Question

Con base en la formula estructural sguiente, Calcule el porcentaje de carbono presente (CH2CO)2C6H3(COOH)

Answer 1

Mm(CH2CO)2C6H3(COOH)

C:11 X 12 = 132 g/mol
H: 8 x 1 = 8 g/mol
O: 4 x 16 = 64 g/mol
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       Mm = 204 g/mol

CARBONO

204 g ---------- 100 %
132 g ----------  x
x = 64.70 %
 
HIDROGENO

204 g ---------- 100 %
    8 g ----------  x

x =3.92 %
          
OXIGENO

204 g ---------- 100 %
  64g ----------  x
x = 31.37 %