Two elements A and B which form $$0.15$$ moles of $$A_2B$$ and $$AB_3$$ type compounds. If both $$A_2B$$ and $$AB_3$$ weigh equally, then the atomic weight of A is ______ times of atomic weight of B.

Let the atomic weight of A be $$M_A$$ and the atomic weight of B be $$M_B$$.

The molecular weight of $$A_2B = 2M_A + M_B$$

The molecular weight of $$AB_3 = M_A + 3M_B$$

Given: 0.15 moles of $$A_2B$$ and 0.15 moles of $$AB_3$$ weigh equally.

Weight of $$A_2B = 0.15 \times (2M_A + M_B)$$

Weight of $$AB_3 = 0.15 \times (M_A + 3M_B)$$

Since they weigh equally:

$$0.15(2M_A + M_B) = 0.15(M_A + 3M_B)$$

Dividing both sides by 0.15:

$$2M_A + M_B = M_A + 3M_B$$

$$2M_A - M_A = 3M_B - M_B$$

$$M_A = 2M_B$$

Therefore, the atomic weight of A is 2 times the atomic weight of B.