# How do you convert 2.54 moles magnesium carbonate to formula units?

By use of $\text{Avogadro's Number} , {N}_{A} = 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$
It is a fact that in $1$ $m o l$ of stuff there are $6.022 \times {10}^{23}$ individual items of that stuff.
Thus in $2.54$ $m o l$ $M g C {O}_{3}$, there are $2.54$ $\times$ ${N}_{A}$ individual formula units. What is the mass of this quantity?