Propane carbon(IV)oxide diffuse at the same rate because [H = 1.00, C = 12.0, O = 16.0]

Analyze the question: Propane (C₃H₈) and carbon(IV) oxide (CO₂) are said to diffuse at the same rate. We need the reason why.

Concept: Graham's Law of Diffusion

The rate of diffusion of a gas is inversely proportional to the square root of its relative molecular mass (or density), at the same temperature and pressure:

Rate ∝ 1/√M

So two gases diffuse at the same rate only if their relative molecular masses are equal.

Solution:

• Propane, C₃H₈: (3 × 12.0) + (8 × 1.00) = 36 + 8 = 44
• Carbon(IV) oxide, CO₂: 12.0 + (2 × 16.0) = 12 + 32 = 44

Both have a relative molecular mass of 44, so by Graham's Law:

Rate(C₃H₈) / Rate(CO₂) = √(44/44) = 1

They therefore diffuse at the same rate.

Checking the other options:

• A. Being gases alone does not guarantee equal rates. Hydrogen and oxygen are both gases but diffuse at very different rates.
• B. Containing carbon has no effect on diffusion rate.
• D. Being denser than air only compares them to air. Two gases can both be denser than air and still have different molar masses and rates.

**Answer: C. They have the same relative molecular mass.

Exam tip: When a question asks about diffusion rates, calculate the molar masses first. Equal molar mass means equal rate. A lighter gas diffuses faster, and a heavier one diffuses slower. Remember: "Same M, same rate."