Un compuesto tiene la siguiente composición centesimal C=92,35 y H=7,7%¿cual es su fórmula molecular si su peso molecular es 78g/mol? Pesos atómicos, C=12g/mol,H=1,00g/mol
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C 92.35/12 = 7.695
H 7.7/1 = 7.7
C 7.695/7.695 = 1
H 7.7/7.695= 1
Relación C1H1
Peso molecular: 13
78÷13=6
Fórmula: C6H6
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C = 92,35 g / 12 g/mol = 7.695 mol
H= 7,7, g / 1 g/mol = 7,7 mol
dividir cada elemento entre el menor de los resultados
C = 7.695 mol / 7.695 mol = 1
H = 7.7 mol / 7.695 mol = 1
FE = (C1H1)
MASA MOLECULAR DE LA FORMULA EMPIRICA
C: 1 x 12 g = 12 g/mol
H: 1 x 1 g = 1 g / mol
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Mm = 13 g/mol
n(MOLES) = 78 g/mol = 6
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13 g/mol
n = 6
FORMULA MOLECULAR = (C1H1)n = (C1H1)6 = C6H6
H= 7,7, g / 1 g/mol = 7,7 mol
dividir cada elemento entre el menor de los resultados
C = 7.695 mol / 7.695 mol = 1
H = 7.7 mol / 7.695 mol = 1
FE = (C1H1)
MASA MOLECULAR DE LA FORMULA EMPIRICA
C: 1 x 12 g = 12 g/mol
H: 1 x 1 g = 1 g / mol
``````````````````````````````````````
Mm = 13 g/mol
n(MOLES) = 78 g/mol = 6
````````````````
13 g/mol
n = 6
FORMULA MOLECULAR = (C1H1)n = (C1H1)6 = C6H6
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