Question

Determinar la formula molecular de un compuesto que tiene 41,4% de carbono 3,5% de hidrogeno 55,1%y una masa molecular experimental de 116 uma

Answer 1

C: 41.4 g/ 12 g/mol = 3.45 mol
H: 3.5 g/ 1 g/mol = 3.5 mol
O: 55.1 g / 16 g/mol = 3.44 mol

dividir

C: 3.45 mol / 3.44 mol = 1
H: 3.5 mol / 3.44 mol = 1
O: 3.44 mol / 3.44 mol = 1

FE: CHO
Mm FE

C: 1 x 12 = 12 g/mol
H: 1 x 1 = 1 g/mol
O: 1 x 16 = 16 g/mol
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       Mm = 29 g/mol

n = 116 uma / 29 uma
n = 4

FM: (CHO)n = (CHO)4