A gas with an empirical formula of CH2 has a density of 1.88 g/L at 0.00 degrees Celsius and 1.00 atm. What is it's chemical formula?

Method 1 :

For the gas (CH₂)ₙ :
Density, m/V = 1.88 g/L
Pressure, P = 1 atm
Temperature, T = 273 K
Gas constant, R = 0.08206 L atm / (mol K)

PV = nRT and n = m/M
Then, PV = (m/M)RT
Molar mass, M = (m/V)RT/P = 1.88 × 0.08206 × 273 / 1 g/mol = 42.1 g/mol

Molar mass of the gas.
(12.0 + 1.0×2) × n = 42.1
n ≈ 3

Hence, the molecular formula = C₃H₆


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Method 2 :

Each mole of gas occupies a volume of 22.4 L at 0.00°C and 1.00 atm.
Molar mass of the gas = (1.88 g/L) × (22.4 L/mol) = 42.1 g/mol

Molar mass of the gas.
(12.0 + 1.0×2) × n = 42.1
n ≈ 3

Hence, the molecular formula = C₃H₆

we can simplify a bit for this particular problem because
.. for an ideal gas.. .. V/mol = 22.41 L/mol @ STP
and
.. STP means T = 273.15K (0.00°C) and P = 1.00 atm

so that
.. mw = (1.88 g/L) * (22.41 L/mol) = 42.12 g/mol

then
.. molar mass / empirical unit mass = # empirical units / molecule
.. 42.12 / 14.03 = 3

so the molecular formula = 3x empirical formula = C3H6