When burning 4.3 g of organic matter, the vapor density of which in the air is 2.966, 6.3 g of water and 6.72

When burning 4.3 g of organic matter, the vapor density of which in the air is 2.966, 6.3 g of water and 6.72 liters of carbon dioxide were obtained. deduce the molecular formula of the substance.

CxHy + O2 = CO2 + H2O;
Given: m (CxHy) = 4.3g;
V (CO2) = 6.3 g;
m (H2O) = 6.72L; DCxHy (by air) = 2.996;
Determine: formula CxNu
Solution:
Let’s make the equation: CxHy + O2 = CO2 + H2O;
m (CO2) = V * M;
M (CO2) = 12 + (16 * 2) = 44 g / mol;
1 mol of gas – 22.4 liters (according to Avogadro’s law);
X mol (CO2) – 6.72 L;
V (CO2) = 6.72 / 22.4 = 0.3 mol;
m (CO2) = O, 3 * 44 = 13.2g;
m (H) = 2.18 * 6.3 = 0.7;
X / Y = 3.6 / 12: 0.7 / 1;
X / Y = 0.3 / 0.7 = 3: 7;
С3Н7 – the simplest formula;
M (CxHy) = 2.966 * 29 = 86.8;
M (post) / M (CxHy) = 86.8 / 44 = 1.9 = 2;
True formula: C6H14 – hexane;
Answer: С6Н14