calcular la composicion de porcental o centesimal e la siguientes sustancias C2H6O , C3H8O3 , C4H10O
Respuestas
Respuesta dada por:
14
Mm C2H6O
C: 2 x 12 = 24 g/mol
H: 6 x 1 = 6 g/mol
O:1 X 16 = 16 g/mol
``````````````````````````````
Mm = 46 g/mol
C: 46 g -------100 %
24 g ------- x
x = 52.69 % C
H: 46 g -------100 %
6 g ------- x
x = 13.04 % H
O: 46 g -------100 %
16 g ------- x
x = 34.78 % O
2.
C3H8O3
C: 3 x 12 = 36 g/mol
H: 8 x 1 = 8 g/mol
O:3 X 16 = 48 g/mol
``````````````````````````````````
Mm = 92 g/mol
C: 92 g -------100 %
36 g ------- x
x = 39.13 % C
H: 92 g -------100 %
8 g ------- x
x = 8069 % H
O: 92 g -------100 %
48 g ------- x
x = 52.17 % O
3.
C4H10O
C: 4 x 12 = 48 g/mol
H: 10 x 1 = 10 g/mol
O:1 X 16 = 16 g/mol
``````````````````````````````````
Mm = 74 g/mol
C: 74g ----- 100%
48 g ----- x
x = 64.86 % C
H: 74g ----- 100%
10 g ----- x
x =13.51 % H
O: 74g ----- 100%
16 g ----- x
x =21.62 % O
C: 2 x 12 = 24 g/mol
H: 6 x 1 = 6 g/mol
O:1 X 16 = 16 g/mol
``````````````````````````````
Mm = 46 g/mol
C: 46 g -------100 %
24 g ------- x
x = 52.69 % C
H: 46 g -------100 %
6 g ------- x
x = 13.04 % H
O: 46 g -------100 %
16 g ------- x
x = 34.78 % O
2.
C3H8O3
C: 3 x 12 = 36 g/mol
H: 8 x 1 = 8 g/mol
O:3 X 16 = 48 g/mol
``````````````````````````````````
Mm = 92 g/mol
C: 92 g -------100 %
36 g ------- x
x = 39.13 % C
H: 92 g -------100 %
8 g ------- x
x = 8069 % H
O: 92 g -------100 %
48 g ------- x
x = 52.17 % O
3.
C4H10O
C: 4 x 12 = 48 g/mol
H: 10 x 1 = 10 g/mol
O:1 X 16 = 16 g/mol
``````````````````````````````````
Mm = 74 g/mol
C: 74g ----- 100%
48 g ----- x
x = 64.86 % C
H: 74g ----- 100%
10 g ----- x
x =13.51 % H
O: 74g ----- 100%
16 g ----- x
x =21.62 % O
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