ($375 - 75$) consists of inert gases, primarily nitrogen. These gases do not participate in the combustion; they simply act as spectators and will remain in the chamber until the very end.

### The Master Equation: Combustion Stoichiometry

To relate the volumes of the gases, we must rely on the general balanced equation for the combustion of any hydrocarbon:
$$C_xH_y(g) + \left(x + \frac{y}{4}ight)O_2(g) ightarrow xCO_2(g) + \frac{y}{2}H_2O(l)$$

According to Gay-Lussac's law of combining volumes, gases react in simple whole-number ratios of their volumes, provided temperature and pressure remain constant. Since $15	ext{ mL}$ of the hydrocarbon reacts completely, the volume of oxygen consumed must be $15$ times its stoichiometric coefficient:
$$V_{O_2 	ext{ reacted}} = 15 \left(x + \frac{y}{4}ight)$$

We already know that exactly $75	ext{ mL}$ of oxygen was used. Equating the two gives us our first master equation:
$$15 \left(x + \frac{y}{4}ight) = 75$$
$$x + \frac{y}{4} = 5$$

### The Final State: Where Did the Volume Go?

After the combustion is complete, the problem states that the remaining gases occupy $330	ext{ mL}$. What exactly makes up this final volume?

**Crucial Detail:** The water formed is in the liquid state ($H_2O(l)$). The volume occupied by a few drops of liquid water is practically zero compared to the volume of the gases, so we completely ignore it.

The final gas mixture consists solely of the carbon dioxide produced and the inert air that never reacted. From our balanced equation, the volume of $CO_2$ produced is $15x	ext{ mL}$.

Setting up the final volume equation:
$$V_{	ext{final}} = V_{CO_2} + V_{	ext{inert}}$$
$$330 = 15x + 300$$

Solving for $x$ is now a breeze:
$$15x = 30$$
$$x = 2$$

We have discovered that our mystery hydrocarbon contains exactly $2$ carbon atoms.

### The Anomaly: When Math Defies Chemistry

Now that we have $x = 2$, let's substitute it back into our first master equation to find the number of hydrogen atoms ($y$):
$$2 + \frac{y}{4} = 5$$
$$\frac{y}{4} = 3$$
$$y = 12$$

Mathematically, the formula of the hydrocarbon is **$C_2H_{12}$**.

However, there is a massive catch here. In the real world of organic chemistry, a hydrocarbon with two carbon atoms can hold a maximum of six hydrogen atoms (Ethane, $C_2H_6$, following the alkane general formula $C_nH_{2n+2}$). A molecule like $C_2H_{12}$ is structurally and chemically impossible.

Because the mathematical data provided in the question leads to an anomalous, non-existent compound, none of the given options ($C_3H_8$, $C_4H_8$, $C_4H_{10}$, $C_3H_6$) are correct. This is a fascinating example of how a perfectly executed mathematical derivation can sometimes reveal flaws in the premise of the question itself!

Select Answer:

Visualized Solution

\text{Combustion Reaction Setup}

\text{General Combustion Equation}

\text{Volume of Oxygen Required}

\text{Calculating } V_{O_2}

\text{Applying Gay-Lussac's Law}

\text{Simplifying the Oxygen Equation}

\text{Analyzing Final Gas Volume}

\text{Solving for Carbon Atoms (x)}

\text{Solving for Hydrogen Atoms (y)}

\text{Conclusion}

The Sigma Insight: Stoichiometric and Volumetric Calculations

The Setup

The Master Equation

The Final State

The Anomaly

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