# How many moles of atoms are in 3.00 g of ""^13C?

There are $\frac{3}{13} \times {N}_{A} \text{ carbon atoms}$, where ${N}_{A} = \text{Avogadro's Number, } 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$
In $13 \cdot g$ of ""^(13)C there are precisely $\text{Avogadro's number of atoms}$.You have specified $3.00 \cdot g$, thus there are $\frac{3.00 \cdot g}{13.00 \cdot g \cdot m o l} \times 6.022 \times {10}^{23} \cdot m o {l}^{-} 1$ $=$ ??""^13C " atoms"