Question

Two oxides of metal were found to contain
31 6% and 48% of oxygen, respectively. The
formula of first oxide is represented by
M,O, Calculate the formula of the second
oxide​

Answer 1

Answer:

First methor

First oxideM2O3O=31.6%M=68.4%second oxideM2OxO=48%M=50%

Let the atomic weight of the metal =Mg

(2M+48)gofM2O3 contains =2Mg of metal

100gofM2O3 contains =(2M×1002M+48)=68.4

Solve for M=52

Atomic weight of metal = 52

Therefore, in second oxide:

O=4816=3 M=5252=1

Therefore, formula =MO3

second method:

First oxide:

68.4gofM contains =31.6g of oxygen

52gofM contains =31.6×5268.4=24gofO2

Ratio of O in M2O3 and M2Ox=24:48 (given) =1:2

Therefore, the ratio of O in second oxide should be twice the oxygen in first oxide and the % of M is fixed.

Therefore, first oxide =M2O3

Second oxide =M2O6 or MO3

Third method:

a. 68.4gofM in first oxide = 2 atom of M

50g of oxygen in first oxide contains = 3 atom of oxygen

48g of oxygen contains =3×4831.6

= 4.5 atom of oxygen

Ratio of M:O=1.5:4.5=1:3

Thus, the formula is MO3