Question

two oxide of metal contain 27.6% and 30% oxygen.if the formula of first oxide is M3O4.then formula of second oxide is
a mo
b m2o
c m203
d mo2
solve it...................by explanation??????????

Answer 1

Percentage of oxygen=27.6 Precentage of metal=100-27.6=72.4 Formula of the metal oxide=M3O4 Let the atomic mass of the metal= x Percentage by weight of the metal in the oide M3O4=3x *100/3x+64=72.4 3x=0.724*3x+0.724*64 0.828x=46.336 x=56 The atomic mass of the metal=56 Element Percentage Atomic Mass Atomic ratio Simplest ratio whole no.ratio M 70 56 70/56=1.25 1.25/1.25=1 2 O 30 16 30/16 1.88/1.25=1.5 3 M2O3... Or... In the case of the 1st oxide, Mass of oxygen = 27.6 Therefore, Mass of metal = 100 - 27.6 = 72.4 In the case of the 2nd oxide, Mass of oxygen = 30 Therefore, Mass of metal = 100 - 30 = 70 Also, Formula of the 1st oxide is M3O4 Therefore, Number of atoms of metal in the 2nd oxide = (3/72.4) x 70 = 2.9 Number of atoms of oxygen in the 2nd oxide = (4/27.6) x 30 = 4.35 Thus, Ratio of metal : oxygen in the 2nd oxide = 2.9 : 4.35 or ≈ 2 : 3 Hence, The Formula of the 2nd oxide is M2O3. Hope I helped u.. Plz mark as the BRAINLIEST answer if helpful..

Answer 2

let atomic mass of  Metal = m
    M3  O4  =>  molecular mass = 3m+4*16
  % of Oxygen:  64/(3m+64)  = 0.276            =>   m = 55.96 (nearly 56)

Let 2nd Oxide be:   Mx Oy
      % of oxygen = 16 y / (56 x + 16 y)  = 0.30 
          => 11.2 y = 16.8 x   
          => 2  y = 3 x

2nd oxide =  M2 O3