Question

determina la formula del compuestos xuya composicion es Mn=34,81°/. k=24,68°/. o=40,51°/. siendo su peso molecular 158=

Answer 1

1. calcular moles

K: 24.68 g / 39.09 g/mol = 0.631 mol
Mn: 34.81 g / 54.93 g/mol = 0.634 mol
O: 40.51 g / 16 g/mol = 2.531

2. dividir entre el menor de los resultados

K: 0.631 mol / 0.631 mol = 1
Mn: 0.634 mol / 0.631 mol = 1
O:  2.531 mol / 0.631 mol = 4

3. FE: KMnO4

4. Calcular la Mm de la FE

K: 1 x 39.09 = 39.09 g/mol
Mn: 1 x 54.93 = 54.93 g/mol
O: 4 x 16 = 64 g/mol
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              Mm = 158.02 g/mol

5. calcular n = Mm del compuesto /Mm de la FE

n =  158 g/mol
     ``````````````````
      158.02 g/mol

n = 1

6. FM: (KMnO4)n = (KMnO4)₁ = KMnO4