Question

hallar la composición centesimal del propano 1-penteno 2-exino

Answer 1

PROPANO: C3H8

Mm:
C: 3  x 12 = 36 g/mol
H: 8 x 1 = 8 g/mol
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           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
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         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