Question

Un compuesto orgánico tiene la siguiente composición centesimal:C = 24,24 %, H = 4,05 % y Cl = 71,71 % .Calcula su fórmula empírica y su fórmula molecular, sabiendo que 0,942 g de dicho compuesto ocupan un volumen de 213 mL a 1 atmósfera y 0 ºC.

Answer 1

Calcular moles.

C: 24.24 g/ 12 g/mol = 2.02 mol

H:  4.05 g/ 1 g/mol 0 4.05 mol

71.71 g/ 35.45 g/mol = 2.02 mol

Dividir entre el menor de los resultados

C: 2.02 mol / 2.02 mol = 1

H: 4.05 mol / 25.02 mol = 2

Cl: 2.02 mol / 2.02 mol = 1

FE: CH2Cl

calcular Mm FE

C: 1 x 12 = 12 g/mol

H: 2 x 1 = 2 g/mol

Cl: 1 x 35.45 = 35.45 g/mol

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              Mm = 49.45

Calcular moles aplicando la Ley de los gases ideales:

T = 273 K  y  P =  1 atm  en CN

R = 0.0821 (L atm/mol K)

V = 213 mL / 1000 = 0.213 L

n =       1 atm  x  0.213 L
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       0.0821(L atm/molK) 273

n = 0.0095 mol

calcular Mm

n = masa/Mm  despejar Mm

Mm =  0.942 g / 0.0095 mol

Mm = 99.2 g/mol

calcular n

n = Mm del Compuesto
     `````````````````````````````
                  Mm FE

n = 99.2 g/mol / 49.45 g/mol

n = 2

FM: (CH2Cl)2 = C2H4Cl2