Apply Avogardo's hypothesis to derive the relationship between vapour density and molecular mass of a gas. An oxide of nitrogen contains same of its own volume of nitrogen whose vapour density is…

Relationship between Vapour Density and Molecular Mass

Derivation using Avogadro’s Hypothesis

Avogadro’s hypothesis states that equal volumes of all gases, under the same conditions of temperature and pressure, contain the same number of molecules.

• Let the molecular mass of a gas be $M$.

• The vapour density (VD) of a gas is defined as the mass of a certain volume of the gas compared to the mass of the same volume of hydrogen gas at the same temperature and pressure.

• Mathematically, vapour density is given by: $$\text{VD}=\frac{\text{Mass of gas}}{\text{Mass of hydrogen}}=\frac{M}{2}$$ (since the molecular mass of hydrogen gas, $\mathrm{H}{}_2$, is 2).

• Therefore, the molecular mass can be expressed in terms of vapour density as: $$M=2\times \text{VD}$$

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Determination of the Molecular Formula of Nitrogen Oxide

Given:

• An oxide of nitrogen ($\mathrm{NxOy}$) contains same volume of nitrogen (i.e., 1 volume of $\mathrm{NxOy}$ contains 1 volume of $\mathrm{N}{}_2$).
• Vapour density of nitrogen ($\mathrm{N}{}_2$) = 54.

Step 1: Calculate the Molecular Mass of $\mathrm{N}{}_2$

Using the relationship between vapour density and molecular mass: $$M_{\mathrm{N}{}_2}=2\times \text{VD}=2\times 54=108\text{g/mol}$$

Step 2: Determine the Molecular Mass of the Oxide ($\mathrm{NxOy}$)

Since the oxide contains equal volumes of nitrogen, by Avogadro’s law, 1 mole of $\mathrm{NxOy}$ contains 1 mole of $\mathrm{N}{}_2$.

• Let the molecular formula of the oxide be $\mathrm{NxOy}$.
• The mass of nitrogen in the oxide = mass of $\mathrm{N}{}_2$ = 108 g/mol.
• The mass of oxygen in the oxide = $16y$ g/mol (since atomic mass of oxygen = 16).

Thus, the total molecular mass of the oxide is: $$M_{\mathrm{NxOy}}=14x+16y$$

But since 1 mole of $\mathrm{NxOy}$ contains 1 mole of $\mathrm{N}{}_2$ (i.e., 2 moles of nitrogen atoms), we have: $$14x=108⟹x=\frac{108}{14}=7.714$$ Since $x$ must be an integer, we multiply by 2 to get whole numbers: $$x=14,\text{and the molecular mass doubles to }216\text{g/mol}$$

Step 3: Find the Value of $y$

The oxide contains only nitrogen and oxygen, so: $$14x+16y=216$$ Substituting $x=14$: $$14(14)+16y=216⟹196+16y=216⟹16y=20⟹y=\frac{20}{16}=1.25$$ Again, $y$ must be an integer. Multiply by 4 to get whole numbers: $$x=56,y=5$$ Now, the molecular mass becomes: $$M=14(56)+16(5)=784+80=864\text{g/mol}$$ But this is inconsistent. Instead, let’s consider the empirical formula approach:

Alternative Approach (Empirical Formula)

Given that 1 volume of $\mathrm{NxOy}$ contains 1 volume of $\mathrm{N}{}_2$, the oxide must contain 2 moles of nitrogen atoms (since $\mathrm{N}{}_2$ has 2 atoms).

Thus, the empirical formula is: $$\mathrm{N}{}_2\mathrm{Oy}$$ The mass of nitrogen in the oxide = 28 g/mol (since $2\times 14=28$).

But the oxide contains same volume of nitrogen, meaning the molecular mass of the oxide is: $$M_{\mathrm{N}{}_2\mathrm{Oy}}=28+16y$$

But from Avogadro’s law, 1 mole of oxide contains 1 mole of $\mathrm{N}{}_2$, so: $$M_{\mathrm{N}{}_2\mathrm{Oy}}=108\text{g/mol}$$ Thus: $$28+16y=108⟹16y=80⟹y=5$$

Therefore, the molecular formula of the oxide is: $$\boxed{\mathrm{N}{}_2\mathrm{O}{}_5}$$

Verification:

• Molecular mass of $\mathrm{N}{}_2\mathrm{O}{}_5=2(14)+5(16)=28+80=108\text{g/mol}$.
• Vapour density of $\mathrm{N}{}_2\mathrm{O}{}_5=\frac{108}{2}=54$, which matches the given vapour density of nitrogen.

Thus, the oxide is dinitrogen pentoxide ($\mathrm{N}{}_2\mathrm{O}{}_5$).