Question

Calcule la composición porcentual del cloruro de etilo, C2H5Cl.

Answer 1

Respuesta:

1.   Mm C2H5Cl

C: 2  x 12 = 24 g/mol

H: 5 x 1 = 5 g/mol

Cl: 1 x 35.5 = 35.5 g/mol

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          Mm = 64.5 g/mol

% C

64.5 g ---- 100 %

  24 g -----   x

X = 37.21 % C

% H

64.5 g ---- 100 %

    5 g -----   x

X = 7.75  % H

%Cl

64.5 g ---- 100 %

35.5 g -----   x

X = 55.03 % Cl

Explicación: