Question

un hidrocarburo (formado por C e H) contiene 85.71% de C y su densidad en condiciones normales es 1.249 g/l. hallar su formula molecular

Answer 1

1. calcular  % de hidrógeno

%H = 100 % - 85.71 % = 14.29 % H

2. Calcular moles de cada elemento

C: 85.71 g / 12 g/mol = 7.14 mol

H: 14.29 g / 1 g/mol = 14.29 mol

3. dividir entre el menor de los resultados

C: 7.14 mol / 7.14 mol = 1

H: 14.29 mol / 7.14 mol = 2

4. FE: CH2

5. calcular Mm de la FE

C: 1 x 12 = 12 g/mol
H: 2 x 1 = 2 g/mol
````````````````````````````````
      Mm = 14 g/mol

6. APLICAR LA FÓRMULA DE LOS GASES IDEALES

P x V = n x R X T

d = m/v ;  n = masa/Mm   sustituir

P x V = m   x R x T
            ````
             Mm

P x Mm = d x R x T

Mm = d x R x T
         ```````````````
               P

R = 0.0821 (L atm / mol K)
T = 273 K
d = 1.249 g/L
P = 1 atm

Mm = 1.249 g/L  x  0.0821 (L atm / mol K) x 273 K
         ````````````````````````````````````````````````````````````
                                   1 atm

Mm = 28 g/mol

7. calcular n = Mm compuesto/Mm FE

n = 28 g/mol
     `````````````
      14 g/mol

n = 2

8. FM: (CH2)2 = C2H4