During the combustion of organic matter with a mass of 2.1 g, 6.6 g of carbon monoxide (4) and 2.7 g

During the combustion of organic matter with a mass of 2.1 g, 6.6 g of carbon monoxide (4) and 2.7 g of water were released. The density of this substance in air is equal to 2.9 determine the molecular formula of the substance

N (CO2) = 6.6 g / 44 g / mol = 0.15 mol
n (C) = 0.15 mol
m (C) = 0.15 mol * 12 g / mol = 1.8 g.
n (H2O) = m / M = 2.7g. / 18 g / mol = 0.15 mol
n (H) = 2 * n (H2O) = 0.15 * 2 = 0.3 mol
m (H) = 0.3 * 1 g / mol = 0.3
m (H) + m (C) = 0.3 g + 1.8 = 2.1 g – that is, there are no other elements
C: H = 0.15: 0.3 or 1: 2
M (CxH₂y) = 2.9 * D = 2.9 * 29g / mol = 84 g / mol
12n + 2n = 84
14n = 84
n = 6,
C₆H₁₂ is hexene