Final answer:
The hydrocarbon gas adheres to Graham's law of effusion. The calculated molar mass based on the given diffusion rate information is 72 g/mol. The correct formula for the hydrocarbon gas with this molar mass is C5H12 (pentane).
Step-by-step explanation:
The question pertains to the application of Graham's law of effusion, which states that the rate of effusion or diffusion of a gas is inversely proportional to the square root of the mass of its particles. In the case described, we're given that a hydrocarbon gas with formula CnH2n+2 diffuses at a rate one sixth that of hydrogen gas (H2) under similar conditions. To find the formula of the gas, we can use Graham's law.
According to Graham's law, we have the relationship:
Rategas / RateH2 = sqrt(Molar massH2 / Molar massgas)
Given that the rate of diffusion of the hydrocarbon is one sixth of hydrogen, we can substitute the respective values into the equation:
1/6 = sqrt(2 / Molar massgas)
Squaring both sides to remove the square root gives:
1/36 = 2 / Molar massgas
Multiplying both sides by the molar mass of the gas and then by 36 gives:
Molar massgas = 72 g/mol
The only hydrocarbon with the formula CnH2n+2 and a molar mass of 72 g/mol is C5H12 (pentane). Thus, the formula of the gas is C5H12.