Eudiometry: Gas Analysis and Formula Determination

Question: A gaseous hydrocarbon measuring $10.0\text{ mL}$ is mixed with $70.0\text{ mL}$ of oxygen gas in a eudiometer tube and exploded. After cooling to room temperature, the residual gas volume is found to be $55.0\text{ mL}$. When the gas mixture is shaken with an aqueous solution of potassium hydroxide, the volume contracts further by $30.0\text{ mL}$. Determine the molecular formula of the hydrocarbon.

Solution:

Step 1: Write down the known volumes.

$$\text{Volume of hydrocarbon } V = 10.0\text{ mL}$$

$$\text{Initial volume of } \text{O}_2 = 70.0\text{ mL}$$

$$\text{Total initial volume} = 10.0\text{ mL} + 70.0\text{ mL} = 80.0\text{ mL}$$

Step 2: Use the contraction with KOH to determine the carbon count $x$.

Aqueous KOH absorbs $\text{CO}_2$. Therefore, the second contraction is equal to the volume of $\text{CO}_2$ formed:

$$\text{Volume of } \text{CO}_2 = 30.0\text{ mL}$$

From the stoichiometric relation:

$$\text{Volume of } \text{CO}_2 = V \cdot x \implies 10.0 \cdot x = 30.0 \implies x = \frac{30.0}{10.0} = 3$$

Step 3: Analyze the residual gas after combustion and cooling.

The residual gas ($55.0\text{ mL}$) consists solely of produced $\text{CO}_2$ and unreacted excess $\text{O}_2$ (since water has condensed to liquid):

$$\text{Residual volume} = V_{\text{CO}_2} + V_{\text{O}_2\text{, unreacted}} = 55.0\text{ mL}$$

$$30.0\text{ mL} + V_{\text{O}_2\text{, unreacted}} = 55.0\text{ mL} \implies V_{\text{O}_2\text{, unreacted}} = 55.0 - 30.0 = 25.0\text{ mL}$$

Step 4: Determine the volume of oxygen consumed and find $y$.

$$\text{Volume of } \text{O}_2 \text{ consumed} = \text{Initial } \text{O}_2 - \text{Unreacted } \text{O}_2 = 70.0\text{ mL} - 25.0\text{ mL} = 45.0\text{ mL}$$

Using the combustion stoichiometry formula:

$$\text{Volume of } \text{O}_2 \text{ consumed} = V\left(x + \frac{y}{4} ight)$$

$$45.0 = 10.0\left(3 + \frac{y}{4} ight)$$

$$\frac{45.0}{10.0} = 4.5 = 3 + \frac{y}{4}$$

$$\frac{y}{4} = 4.5 - 3 = 1.5 \implies y = 1.5 \times 4 = 6$$

Verification via first contraction route:

$$\text{Total initial volume} = 80.0\text{ mL}$$

$$\text{Volume after cooling} = 55.0\text{ mL}$$

$$\text{First contraction } \Delta V_1 = 80.0 - 55.0 = 25.0\text{ mL}$$

Theoretical first contraction: $V(1 + y/4) = 10.0(1 + 6/4) = 10.0(2.5) = 25.0\text{ mL}$. This confirms $y = 6$.

Thus, $x = 3$ and $y = 6$, corresponding to propene ($\text{C}_3\text{H}_6$) or cyclopropane.

Answer: C3H6

Question: A $20.0\text{ mL}$ mixture of carbon monoxide ($\text{CO}$) and methane ($\text{CH}_4$) is exploded with $30.0\text{ mL}$ of oxygen gas. After cooling to room temperature, the resulting volume of gas is $34.0\text{ mL}$. Find the volume percentage of methane in the original mixture.

Solution:

Step 1: Assign variables to the components of the mixture.

Let the volume of $\text{CH}_4$ in the initial mixture be $a\text{ mL}$.

Then the volume of $\text{CO}$ in the initial mixture is $(20.0 - a)\text{ mL}$.

Step 2: Write the balanced combustion reactions and their volumetric requirements.

For methane:

$$\text{CH}_4\text{ (g)} + 2\text{O}_2\text{ (g)} \longrightarrow \text{CO}_2\text{ (g)} + 2\text{H}_2\text{O (l)}$$

From $a\text{ mL}$ of $\text{CH}_4$:

$$\text{O}_2 \text{ consumed} = 2a\text{ mL}$$

$$\text{CO}_2 \text{ produced} = a\text{ mL}$$

For carbon monoxide:

$$\text{CO (g)} + \frac{1}{2}\text{O}_2\text{ (g)} \longrightarrow \text{CO}_2\text{ (g)}$$

From $(20.0 - a)\text{ mL}$ of $\text{CO}$:

$$\text{O}_2 \text{ consumed} = 0.5(20.0 - a) = 10.0 - 0.5a\text{ mL}$$

$$\text{CO}_2 \text{ produced} = (20.0 - a)\text{ mL}$$

Step 3: Calculate total oxygen consumed and total $\text{CO}_2$ produced.

$$\text{Total } \text{O}_2 \text{ consumed} = 2a + (10.0 - 0.5a) = 10.0 + 1.5a\text{ mL}$$

$$\text{Total } \text{CO}_2 \text{ produced} = a + (20.0 - a) = 20.0\text{ mL}$$

Step 4: Express the residual volume and solve for $a$.

