Question

( 51. An hypothetical element (Y) has two oxides. If first
oxide (YO) contains 40% oxygen and second oxide
contains 20% oxygen then the formula of second
oxide is
(1) YOZ
- (2) Y,02
(3) Y₂0
(4) Y209

Answer 1

Answer:

Explanation:

Well hope you get the solution you are looking for

Answer 2

Answer:

The molecular formula of second oxide is;$Y_{2}O_1$

Explanation:

1) Molar mass of YO = M

Atomic mass of oxygen =  16 g/mol

Percentage of O in YO = 40%

$40\%=\frac{16 g/mol}{M}\times 100$

$M = 40 g/mol$

M = atomic mass of Y + 16 g/mol

Atomic mass of Y = M - 16 g/mol = 40 g/mol - 16 g/mol = 24 g/mol

In second oxide contains 20% oxygen and 80% of Y. that is in 100 g of second oxide we have 20 grams of oxygen and 80 grams of Y.

Moles of oxygen = $\frac{20 g}{16 g/mol}=1.25 mol$

Moles of Y = $\frac{80 g}{24 g/mol}= 3.33 mol$

For oxygen = $\frac{1.25 mol}{1.25 mol}=1$

For Y = $\frac{3.33 mol}{1.25 mol}=2.66\approx 2$

The molecular formula of second oxide is;$Y_{2}O_1$