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Grade 11Physical Chemistry

two oxides oa a metal contain 72.4% and 70% of metal respectively? if formula of 2nd oxide isM2O3, find that of the first

Profile image of subhya ranjan
8 Years agoGrade 11
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2 Answers

Profile image of Arun
8 Years ago
Dear student
 
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) 
 
 
Regards
Arun
Profile image of Kavya
5 Years ago
As, For 2 metal oxide (M2O3) 
Mass % of O = 100 - mass % of Metal (M)
                        = 100 - 70 
                         = 30% 
Atomic mass of O = 16
Then, According to the Law of Constant Proportion 
70/x  : 30/16 = 2:3
On solving x = 56
And 56 is the mass of Fe (iron) atom.
 For 1 Metal Oxide 
Mass % of Oxygen = 100 - 72.4
                                 = 31.6
According to Law of Constant proprtions 
72.4/56 : 31.6/16 = 3:4
Answer: M3O4

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