Question

при сжигании 4,4 г. углеводорода получили 13,2 г. оксида углерода (IV). относительная плотность вещества по воздуху равна 1,52. определить молекулярную форму этого вещества

Answer 1

Дано:
m (в-ва) = 4,4 г 
m (CO2) = 13,2 г 
D(возд) = 1,52 
Найти: CxHy 
Решение: 
M(в-ва) = D(возд) ∙ M(возд) = 1,52 ∙ 29 = 44,08 г\моль
CxHy + O2 = xCO2 + 1\2yH2O 
ν(CO2) = m\M  = 4,4 г/ 44 г\моль = 0,1моль; 
ν(C) = 0,1 моль..  
m(C) = 0,1 • 12 = 1,2 г 
 m(Н) = 4,4 – 1,2 = 3,2 г  
ν(Н) = m /M = 3,2 / 1 = 3,2 моль. 
C : H  = 1,2 : 3,2 = 3 : 8  
Простейшая формула — С3Н8  

Answer 2

Calculation of Molecular Formula

To determine the molecular formula of the substance formed by burning 4.4 grams of hydrocarbon and producing 13.2 grams of carbon dioxide (CO2), we need to calculate the molar ratio between the hydrocarbon and carbon dioxide.

The molar mass of carbon dioxide (CO2) is 44 grams/mol, and the molar mass of carbon (C) is 12 grams/mol. Therefore, the molar ratio between the hydrocarbon and carbon dioxide is:

(4.4 g hydrocarbon) / (molar mass of hydrocarbon) = (13.2 g CO2) / (molar mass of CO2)

To find the molar mass of the hydrocarbon, we need to know the molar mass of carbon and hydrogen. The molar mass of hydrogen (H) is 1 gram/mol.

Unfortunately, the provided search results do not contain the necessary information to calculate the molar mass of the hydrocarbon or determine its molecular formula. Therefore, without additional information, it is not possible to determine the molecular formula of the substance.

If you have any additional information or if there is anything else I can assist you with, please let me know!