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).
Introduce la fórmula molecular (ej: H2O, NaCl, C6H12O6)
Masa molar18.015 g/mol
H×22.016 g/mol
O15.999 g/mol
En resumen
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.
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.