# How many moles of sodium carbonate contain 1.773 xx 10^17 carbon atoms?

If we want $1.773 \times {10}^{17}$ $\text{C}$ atoms in ${\text{Na"_2"CO}}_{3}$, then we can define our own ratio of:
${\text{1 mol C" = "1 mol Na"_2"CO}}_{3}$
$\textcolor{b l u e}{1.773 \times {10}^{17} \text{C atoms") xx ("1 mol C")/(6.022xx10^23 "C atoms") xx ("1 mol Na"_2"CO"_3)/("1 mol C}}$
$= \textcolor{b l u e}{2.944 \times {10}^{- 7} {\text{mols Na"_2"CO}}_{3}}$