Question

A 44.39 gram sample of iron is heated in the presence of excess oxygen. A metal oxide is formed with a mass of 63.46 g. Determine the empirical formula of the metal oxide. Enter the elements in the order Fe, O empirical formula =

Answer 1

1) Mass or iron = 44.39 g.

2) Mass of Metal oxide ( iron oxide ) = Mass of iron + Mass of oxygen

63.46 g = 44.39 g + Mass of oxygen.

thetefore,

Mass of oxygen = 63.46 g - 44.39 g = 19.07 g oxygen.

3)Moles of iron = Mass(g)/Molar mass = 44.39 g/55.84 g/mol

= 0.795 mole

4) Moles of oxygen = 19.07 g / 15.999 g / mole

= 1.191 mole.

5) To get whole number, we divide each mole by smallest number of mole.

moles of iron = 0.795 / 0.795 = 1 mole x 2 = 2 mole

mole of oxygen = 1.191 / 0.795 = 1.50 mole x 2 = 3 mole.

That is, empirical formula of oxide contains 2 Fe and 3 O.

Therefore impirical formula of metal oxide = Fe₂O₃.