Question

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An oxide of metal have 20% oxygen, the eq. wt. ofmetal oxide is -(A) 32(B) 40(C) 48(D) 52

Answer 1

In a metal oxide, 20% oxygen is present by weight. What is the equivalent weight of metal sulphate?

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2 ANSWERS



Pratyush Jain, Master's Chemical Engineering, Indian Institute of Technology, Kanpur

Answered May 4, 2018

This question can have different answers depending upon the stoichiometry of the elements in formula of metal oxide. I am assuming metal in metal sulphate can have different oxidation state than that in metal oxide.

If 20% mass is of oxygen and remaining 80% mass is contributed by metal, metal weight is 4 times (80/20=4) the weight of oxygen.

START WITH TAKING M1, where O refers to oxide

MO; O=16, M=64, Copper having mass 63.5 amu can fit for this combination and metal oxide is Copper peroxide and metal sulphate is Copper(II) Sulphate having 160 g/mol weight.

MO2; M=128; Tellurium is a metalloid.

MO3; M = 192; Iridium. IrO2 can fit here. But not sure if Iridium sulphate do exist.

MO4; M = 256. No metal metal having 256 amu weight; although Fermium is close.

FOR M2

M2O; M = 32; Sulphur is not a metal.

M2O2; M=64; that is again copper.

M2O3; M = 96; Molybdenum may fit here. But again problem is to find its sulphate.

M2O5; No element having 160 amu.

For M3

M3O; M = 32. M3O2; M = 43. M3O4; M=85.3, Rubidium(85.4) can fit here. rubidium sulphate has molar mass 267. Rb2O3 can also be found in some literature.

More combinations can be tried to find other possible answers.

P.S. : Calculation mistakes may occur because I am feeling very sleepy right now.