A 1.17 g sample of an alkane hydrocarbon gas occupies a volume of 674 mL at 28°C and 741 mmHg.Alkanes are known to have the general formula CnH2n+2.What is the molecular formula of the gas in this sample? (R = 0.08206 L.atm/K.mol)

Question

Quizplus · Accepted Answer

First, we need to calculate the number of moles of the gas using the ideal gas law:

PV = nRT

n = (PV)/(RT)
n = (741 mmHg)(0.674 L)/(0.08206 atm·L/mol·K)(301 K)
n = 0.0272 mol

Next, we can calculate the molar mass of the alkane hydrocarbon:

molar mass = (mass of sample)/(number of moles)
molar mass = 1.17 g/0.0272 mol
molar mass = 43.0 g/mol

Now, we can use the formula for alkane hydrocarbons (CnH2n+2) to determine the number of carbon atoms in the molecule:

molar mass = (12n + 2(1n + 2)) g/mol
43.0 g/mol = 14n + 30
14n = 13
n ≈ 0.93

Since we cannot have a fractional number of carbon atoms, we must round up to 1. Therefore, the molecular formula of the gas is C1H4, which corresponds to choice C.

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