Calcular masa molecular y composición porcentual de
A) Mg3 (PO4)2
B) Cu(NO3)2
C) Mg SO4
D) Fe(OH)3
E) CO2
Respuestas a la pregunta
Contestado por
13
A) Mg3 (PO4)2
Mm Mg: 3 x 24 = 72 g/mol
P: 2 x 31 = 62 g/mol
O: 8 x 16 = 128 g/mol
````````````````````````````````````````````
Mm = 262 g/mol
Mg: 262 g ----- 100 %
72 g ----- x
x = 27.48 %
P: 262 g ----- 100 %
62 g ----- x
x = 23.66 %
O: 262 g ----- 100 %
128 g ----- x
x = 48.85 %
B) Cu(NO3)2
Mm Cu: 1 x 63.5 = 63.5 g/mol
N: 2 x 14 = 28 g/mol
O: 6 x 16 = 96 g/mol
````````````````````````````````````````````
Mm = 187.5 g/mol
Cu: 187.5 g ----- 100 %
63.5 g ----- x
x = 34.20 %
N: 187.5 g ----- 100 %
28 g ----- x
x = 14.93 %
O: 187.5 g ----- 100 %
96 g ----- x
x = 51.2 %
C) Mg SO4
Mm Mg: 1 x 24 = 24 g/mol
S: 1 x 32 = 32 g/mol
O: 4 x 16 = 64 g/mol
````````````````````````````````````````````
Mm = 120 g/mol
Mg: 120 g ----- 100 %
24 g ----- x
x = 20 %
S: 120 g ----- 100 %
32 g ----- x
x = 26.66 %
O: 120 g ----- 100 %
64 g ----- x
x = 53.33 %
D) Fe(OH)3
Mm Fe: 1 x 56 = 56 g/mol
O: 3 x 16 = 48 g/mol
H: 3 x 1 = 3 g/mol
````````````````````````````````````
Mm = 107 g/mol
Fe
107 g -------- 100 %
56 g -------- x
x = 52.33 %
O:
107 g -------- 100 %
48 g -------- x
x = 44.85 %
H:
107 g -------- 100 %
3 g -------- x
x = 2.80 %
E) CO2
Mm C: 1 x 12 = 12 g/mol
O: 2 x 16 = 32 g/mol
`````````````````````````````
Mm = 44 g/mol
C
44 g ---- 100 %
12 g ----- x
x = 27.27 %
O:
44 g ------ 100 %
32 g ------ x
x = 72.72 %
Mm Mg: 3 x 24 = 72 g/mol
P: 2 x 31 = 62 g/mol
O: 8 x 16 = 128 g/mol
````````````````````````````````````````````
Mm = 262 g/mol
Mg: 262 g ----- 100 %
72 g ----- x
x = 27.48 %
P: 262 g ----- 100 %
62 g ----- x
x = 23.66 %
O: 262 g ----- 100 %
128 g ----- x
x = 48.85 %
B) Cu(NO3)2
Mm Cu: 1 x 63.5 = 63.5 g/mol
N: 2 x 14 = 28 g/mol
O: 6 x 16 = 96 g/mol
````````````````````````````````````````````
Mm = 187.5 g/mol
Cu: 187.5 g ----- 100 %
63.5 g ----- x
x = 34.20 %
N: 187.5 g ----- 100 %
28 g ----- x
x = 14.93 %
O: 187.5 g ----- 100 %
96 g ----- x
x = 51.2 %
C) Mg SO4
Mm Mg: 1 x 24 = 24 g/mol
S: 1 x 32 = 32 g/mol
O: 4 x 16 = 64 g/mol
````````````````````````````````````````````
Mm = 120 g/mol
Mg: 120 g ----- 100 %
24 g ----- x
x = 20 %
S: 120 g ----- 100 %
32 g ----- x
x = 26.66 %
O: 120 g ----- 100 %
64 g ----- x
x = 53.33 %
D) Fe(OH)3
Mm Fe: 1 x 56 = 56 g/mol
O: 3 x 16 = 48 g/mol
H: 3 x 1 = 3 g/mol
````````````````````````````````````
Mm = 107 g/mol
Fe
107 g -------- 100 %
56 g -------- x
x = 52.33 %
O:
107 g -------- 100 %
48 g -------- x
x = 44.85 %
H:
107 g -------- 100 %
3 g -------- x
x = 2.80 %
E) CO2
Mm C: 1 x 12 = 12 g/mol
O: 2 x 16 = 32 g/mol
`````````````````````````````
Mm = 44 g/mol
C
44 g ---- 100 %
12 g ----- x
x = 27.27 %
O:
44 g ------ 100 %
32 g ------ x
x = 72.72 %
Otras preguntas