Question

Two oxides of an element contain 57.1% and 72.7% of oxygen. If the first oxide is MO, the second oxide is:

Answer 1

answer : MO2

explanation : it is given that, 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

⇒ 57.1% = 16/(M + 16) × 100

⇒ 57.1(M + 16) = 1600

⇒ 57.1M + 57.1 × 16 = 1600

⇒57.1M = 1600 - 913.6 = 686.4

⇒ M = 686.4/57.1 ≈ 12 amu

hence, atomic mass of element M = 12amu.

Let us assume that second metal oxide is MOy

then, molar mass of MOy = (12 + 16y) = (12 + 16y)

and percentage mas of oxygen = y × 16/(12 + 16y) × 100

⇒72.7% = 16y/(24 + 16y) × 100

⇒ 0.727(12 + 16y) = 16y

⇒ 8.724 + 11.632y = 16y

⇒ y = 8.724/4.368 ≈ 2

hence, metal oxide is MO2

Answer 2

Answer:

Mo2

Explanation: