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Ba: 52.5 g/ 137,32 g/mol = 0.3823 mol
N: 10,7 g / 14 g/mol = 0.7642 mol
O: 36,8 g / 16 g/mol = 2.3 mol
dividir cada elemento entre el menor resultado
Ba: 0.3823 ml / 0.3823 mol = 1
N: 0.7642 mol / 0.3823 mol = 2
O: 2.3 mol / 0.3823 mol = 6
FE ( BaN2O6)n = FÓRMULA EMPIRICA
Masa molecular de la FE
Ba: 1 x 137.32 g = 137.32 g/mol
N: 2 x 14 g = 28 g / mol
O: 6 x 16 g = 96 g/mol
Mm FE = 261.32 g/mol
n = 261 / 261.32 = 1
FM = ( BaN2O6)1 =
BaN2O6 (Nitrato de bario)