The combustion of organic matter with a mass of 6.2 yielded 4.48 liters of carbon dioxide and 5.4 g of water.

The combustion of organic matter with a mass of 6.2 yielded 4.48 liters of carbon dioxide and 5.4 g of water. The vapor density of the substance with respect to nitrogen is 2.214. establish the molecular formula of the substance.

1. Let the substance have the formula CxHyOz, since the density of its vapors by N2 is known, we will find its molecular weight:
Mr (CxHyOz) = 2.214 * Mr (N2) = 2.214 * 28 = 62
2. The equation of combustion of matter:
CxHyOz + O2 = xCO2 + Y / 2H2O
3. When burning 6.2 g of the substance 4.48 l of CO2 is formed, and at 62 g – 22.4 hl (according to the equation).
Then x = 62 * 4.48 / 6.2 * 22.4 = 2 (number C)
4. Find at
When burning, 6.2 g of water – 5.4 g of water, and at 63g – 9y, hence y = 62 * 5.4 / 6.2 * 9 = 6 (number H)
5. Mr (C2H6Oz) = 62, find z
62 – (12 * 2 + 1 * 6) = 32, which means that this substance contains 2 oxygen atoms. С2Н6О2 or СН3СООН – acetic acid.