The relative density of the vapor of the cycloalkane for hydrogen is 28. The mass fraction of carbon

The relative density of the vapor of the cycloalkane for hydrogen is 28. The mass fraction of carbon in the cycloalkane is 85.71%. Find the molecular formula of the cycloalkane.

The relative density of the vapor of the cycloalkane for hydrogen is 28. The mass fraction of carbon in the cycloalkane is 85.71%. Find the molecular formula of the cycloalkane.

Given:

D (H2) = 28;

w% (C) = 85.71%

Find: formula -?

Decision:

1) Find the mass of 1 mole of a substance:

M (cyclane) = D (H2) * M (H2) = 28 * 2 g / mol = 56 / mol.

2) Find the mass of carbon atoms:

56 g – 100%

xg – 85.71%, then x = 85.71% * 56g: 100% = ~ 48g, then the mass of hydrogen accounts for 56g – 48g = 8g.

3) Calculate the number of carbon atoms:

N (C) = m: M = 48g: 12g / mol = 4 carbon atoms.

4) Calculate the number of hydrogen atoms:

N (H) = m: M = 8g: 1g / mol = 8 hydrogen atoms.

Thus, we got the formula C4H8. This is cyclobutane.

Answer: C4H8 – cyclobutane.