un hidrocarburo saturado gaseoso esta formado por un 80% de carbono. ¿Cual es su formula molecular si en condiciones normales si densidad es 1,34 g*L-1
Respuestas
Respuesta dada por:
6
1. Calcular % de H
% H = 100% - 80% = 20 % de H
2. Calcular moles de cada elemento
C: 80 g / 12 g/mol = 6.67 mol
H: 20 g/1 g/mol = 20 mol
3. dividir entre el menor de los resultados:
C: 6.67 mol / 6.67 mol = 1
H: 20 mol / 6.67 mol = 3
4. Fórmula empírica: CH3
5. Calcular Mm FE
C: 1 x 12 = 12 g/mol
H: 3 x 1 = 3 g/mol
```````````````````````````````
Mm = 15 g/mol
6. aplicar la ley general de los Gases ya que la fórmula molecular CN
V · P = n · R · T
T= 273 K
P = 1 atm
R = 0.0821 (L atm/mol K)
n = m/Mm (1) ; d = m/v (2)
sustituir (1) V · P = m/Mm · R · T despejar Mm y V
P · Mm = m/ v · R · T sustituir m/v = d
Mm = d · R · T
`````````````
P
Calcular Mm del gas
Mm = 1.34 g/mL · 0.0821 ( L · atm / mol · K) · 273 K
``````````````````````````````````````````````````````````````
1 atm
Mm = 30.03 g/mol
7. Calcular n = Mm compuesto/Mm FE
n = 30 g/mol / 15 g/mol
n = 2
8. Formula molecula: (CH3)2 = C2H3 / ( CH3 - CH3 ) etano
% H = 100% - 80% = 20 % de H
2. Calcular moles de cada elemento
C: 80 g / 12 g/mol = 6.67 mol
H: 20 g/1 g/mol = 20 mol
3. dividir entre el menor de los resultados:
C: 6.67 mol / 6.67 mol = 1
H: 20 mol / 6.67 mol = 3
4. Fórmula empírica: CH3
5. Calcular Mm FE
C: 1 x 12 = 12 g/mol
H: 3 x 1 = 3 g/mol
```````````````````````````````
Mm = 15 g/mol
6. aplicar la ley general de los Gases ya que la fórmula molecular CN
V · P = n · R · T
T= 273 K
P = 1 atm
R = 0.0821 (L atm/mol K)
n = m/Mm (1) ; d = m/v (2)
sustituir (1) V · P = m/Mm · R · T despejar Mm y V
P · Mm = m/ v · R · T sustituir m/v = d
Mm = d · R · T
`````````````
P
Calcular Mm del gas
Mm = 1.34 g/mL · 0.0821 ( L · atm / mol · K) · 273 K
``````````````````````````````````````````````````````````````
1 atm
Mm = 30.03 g/mol
7. Calcular n = Mm compuesto/Mm FE
n = 30 g/mol / 15 g/mol
n = 2
8. Formula molecula: (CH3)2 = C2H3 / ( CH3 - CH3 ) etano
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