\(1đvC=\dfrac{6,642.10^-23}{40}=1,6605.10^{-24}(g)\)
\(Na=1,6605.10^{-24}. 23=3,81915.10^{-23}(g)\)
\(S=1,6605.10^{-24}.32=5,3136.10^{-23}(g)\)
\(Mg=1,6605.10^{-24}.24=3,9852.10^{-23}(g)\)
\(Zn=1,6605.^{-24}.65=1,079325.10^{-22}(g)\)
\(Fe=1,6605.10^{-24}.56=9,2988.10^{-23}(g)\)