Question

Determinar la composición porcentual en masa del Fe(NO3)3 (Masa molar Fe= 55,85 g/mol; N=14,01 g/mol; O= 16,00 g/mol)

Answer 1

Fe(NO3)3

Fe: 1 x 55.85 g/mol = 55.85 g/mol
 N:  3 x 14.01 g/mol = 42.03 g/mol
  O: 9 x 16 g/mol = 144 g/mol
````````````````````````````````````````````````````
                 Mm = 241.88 g/mol

Hierro

241.88 g ---- 100%
  55.85 g -----   x

x = 23.08 % Fe

Nitrógeno
241.88 g ---- 100%
  42.03 g -----   x

x = 17.38 % N

Oxígeno
241.88 g ---- 100%
     144 g -----   x

x = 59.53 % O

Answer 2

Explicación:

C6H12:

C=carbono-12.0107×6=72.0742

H-hidrogeno-1.007×12=12.084

sumo total= 72.0742+12.084 = 84.1482

res// 84.16

CICLOHEXANO

Fe(NO3)3:

Fe=hierro-55.845

N=nitrógeno-14.006×6=42.0201

O=oxígeno-15.9994×9=143.9946

sumo total=55.846+42.0201+143.9946=241.8597

res// 241.86

NITRATO DE HIERRO