An amount of 23 g of gas (density ρ = 2.05 g dm-3 at STP) when burned, gives 44 g of carbon dioxide — Physical Chemistry Chemistry Question

The unknown gas : X
From the ideal gas law: M(X) = ρ R T / p = 46 g mol^-1
n(X) = 23 g / 46 g mol^-1 = 0.5 mol
n(C) = n(CO₂) = 44 g / 44 g mol^-1 = 1 mol
m(C) = 12 g
n(H₂O) = 27 g / 18 g mol^-1 = 1.5 mol
n(H) = 3 mol
m(H) = 3 g
The compound contains also oxygen, since
m(C) + m(H) = 12 g + 3 g = 15 g < 23 g
m(O) = 23 g - 15 g = 8 g
n(O) = 0.5 mol
n(C) : n(H) : n(O) = 1 : 3 : 0.5 = 2 : 6 : 1
The empirical formula of the compound is C₂H₆O.

Ethanol is liquid in the given conditions and therefore, the unknown gas is dimethyl ether.