Hallar la composición centesimal del propano 1 - penteno 2 - exino?

Mm :

C : 3 x 12 = 36 g / mol

H : 8 x 1 = 8 g / mol

```````````````````````````` Mm : 44 g / mol

C : 44 g - - - - 100 % 36 g - - - - x x = 81.

81 %C

H : 44 g - - - 100 % 8 g - - - x x = 18.

18 %H

1 - penteno CH2 = CH - CH2 - CH2 - CH3 / C5H10

Mm

C : 5 x 12 = 60 g / mol

H : 10 x 1 = 10 g / mol

```````````````````````````` Mm : 70 g / mol

l

C : 70 g - - - - 100 % 60 g - - - - x x = 85.

71 %C

H : 70 g - - - 100 % 10 g - - - x x = 14.

29 %H

2 - hexino CH3 - C ≡ C - CH2 - CH2 - CH3 / C610

Mm

C : 6 x 12 = 72 g / mol

H : 10 x 1 = 10 g / mol

```````````````````````````` Mm : 82 g / mol

l

C : 82 g - - - - 100 % 72 g - - - - x x = 87.

80 %C

H : 82 g - - - 100 % 10 g - - - x x = 12.

20 %H.