Determina la fórmula empírica y la formula molecular de carbono con el cloro C 7.79 CL 92.21 masa molecular 154 ayuda plz ?
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Mm C = 12 g/mol
Cl = 35.4 g/mol
calcular moles de cada elemento:
C: 7.79 g/ 12 g/mol = 0.649 mol
Cl: 92.21 g/ 35.4 g/mol = 2.604 mol
2. dividir cada elemento entre el menor de los resultados:
C: 0.649 mol / 0.649 mol = 1
Cl: 2.604 mol / 0.649 mol = 4
3. FORMULA EMPIRICA: CCL4
4.Calcular Mm de la FE:
C: 1 x 12 = 12 g /mol
Cl: 4 x 35.4 = 141.6 g/mol
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Mm = 153.6 g/mol
n = 154 g/mol / 153.6 g/mol
n = 1
5. FORMULA MOLECULAR:
FM = (CCl4)1 = CCl4 (tetracloruro de carbono
Cl = 35.4 g/mol
calcular moles de cada elemento:
C: 7.79 g/ 12 g/mol = 0.649 mol
Cl: 92.21 g/ 35.4 g/mol = 2.604 mol
2. dividir cada elemento entre el menor de los resultados:
C: 0.649 mol / 0.649 mol = 1
Cl: 2.604 mol / 0.649 mol = 4
3. FORMULA EMPIRICA: CCL4
4.Calcular Mm de la FE:
C: 1 x 12 = 12 g /mol
Cl: 4 x 35.4 = 141.6 g/mol
``````````````````````````````````
Mm = 153.6 g/mol
n = 154 g/mol / 153.6 g/mol
n = 1
5. FORMULA MOLECULAR:
FM = (CCl4)1 = CCl4 (tetracloruro de carbono
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