A non-metal X forms two oxides, A and B. The ratio of weight of the element X to the weight of oxygen in A and B is 7:20 and 7: 16, respectively. If the molecular mass of the oxide, A is 108, identify A and B.

Explanation

In 108 g of `A , (7)/(27) xx 108 = 28g` of A is present .
Since the valency of oxgyen is 2, we can consider the formula of A as `X_2 O_a` .
`:. a=(108 -28)/(16)=5` and the formula of A is `X_2O_5`.
Atomic weight of X = 28/2 =14
`:.` Element X is nitrogen , and A is `N_2O_5`.
Let the formula of B is `N_2O_b` since the ratio of N and O in B is 7:16 , the number of oxygen atoms present in B is 4 [`:.` 28 g of nitrogen combines with 64 g of oxygen]
`:.` Formula of B is `N_2O_4` (or) `NO_2`.