Question

On being heated in oxygen 5.72 g of red metallic oxide A was converted to 6.36 g Black metallic oxide B. When 4.77 g of B was heated in a stream of H₂ gas, 3.81 g of metal M was formed. ( Atomic weight of metal is 63.50 g) Find

The formula of Red metallic oxide A .....

The formula if Black metallic oxide .....

The equivalent weight of metal M in B .....

Answer 1

According to the question

wt of metal obtained on heating 4.77gmof oxide B=3.81gm

wt of oxygen lost=wt of oxide --wt of metal
=4.77-3.81
=0.96gm
Equivalent mass of metal in oxide B=(wt of metal /wt of metal )*8

=3.81*8/0.96
=31.50

now valency of metal in B=atomic mass of metal/equivalent wt
=63.50/31.5=2


thus formula of black metal oxide = M+2+O-2=MO

Now since according to the first Question .......

6.36gm Of black oxide obtained by 5.72gm of A(Red Metallic oxide )

4.77gm of B is obtained by (5.72/6.36)*4.77gm A =4.29gm A

Now we can say that,

3.81gm M obtained by 4.29gm A

wt of oxygen lost form A = wt of A --wt of metal


=4.29-3.81=0.48gm

Equivalent wt of metal in A =(3.81/0.48)*8=63.50
Valency of metal A= Atomic wt / Equivalent wt
=63.50/63.50=1
Formula of oxide A =M+ + O-2=M2​O