Question

A metal forms two oxides. % of metal on them is 72.4 and 70. If first oxide is m3o4, what is the formula of the other oxide?

Answer 1

Answer:

For Oxygen, weigh by percentage= 100 -70 = 30

But Ar(O) = 16

Using law of proportion - [% / Ar], so according to it,

= 70/x : 30/16 = 2:3

= 70/x : 1.875 = 2:3,

where x = 56 (Fe)

If w(O) = 100 – 72.4 = 27.6 (%)

=72.4/56 : 27.6/16

= 1.2928 : 1.725

= 1 : 1.334

Simplifying we get

= 3 : 4

The answer is then- M3O4 (Fe3O4)