Question

two oxides of metal contain 72.4 and 70 of metal respectively if formula of 2nd oxide is m2o3 find that of the first

Answer 1

For Oxygen, we can calculate as w(%) = 100 -70 = 30

But Ar(O) = 16

Based on the law of proportion - [% / Ar], so according to it,

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

= 70/x : 1.875 = 2:3, where x = 56 (Fe)


So, if w(O) = 100 – 72.4 = 27.6 (%)


=72.4/56 : 27.6/16 = 1.2928 : 1.725 = 1 : 1.334 = 3 : 4

The answer should be - M3O4 (Fe3O4) . 

Answer 2

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)