Question

Allotropes are different molecular forms of an element, such as dioxygen \(\left(\mathrm{O}_{2}\right)\) and ozone \(\left(\mathrm{O}_{3}\right) .\) (a) What is the density of each oxygen allotrope at \(0^{\circ} \mathrm{C}\) and 760 torr? (b) Calculate the ratio of densities, \(d_{\mathrm{O}_{3}} / d_{\mathrm{O}_{2}}\) and explain the significance of this number.

Short answer

The density of dioxygen (O₂) is 1.429 g/L. The density of ozone (O₃) is 2.144 g/L. The ratio \( \frac{d_{\text{O}_3}}{d_{\text{O}_2}} \) is 1.5, meaning ozone is 1.5 times denser than dioxygen.

Step-by-step solution

- Determine the molar masses

Calculate the molar mass of dioxygen (O₂) and ozone (O₃).

- Use the Ideal Gas Law

Apply the ideal gas law \[ PV = nRT \] to find the volume of each gas at the given conditions (0°C and 760 torr).

- Convert temperature and pressure

Convert the given temperature to Kelvin (K) and pressure to atmospheres (atm): \[ T = 273.15 \, K \] and \[ P = 1 \, atm \] as 760 torr is equivalent to 1 atm.

- Calculate the density of dioxygen

Using the volume calculated in step 2 for dioxygen, find its density: \[ d_{\text{O}_2} = \frac{m}{V} \] where m is the molar mass of O₂ and V is the volume at 273.15 K and 1 atm.

- Calculate the density of ozone

Similarly, using the volume calculated in step 2 for ozone, find its density: \[ d_{\text{O}_3} = \frac{m}{V} \] where m is the molar mass of O₃ and V is the volume at 273.15 K and 1 atm.

- Calculate the density ratio

Find the ratio of the densities \[ \frac{d_{\text{O}_3}}{d_{\text{O}_2}} \] and discuss its significance, which provides insight into the relative heaviness of ozone compared to dioxygen.

Key concepts

molar mass

To understand the density of different oxygen allotropes, knowing their molar masses is essential. Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol).

For dioxygen \(( \text{O}_2 \)), each oxygen atom has an atomic mass of about 16 g/mol. Since there are two oxygen atoms in dioxygen, the molar mass of \( \text{O}_2 \) is:
\[ M_{\text{O}_2} = 2 \times 16 \text{ g/mol} = 32 \text{ g/mol} \]
For ozone \( \text{O}_3 \), with three oxygen atoms, the molar mass is:
\[ M_{\text{O}_3} = 3 \times 16 \text{ g/mol} = 48 \text{ g/mol} \]
These molar masses help us move to the next steps in calculating densities and understanding their relative differences.

Ideal Gas Law

The Ideal Gas Law is a fundamental equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. The equation is:

\[ PV = nRT \]
Here:

• P is the pressure of the gas (in atm).
• V is the volume of the gas (in L).
• n is the number of moles of the gas.
• R is the ideal gas constant (0.0821 L atm/mol K).
• T is the temperature (in K).

Given that we know the pressure (1 atm) and temperature (273.15 K), we can rearrange the Ideal Gas Law equation to solve for volume (V):
\[ V = \frac{nRT}{P} \]
For simplicity, let’s assume we are calculating for 1 mole of each gas. This approach will give us the volume occupied by 1 mole of dioxygen and 1 mole of ozone under standard conditions. Knowing the volume is crucial to calculate the density.

density calculation

Density is defined as the mass per unit volume of a substance. For gases, it's typically expressed in grams per liter (g/L). Using the mass (m) and volume (V) from previous steps, density (d) can be calculated as:

\[ d = \frac{m}{V} \]
For dioxygen \(\text{O}_2\), with a molar mass of 32 g and assuming the volume for 1 mole from the Ideal Gas Law:
\[ d_{\text{O}_2} = \frac{32 \text{g}}{22.4 \text{L}} = 1.429 \text{ g/L} \]
Similarly, for ozone \(\text{O}_3\), with a molar mass of 48 g and the same volume:

\[ d_{\text{O}_3} = \frac{48 \text{g}}{22.4 \text{L}} = 2.143 \text{ g/L} \]
Finally, we can find the ratio of densities to understand how much heavier ozone is compared to dioxygen:
\[ \frac{d_{\text{O}_3}}{d_{\text{O}_2}} = \frac{2.143 \text{ g/L}}{1.429 \text{ g/L}} \approx 1.5 \]
This ratio indicates that ozone is about 1.5 times denser than dioxygen, showing that it is heavier and could have different physical and chemical behaviors.