1. The problem asks for the formula of the nitride of element X, given that X is part of the compound $X_2(SO_4)_3$. 2. From the formula $X_2(SO_4)_3$, we see that element X forms a compound with sulfate ions $SO_4^{2-}$. 3. The sulfate ion has a charge of $-2$, and there are 3 sulfate ions, so total negative charge is $3 \times (-2) = -6$. 4. To balance the charge, the 2 atoms of element X must have a total positive charge of $+6$. 5. Therefore, the charge on each X atom is $\frac{+6}{2} = +3$, so X is in the +3 oxidation state. 6. The nitride ion is $N^{3-}$. 7. To form a neutral nitride compound with X in the +3 oxidation state, the formula must balance charges: $$\text{Charge of X} \times \text{number of X atoms} + \text{Charge of N} \times \text{number of N atoms} = 0$$ 8. Since X is +3 and N is -3, one atom of X will balance one atom of N. 9. Therefore, the formula of the nitride of X is $XN$. Final answer: The formula of the nitride of element X is $XN$.