Reduction occurs at the cathode in
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The empirical formula of the metal oxide would be
Mass of scandium = 0.502g
Mass of scandium + oxygen = 0.768g
Mass of oxygen = 0.768 - 0.502 = 0.266g
Mole of scandium = 0.502/44.95 = 0.011
Mole of oxygen = 0.266/16 = 0.017
Divide by the lowest:
Scandium = 0.011/0.011 = 1
Oxygen = 0.017.0.011 = 1.5
Thus, the empirical formula of the compound would be or
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