Question
Two oxides of a metal contain 72.4% and 70% of metal respectively. If formula of 2nd oxide is M2O3, find that of the first?
Sir the answer is M3O4. But I don't know the steps for doing this type of question, so pls explain me each step!!!
Joshi sir comment
for second oxide ratio of metal and oxygen = 70:30
let molcular mass of metal = M so 70/30 = 2M/3*16
or 7/3 = 2M/48 so M = 24*7/3 = 56
now for first oxide ratio of metal and oxygen = 72.4/27.6 (By weight)
so ratio by no. of atoms = (72.4/56)/(27.6/16) = 72.4*16/27.6*56 = 1158.4/1545.6 = 3/4