Final answer:
The hydrocarbon with an empirical formula CH2 and a density of 1.3 g/L at 0ºC and 1.00 atm has a molecular formula of C2H4, which is ethene (ethylene). This calculation is based on the molar mass derived from the ideal gas law and the empirical formula mass.
Step-by-step explanation:
To find the possible formula for the hydrocarbon with the empirical formula CH2 and a density of 1.3 g/L at 0ºC and 1.00 atm, we need to calculate its molar mass.
Using the ideal gas law (PV = nRT), where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin, we can solve for the molar mass (MM).
At STP (0ºC and 1.00 atm), one mole of any gas occupies 22.4 L, so we can determine the number of moles in 1.3 g of the gas as follows: n = mass/MM. Given that density (d) = mass/volume (V), we get dV = mass. Substituting the known values in, we have 1.3 g/L × 1 L = mass = 1.3 g. Now, we calculate the molar mass (MM) using the molar volume of a gas at STP (22.4 L/mol): MM = mass/n = 1.3 g ÷ (1 L / 22.4 L/mol) = 29.12 g/mol.
Since the empirical formula mass (EM) of CH2 is 14.03 g/mol, we divide the actual molar mass by the empirical formula mass to find the number of empirical units in the molecular formula: 29.12 g/mol ÷ 14.03 g/mol ≈ 2.
Therefore, multiplying the empirical formula by this number gives us the molecular formula C2H4, which is ethene (ethylene).