Question

In an experiment, 10cm3
of an organic compound, J, in the gaseous state was sparked with an
excess of oxygen. 20cm3
of carbon dioxide and 5cm3
of nitrogen were obtained among the
products. All gas volumes were measured at the same temperature and pressure.
What could be the identity of J?
1 C2H6N2
2 C2H3N
3 C2H7N

Answer 1

Answer on Question #81416 – Chemistry – Organic Chemistry

Task:

In an experiment, 10 cm³ of an organic compound, J, in the gaseous state was sparked with an excess of oxygen (O₂). 20 cm³ of carbon dioxide (CO₂) and 5 cm³ of nitrogen (N₂) were obtained among the products. All gas volumes were measured at the same temperature and pressure. What could be the identity of J?

1 C₂H₆N₂

2 C₂H₃N

3 C₂H₇N

Solution:

The volume of 1 mol of an ideal gas at STP is 22.4 L. (Vₘ = 22.4 L/mol.)

$$n = \frac{V}{V_m};$$ $$n(C_xH_yN_z) = \frac{V(C_xH_yN_z)}{V_m} = \frac{10\, \text{cm}^3}{V_m};$$ $$n(N_2) = \frac{V(N_2)}{V_m} = \frac{5\, \text{cm}^3}{V_m};$$ $$n(CO_2) = \frac{V(CO_2)}{V_m} = \frac{20\, \text{cm}^3}{V_m}.$$

Schematic equation of chemical reaction:

$$C_xH_yN_z + \left( x + \frac{y}{4} \right)O_2 = xCO_2 + \frac{y}{2}H_2O + \frac{z}{2}N_2$$

By the reaction equation:

$$n(C_xH_yN_z) = \frac{n(CO_2)}{x} = \frac{2 \cdot n(N_2)}{z};$$

Then,

$$\frac {n (C O _ {2})}{n (C _ {x} H _ {y} N _ {z})} = \frac {x}{1} = \frac {2 0 c m ^ {3}}{1 0 c m ^ {3}}; \Rightarrow x = 2$$ $$\frac {n (N _ {2})}{n (C _ {x} H _ {y} N _ {z})} = \frac {z}{2} = \frac {5 c m ^ {3}}{1 0 c m ^ {3}}; \Rightarrow z = 1$$

That means there are 2 carbon atoms in J, which also means there is 1 N atom in J.

Then there are two options: $\mathrm{C}_2\mathrm{H}_3\mathrm{N}$ and $\mathrm{C}_2\mathrm{H}_7\mathrm{N}$ .

Answer: $\mathrm{C}_2\mathrm{H}_3\mathrm{N}$ and $\mathrm{C}_2\mathrm{H}_7\mathrm{N}$

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