During the combustion of organic matter with a volume of 1.12 liters, carbon dioxide with a mass of 2.2 g and water

During the combustion of organic matter with a volume of 1.12 liters, carbon dioxide with a mass of 2.2 g and water with a mass of 0.9 g were formed. Vapors of organic matter are 7.5 times heavier than helium. what organic matter was burned?

The volume of organic matter is given, and therefore you can find the amount: n (CxHyOz) = 1.12 l / 22.4 l / mol = 0.05 mol. Given D (h2) CxHyOz = 7.5-> M (CxHyOz) = 7.5 × 4 g / mol = 30g / mol. Let’s find the mass of organic matter: 30 g / molt × 0.05 mol = 1.5 g (I find this in order to check later if there is oxygen in organic matter) n (C) = n (CO2) = 2 , 2g / 44 g / mol = 0.05 mol. m (C) = 12 g / mol × 0.05 mol = 0.6 g. n (H) = 2 n (h2o) = 2 × 0.9 g / 18 g / mol = 0.1 mol. m (H) = 1 g / mol × 0.1 mol = 0.1 g m (o) = m (CxHyOz) -m (C) -m (H) = 1.5 g-0.6g-0, 1 g = 0.8 g (there is still oxygen in the organic matter). Let’s find its amount: n (O) = 0.8 g / 16 g / mol = 0.05 mol. n (c): n (H): n (O) = 0.05: 0.1: 0.05 = 1: 2: 1 CH2O is the simplest formula. K = M (CH2O) / M (CxHyOz) = 30 g / mol: 30 g / mol = 1, -> CH2O – molecular formula.
This is formaldehyde: H-CHO