The residual gases at room temperature consist of produced $\text{CO}_2$ and unreacted excess $\text{O}_2$:

$$\text{Residual } \text{O}_2 = 30.0 - (10.0 + 1.5a) = 20.0 - 1.5a\text{ mL}$$

$$\text{Total residual volume} = V_{\text{CO}_2} + V_{\text{O}_2\text{, unreacted}} = 20.0 + (20.0 - 1.5a) = 40.0 - 1.5a$$

Given that the residual volume is $34.0\text{ mL}$:

$$40.0 - 1.5a = 34.0$$

$$1.5a = 40.0 - 34.0 = 6.0 \implies a = \frac{6.0}{1.5} = 4.0\text{ mL}$$

Step 5: Calculate the percentage of methane.

$$\text{Volume of } \text{CH}_4 = 4.0\text{ mL}$$

$$\text{Volume of } \text{CO} = 20.0 - 4.0 = 16.0\text{ mL}$$

$$\text{Percentage of } \text{CH}_4 = \left(\frac{4.0\text{ mL}}{20.0\text{ mL}}\right) \times 100 = 20.0\%$$

Second route verification: Contraction during combustion equals $(V_{\text{reactants}} - V_{\text{products}})$. Initial total volume is $20.0 + 30.0 = 50.0\text{ mL}$. Final volume is $34.0\text{ mL}$. Contraction is $50.0 - 34.0 = 16.0\text{ mL}$. Contraction from $\text{CH}_4$ combustion is $a + 2a - a = 2a$. Contraction from $\text{CO}$ combustion is $(20 - a) + 0.5(20 - a) - (20 - a) = 0.5(20 - a) = 10 - 0.5a$. Total contraction $= 2a + 10 - 0.5a = 10 + 1.5a = 16.0\text{ mL} \implies 1.5a = 6.0 \implies a = 4.0\text{ mL}$. The answers match identically.

Answer: 20.0%

Question: A volume of $15.0\text{ mL}$ of a gaseous organic compound containing carbon, hydrogen, and oxygen is mixed with $60.0\text{ mL}$ of oxygen gas and sparked in a eudiometer. After cooling, the volume is $45.0\text{ mL}$. On treatment with caustic potash (KOH) solution, the volume contracts to $15.0\text{ mL}$. The remaining $15.0\text{ mL}$ of gas is completely absorbed by alkaline pyrogallol. Determine the empirical molecular formula of the compound.

Solution:

Step 1: Identify the components of each contraction.

$$\text{Volume of fuel compound } V = 15.0\text{ mL}$$

$$\text{Initial volume of } \text{O}_2 = 60.0\text{ mL}$$

$$\text{Volume after cooling} = 45.0\text{ mL}$$

After treatment with KOH, the volume drops from $45.0\text{ mL}$ to $15.0\text{ mL}$. The volume absorbed by KOH is:

$$V_{\text{CO}_2} = 45.0\text{ mL} - 15.0\text{ mL} = 30.0\text{ mL}$$

Because the remaining $15.0\text{ mL}$ is entirely absorbed by alkaline pyrogallol, it must be excess oxygen:

$$V_{\text{O}_2\text{, unreacted}} = 15.0\text{ mL}$$

Step 2: Determine carbon subscript $x$.

$$V \cdot x = V_{\text{CO}_2} \implies 15.0 \cdot x = 30.0 \implies x = 2$$

Step 3: Determine oxygen consumed and express hydrogen and oxygen balance.

$$\text{Volume of } \text{O}_2 \text{ consumed} = 60.0\text{ mL} - 15.0\text{ mL} = 45.0\text{ mL}$$

From the combustion equation of $\text{C}_x\text{H}_y\text{O}_z$:

$$\text{Volume of } \text{O}_2 \text{ consumed} = V\left(x + \frac{y}{4} - \frac{z}{2}\right)$$

$$45.0 = 15.0\left(2 + \frac{y}{4} - \frac{z}{2}\right) \implies 2 + \frac{y}{4} - \frac{z}{2} = 3 \implies \frac{y}{4} - \frac{z}{2} = 1$$

Multiplying across by 4 gives: $y - 2z = 4$.

Step 4: Use the first contraction upon cooling to resolve $y$ and $z$.

$$\text{Initial total volume} = 15.0\text{ mL (fuel)} + 60.0\text{ mL (}\text{O}_2\text{)} = 75.0\text{ mL}$$

$$\text{Volume after cooling} = 45.0\text{ mL}$$

$$\text{First contraction } \Delta V_1 = 75.0\text{ mL} - 45.0\text{ mL} = 30.0\text{ mL}$$

Theoretical contraction upon combustion and water condensation:

$$\Delta V_1 = V_{\text{fuel}} + V_{\text{O}_2\text{ consumed}} - V_{\text{CO}_2} = 15.0 + 45.0 - 30.0 = 30.0\text{ mL}$$

In terms of formula coefficients:

$$\Delta V_1 = V\left(1 + \frac{y}{4} - \frac{z}{2}\right)$$

$$30.0 = 15.0\left(1 + \frac{y}{4} - \frac{z}{2}\right) \implies 1 + \frac{y}{4} - \frac{z}{2} = 2 \implies \frac{y}{4} - \frac{z}{2} = 1$$

This yields the same relation $y - 2z = 4$. Because the compound is a gas at room temperature, let us check chemically stable organic compounds with $x = 2$:

Case 1: If $z = 0$, then $y = 4$, which gives ethylene ($\text{C}_2\text{H}_4$).

Case 2: If $z = 1$, then $y = 6$, which gives dimethyl ether or ethanol ($\text{C}_2\text{H}_6\text{O}$). Dimethyl ether is a gas at room temperature (boiling point $-24^{\circ}\text{C}$).

Case 3: If $z = 2$, then $y = 8$, which violates standard tetravalency of saturated carbon compounds (maximum $2x + 2 = 6$).

For an oxygenated compound explicitly stated in the problem ($z \ge 1$), $z = 1$ and $y = 6$. The molecular formula is $\text{C}_2\text{H}_6\text{O}$.

Answer: C2H6O