\(a.\)
\(M_{MgCO_3}=84\left(\dfrac{g}{mol}\right)\)
\(\%Mg=\dfrac{24}{84}\cdot100\%=28.57\%\)
\(\%C=\dfrac{12}{84}\cdot100\%=14.28\%\)
\(\%O=100-28.57-14.28=57.15\%\)
\(b.\)
\(M_{NaCl}=58.5\left(\dfrac{g}{mol}\right)\)
\(\%Na=\dfrac{23}{58.5}\cdot100\%=39.31\%\)
\(\%Cl=\dfrac{35.5}{58.5}\cdot100\%=60.69\%\)
\(a,M_{MgCO_3}=84(g/mol)\\ \begin{cases} \%_{Mg}=\dfrac{24}{84}.100\%=28,57\%\\ \%_{C}=\dfrac{12}{84}.100\%=14,29\%\\ \%_{O}=100\%-28,57\%-14,29\%=57,14\% \end{cases}\)
\(a,M_{NaCl}=58,5(g/mol)\\ \begin{cases} \%_{Na}=\dfrac{23}{58,5}.100\%=39,32\%\\ \%_{C}=100\%-39,32\%=60,68\% \end{cases}\)
