1. Let the atomic weights of elements $A$ and $B$ be: $M_A=a, \qquad M_B=b$

2. We are given that there are $0.15$ moles each of compounds $A_2B$ and $AB_3$, and both samples weigh equally.

3. Molar masses of the compounds:

   • For $A_2B$: $M(A_2B)=2a+b$
   • For $AB_3$: $M(AB_3)=a+3b$

4. Since equal moles ($0.15$ mol) of both compounds have equal mass: $0.15(2a+b)=0.15(a+3b)$

5. Cancel $0.15$ from both sides: $2a+b=a+3b$

6. Solve for the ratio: $2a-a=3b-b$ $a=2b$

7. Therefore, $\frac{M_A}{M_B}=2$

So, the atomic weight of $A$ is 2 times the atomic weight of $B$.