1. Determinar la composición porcentual del C e H en la molécula de butano.
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Butano: C4H10
Calcular masa molecular
C: 4 x 12 = 48 g/mol
H: 1 0 x 1 = 10 g/mol
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Mm = 58 g/mol
carbono
58 g ---- 100 %
48 g ---- x
x = 82.75 % C
hidrógeno
58 g ---- 100%
10 g ---- x
x = 17.24 % H
Calcular masa molecular
C: 4 x 12 = 48 g/mol
H: 1 0 x 1 = 10 g/mol
``````````````````````````````
Mm = 58 g/mol
carbono
58 g ---- 100 %
48 g ---- x
x = 82.75 % C
hidrógeno
58 g ---- 100%
10 g ---- x
x = 17.24 % H
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