Question

el análisis de dos hidrocarburos dan la siguiente composición porcentual C=92'26% H=7,74%. Si los pesos moleculares de los hidrocarburos, calculados por otro metodo son: 26,0g y 78,0g ¿cuales son las formulas moleculares?

Answer 1

1. calcular moles

C: 92.26 g / 12 g/mol = 7.68 moles
H: 7.74 g / 1 g/mol = 7.74 mol

2. dividir entre el menor de los resultados

C: 7.68 mol / 7.68 mol = 1
H: 7.74 mol / 7.68 mol = 1

3. FE: CH

4. Calcular Mm FE

C: 1 x 12 = 13 g/mol
H: 1 x 1 = 1 g/mol
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       Mm = 13 g/mol

5 calcular n utilizando 26 g/mol

n = Mm Compuesto / Mm FE

n = 26 g/mol / 13 g/mol

n = 2

6. FM: (CH)2 = C2H2  (etino)

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calcular n utilizando 78 g/mol

n = Mm Compuesto / Mm FE

n = 78 g/mol / 13 g/mol

n = 6

6. FM: (CH)6 = C6H6  (benceno)