chem Q

Mass fraction of O in H2O = 16/(1x2 + 16) = 16/18

In 25.4 g of the oxide of metal:
Mass of O = 3.6 x (16/18) = 3.2 g
Mass of metal Y = 25.4 - 3.2 = 22.2 g

Mole ratio Y : O
= (22.2/63.5) : (3.2/16)
= 0.350 : 0.2
= 7 : 4

Empirical formula = Y7O4
=

2009-04-06 23:54:49 補充：
The above answer seems not reasonable.
But if the mass of metal Y is 25.4 g,
molar ratio Y : O
= (25.4/63.5) : (3.2/16)
= 0.4 : 0.2
= 2 : 1

Empirical formula = Y2O
It looks reasonable.

2009-04-06 23:58:50 補充：
sandy's answer is contradictory.
She/He stated that:
"no. of mole of an oxide of metal=25.4/(63.5+16)"
This means that the molar mass of the oxide is (63.5+16) and the oxide is thus YO.
This is contradictory to the answer Y3O2 (Cu3O2).

YO + H2 ---- > Y + H2O
25.4g 3.6g


no. of mole of water=3.6/(2+16)

= 0.2 mol


no. of mole of an oxide of metal=25.4/(63.5+16)

=0.319

mole ratio oxide of metal : water

0.319 : 0.2

3 : 2
Cu3O2