Question

2 oxides of element contain 55% and 45% of oxygen if 1st has formula NO2 what is formula of second oxide

Answer 1

Answer:

metal oxide MO contains 57.1% of oxygen.

atomic mass of Oxygen = 16 amu

so, percentage mass of O in MO = atomic mass of O/molar mass of MO × 100

or, 57.1% = 16/(M + 16) × 100

or, 57.1(M + 16) = 1600

or, 57.1M + 57.1 × 16 = 1600

or, 57.1M = 1600 - 913.6 = 686.4

or, M = 686.4/57.1 ≈ 12 amu

hence, atomic mass of element M = 12amu.

Let us assume that second metal oxide is M2Oy

then, molar mass of M2Oy = (2 × 12 + 16y) = (24 + 16y)

and percentage mas of oxygen = y × 16/(24 + 16y) × 100

or, 72.7% = 16y/(24 + 16y) × 100

or, 0.727(24 + 16y) = 16y

or, 17.448 + 11.632y = 16y

or, y = 17.448/4.368 ≈ 4

hence, metal oxide is M2O4