A rock is 5.0% by mass fayalite (Fe₂SiO₄), 7.0% by mass forsterite (Mg₂SiO₄), and the remainder silicon dioxide. What is the mass percent of each element in the rock?

If the rock is of 100 g then it will have 5 g of ${\mathrm{Fe}}_2{\mathrm{SiO}}_4$, 7 g of ${\mathrm{Mg}}_2{\mathrm{SiO}}_4$and remaining 88 g of ${\mathrm{SiO}}_2$.

$$\begin{gathered} 1\mathrm{mole}\mathrm{of}{\mathrm{SiO}}_2(60\mathrm{g})\mathrm{contains}28\mathrm{g}\mathrm{of}\mathrm{Si} \\ \mathrm{Then}88\mathrm{g}\mathrm{of}{\mathrm{SiO}}_2\mathrm{will}\mathrm{contain}\frac{28\mathrm{g}}{60\mathrm{g}}\times 88\mathrm{g}=41\mathrm{g} \\ 1\mathrm{mole}\mathrm{of}{\mathrm{SiO}}_2(60\mathrm{g})\mathrm{contains}32\mathrm{g}\mathrm{of}\mathrm{O} \\ \mathrm{Then}88\mathrm{g}\mathrm{of}{\mathrm{SiO}}_2\mathrm{will}\mathrm{contain}\frac{32\mathrm{g}}{60\mathrm{g}}\times 88\mathrm{g}=47\mathrm{g} \end{gathered}$$

On adding the atomic masses of constituent atoms,

$$\begin{gathered} \mathrm{molecular}\mathrm{mass}=2\mathrm{Mg}+\mathrm{Si}+4\mathrm{O} \\ =2(24.31)+28.09+\left(16\right)=140.71\mathrm{amu} \end{gathered}$$

On solving the percentage,

$$\begin{gathered} 1\mathrm{mole}\mathrm{of}{\mathrm{Mg}}_2{\mathrm{SiO}}_4(140.7\mathrm{g})\mathrm{contains}48\mathrm{g}\mathrm{of}\mathrm{Mg} \\ \mathrm{Then}7\mathrm{g}\mathrm{of}{\mathrm{Mg}}_2{\mathrm{SiO}}_4\mathrm{will}\mathrm{contain}\frac{48\mathrm{g}}{140.7\mathrm{g}}\times 7\mathrm{g}=2.39\mathrm{g} \\ 1\mathrm{mole}\mathrm{of}{\mathrm{Mg}}_2{\mathrm{SiO}}_4(140.7\mathrm{g})\mathrm{contains}28\mathrm{g}\mathrm{of}\mathrm{Si} \\ \mathrm{Then}7\mathrm{g}\mathrm{of}{\mathrm{Mg}}_2{\mathrm{SiO}}_4\mathrm{will}\mathrm{contain}\frac{28\mathrm{g}}{140.7\mathrm{g}}\times 7\mathrm{g}=1.39\mathrm{g} \\ 1\mathrm{mole}\mathrm{of}{\mathrm{Mg}}_2{\mathrm{SiO}}_{-4}(140.7\mathrm{g})\mathrm{contains}64\mathrm{g}\mathrm{of}\mathrm{O} \\ \mathrm{Then}7\mathrm{g}\mathrm{of}{\mathrm{Mg}}_2{\mathrm{SiO}}_4\mathrm{will}\mathrm{contain}\frac{64\mathrm{g}}{140.7\mathrm{g}}\times 7\mathrm{g}=3.18\mathrm{g} \end{gathered}$$

On solving the percentage,

$$\begin{gathered} 1\mathrm{mole}\mathrm{of}F{e}_2{\mathrm{SiO}}_4(203.77\mathrm{g})\mathrm{contains}111.7\mathrm{g}\mathrm{of}Fe \\ \mathrm{Then}5\mathrm{g}\mathrm{of}F{e}_2{\mathrm{SiO}}_4\mathrm{will}\mathrm{contain}\frac{111.7\mathrm{g}}{203.77\mathrm{g}}\times 5\mathrm{g}=2.74\mathrm{g} \\ 1\mathrm{mole}\mathrm{of}F{e}_2{\mathrm{SiO}}_4(203.77\mathrm{g})\mathrm{contains}28\mathrm{g}\mathrm{of}\mathrm{Si} \\ \mathrm{Then}5\mathrm{g}\mathrm{of}F{e}_2{\mathrm{SiO}}_4\mathrm{will}\mathrm{contain}\frac{28\mathrm{g}}{203.77\mathrm{g}}\times 5\mathrm{g}=0.687\mathrm{g} \\ 1\mathrm{mole}\mathrm{of}F{e}_2{\mathrm{SiO}}_4(203.77\mathrm{g})\mathrm{contains}64\mathrm{g}\mathrm{of}\mathrm{O} \\ \mathrm{Then}5\mathrm{g}\mathrm{of}F{e}_2{\mathrm{SiO}}_4\mathrm{will}\mathrm{contain}\frac{64\mathrm{g}}{203.77\mathrm{g}}\times 5\mathrm{g}=1.57\mathrm{g} \end{gathered}$$

On adding all the percentages,

$\begin{array}{l}\mathrm{Fe}=2.74\%\\ {\mathrm{M}}_{\mathrm{g}}=2.39\%\\ \mathrm{Si}=0.69\%+1.4\%+41.0\%=43.09\%\\ \mathrm{O}=1.57\%+3.18\%+47\%=51.75\%\end{array}$