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

There are ##3/13xxN_A" carbon atoms"##, where ##N_A="Avogadro's Number, "6.022xx10^23*mol^-1##

In ##13*g## of ##""^(13)C## there are precisely ##"Avogadro's number of atoms"##.You have specified ##3.00*g##, thus there are ##(3.00*g)/(13.00*g*mol)xx6.022xx10^23*mol^-1## ##=## ##??""^13C " atoms"##