A 1.17 g sample of an alkane hydrocarbon gas occupies a volume of 674 mL at 28^$\circ$C and 741 mmHg. Alkanes are known to have the general formula C_nH_2n+2. What is the molecular formula of the gas in this sample (R = 0.08206 L * atm/K * mol)

Question

Quizplus · Accepted Answer

Using the Ideal Gas Law, we can calculate the number of moles of the gas present:
n = PV/RT = (741 mmHg)(0.674 L)/(0.08206 atm L/mol K)(301 K) = 0.0279 mol
Next, we can use the given mass of the gas to calculate its molar mass:
molar mass = 1.17 g / 0.0279 mol = 42.0 g/mol
Finally, we can use the molar mass and the fact that the gas is an alkane to deduce its molecular formula. The molecular formula of an alkane is simply CnH2n+2. To get from the molar mass of 42.0 g/mol to a whole number multiple of the molar mass of the simplest alkane (methane, CH4), we can divide by 14 (the molar mass of methane is 16 g/mol). Dividing 42.0 by 14 gives 3, so the molecular formula of the gas is C3H8. This matches choice C.

A 117 G Sample of an Alkane Hydrocarbon Gas Occupies a a Volume