Que indica la formula empírica del Bromuro de Potasio (KBr)?
Que indica la formula empírica del Cloruro de Magnesio (MgCl2)?
Respuestas
Respuesta dada por:
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KBr
Mm
K: 1 x 39 = 39 g/mol
Br: 1 x 80 = 80 g/mol
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Mm = 119 g/mol
K:
119 g ---- 100 %
39 g ----- x
x = 32.77 %
Br
119 g ---- 100 %
80 g ----- x
x = 67.22 %
K: 32.77 g / 39 g/mol = 0.84 mol
Br: 67.22 g/ 80 g/mol = 0.84 mol
dividir
K: 0.84 mol / 0.84 mol = 1
Br: 0.84 mol / 0.84 mol = 1
FE: KBr
````````````````````````````````````````````````````````````````````````````
Mm: MgCl2
Mg: 1 x 24 = 24 g/mol
Cl: 2 x 35.4 = 70.8 g/mol
````````````````````````````````
Mm = 94.8 g/mol
Mg:
94.8 g ---- 100 %
24 g ----- x
x = 25.31 %
Cl
94.8 g ---- 100 %
70.8 g ----- x
x = 74.68 %
Mg: 25.31 g / 24 g/mol = 1.054 mol
Cl: 74.68 g/ 35.4 g/mol = 2.109 mol
dividir
Mg: 1.054 mol / 1.054 mol = 1
Cl: 2.109 mol / 1.054 mol = 2
FE:MgCl2
Mm
K: 1 x 39 = 39 g/mol
Br: 1 x 80 = 80 g/mol
````````````````````````````````
Mm = 119 g/mol
K:
119 g ---- 100 %
39 g ----- x
x = 32.77 %
Br
119 g ---- 100 %
80 g ----- x
x = 67.22 %
K: 32.77 g / 39 g/mol = 0.84 mol
Br: 67.22 g/ 80 g/mol = 0.84 mol
dividir
K: 0.84 mol / 0.84 mol = 1
Br: 0.84 mol / 0.84 mol = 1
FE: KBr
````````````````````````````````````````````````````````````````````````````
Mm: MgCl2
Mg: 1 x 24 = 24 g/mol
Cl: 2 x 35.4 = 70.8 g/mol
````````````````````````````````
Mm = 94.8 g/mol
Mg:
94.8 g ---- 100 %
24 g ----- x
x = 25.31 %
Cl
94.8 g ---- 100 %
70.8 g ----- x
x = 74.68 %
Mg: 25.31 g / 24 g/mol = 1.054 mol
Cl: 74.68 g/ 35.4 g/mol = 2.109 mol
dividir
Mg: 1.054 mol / 1.054 mol = 1
Cl: 2.109 mol / 1.054 mol = 2
FE:MgCl2
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