Question

10 mL of gaseous organic compound containing C, H and O only, was mixed with 100 mL of $O_2$ and exploded under identical conditions and then cooled. The volume left after cooling was 90 mL.

On treatment with KOH a contraction of 20 mL was observed. If vapour density of compound is 23, if molecular formula of the compound is $C_xH_yO_z$, then find (x + y + z).

Answer 1

9

Answer 2

The problem involves applying Gay-Lussac's Law of Gaseous Volumes and the concept of vapour density to determine the molecular formula of an organic compound.

Let the molecular formula of the organic compound be $C_xH_yO_z$. The balanced chemical equation for the combustion of the compound is: $C_xH_yO_z(g) + (x + y/4 - z/2)O_2(g) \rightarrow xCO_2(g) + y/2 H_2O(l)$ Note: Since the reaction is cooled, $H_2O$ is in liquid form, and its volume is negligible.

Given data:

1. Volume of gaseous organic compound = 10 mL
2. Volume of $O_2$ mixed = 100 mL
3. Volume left after cooling = 90 mL
4. Contraction on treatment with KOH = 20 mL
5. Vapour density (VD) of compound = 23

Step 1: Determine the volume of $CO_2$ produced. KOH absorbs acidic gases like $CO_2$. The contraction observed on treatment with KOH is due to the absorption of $CO_2$. Volume of $CO_2$ produced = 20 mL.

Step 2: Determine the value of 'x'. According to Gay-Lussac's Law, the volumes of gaseous reactants and products are in a simple whole-number ratio, which corresponds to their stoichiometric coefficients in the balanced equation. From the balanced equation, 1 volume of $C_xH_yO_z$ produces x volumes of $CO_2$. Given that 10 mL of $C_xH_yO_z$ produced 20 mL of $CO_2$: $\frac{\text{Volume of } C_xH_yO_z}{\text{Volume of } CO_2} = \frac{1}{x}$ $\frac{10 \text{ mL}}{20 \text{ mL}} = \frac{1}{x}$ $10x = 20 \Rightarrow x = 2$.

Step 3: Determine the volume of $O_2$ reacted. The volume left after cooling (90 mL) consists of the $CO_2$ produced and the unreacted $O_2$. Volume left = Volume of $CO_2$ + Volume of unreacted $O_2$ 90 mL = 20 mL + Volume of unreacted $O_2$ Volume of unreacted $O_2$ = 90 - 20 = 70 mL.

Initial volume of $O_2$ taken = 100 mL. Volume of $O_2$ reacted = Initial volume of $O_2$ - Volume of unreacted $O_2$ Volume of $O_2$ reacted = 100 mL - 70 mL = 30 mL.

Step 4: Formulate an equation for 'y' and 'z' using the volume of $O_2$ reacted. From the balanced equation, 1 volume of $C_xH_yO_z$ reacts with $(x + y/4 - z/2)$ volumes of $O_2$. Given that 10 mL of $C_xH_yO_z$ reacted with 30 mL of $O_2$: $\frac{\text{Volume of } C_xH_yO_z}{\text{Volume of } O_2 \text{ reacted}} = \frac{1}{(x + y/4 - z/2)}$ $\frac{10 \text{ mL}}{30 \text{ mL}} = \frac{1}{(x + y/4 - z/2)}$ $10(x + y/4 - z/2) = 30$ $x + y/4 - z/2 = 3$ Substitute $x=2$: $2 + y/4 - z/2 = 3$ $y/4 - z/2 = 1$ Multiplying by 4 to clear denominators: $y - 2z = 4$ (Equation 1)

Step 5: Formulate an equation for 'y' and 'z' using vapour density. Molecular mass (M) = 2 $\times$ Vapour density (VD) M = 2 $\times$ 23 = 46 g/mol. The molecular mass of $C_xH_yO_z$ is given by: $12x + 1y + 16z = \text{Molecular mass}$ $12x + y + 16z = 46$ Substitute $x=2$: $12(2) + y + 16z = 46$ $24 + y + 16z = 46$ $y + 16z = 46 - 24$ $y + 16z = 22$ (Equation 2)

Step 6: Solve the system of equations for 'y' and 'z'. We have two linear equations:

1. $y - 2z = 4$
2. $y + 16z = 22$

From Equation 1, express y in terms of z: $y = 4 + 2z$ Substitute this expression for y into Equation 2: $(4 + 2z) + 16z = 22$ $4 + 18z = 22$ $18z = 22 - 4$ $18z = 18$ $z = 1$

Now substitute the value of z back into the expression for y: $y = 4 + 2(1)$ $y = 4 + 2$ $y = 6$

Step 7: Determine the molecular formula and calculate (x + y + z). We found: $x = 2$ $y = 6$ $z = 1$ The molecular formula of the compound is $C_2H_6O$.

Finally, calculate (x + y + z): $x + y + z = 2 + 6 + 1 = 9$.