Chemistry, asked by bhuvanav245, 4 months ago

if the oxides of a metal contain 78.7% and 64.5% of oxygen and if the formula of the first oxide is MO, find the formula of other oxide, the ration of oxygen in two oxide is 1:3

Answers

Answered by abhi178

Given info : if the oxides of a metal contain 78.7% and 64.5% of oxygen and if the formula of the first oxide is MO.

To find : the formula of the other oxide. The ratio of oxygen in two oxides is 1 : 3.

Case 1 : oxygen = 78.7 % so metal = 100 - 78.7 = 21.3 %

Let atomic mass of metal is M

Percentage mass of metal = M/(M + 16) × 100

⇒21.3 = 100M/(M + 16)

⇒21.3M + 21.3 × 16 = 100M

⇒21.3 × 16 = 78.7 M

⇒M = 4.33

Now, let formula of the other oxide is MxOy then,

Percentage of metal = xM/(xM + yO) × 100

⇒100 - 64.5 = x4.33/(x4.33 + y16) × 100

⇒35.5 = 433x/(4.33x + 16y)

⇒4.33x × 35.5 + 16y × 35.5 = 433x

⇒153.7x + 568y = 433x

⇒568y = 279.3x

⇒y/x = 279.3/568 ≈ 0.5 = 1/2

Therefore the formula of the other is M₂O

Answered by Anonymous

Given info : if the oxides of a metal contain 78.7% and 64.5% of oxygen and if the formula of the first oxide is MO.

To find : the formula of the other oxide. The ratio of oxygen in two oxides is 1 : 3.