Q130CP - Chemistry: Molecular Nature Of Matter And Change

Question

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?

Step-by-Step Solution

Answer

The mass percent of each element in the rock as follows:

 

$$\begin{gathered} \mathrm{Fe}=2.74\% \\ \mathrm{Mg}=2.39\% \\ \mathrm{Sl}=43.09\% \\ \mathrm{O}=51.75\% \end{gathered}$$

Step 1: Finding the mass percent for SiO 2

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 ${\mathrm{SiO}}_2$.

 $$\begin{gathered} 1\mathrm{mole} \mathrm{of} {\mathrm{SiO}}_2(60 \mathrm{g}) \mathrm{contains} 28 \mathrm{g} \mathrm{of} \mathrm{Si} \\ \mathrm{Then} 88 \mathrm{g} \mathrm{of} {\mathrm{SiO}}_2\mathrm{will} \mathrm{contain} \frac{28 \mathrm{g}}{60 \mathrm{g}}\times 88 \mathrm{g}=41 \mathrm{g} \\ 1\mathrm{mole} \mathrm{of} {\mathrm{SiO}}_2(60 \mathrm{g}) \mathrm{contains} 32 \mathrm{g} \mathrm{of} \mathrm{O} \\ \mathrm{Then} 88 \mathrm{g} \mathrm{of} {\mathrm{SiO}}_{2 }\mathrm{will} \mathrm{contain} \frac{32 \mathrm{g}}{60 \mathrm{g} }\times 88 \mathrm{g}=47 \mathrm{g} \end{gathered}$$

Step 2: Finding the mass percent for Mg 2 SiO 4

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,

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

Step 3: Finding the mass percent for Fe 2 SiO 4

On solving the percentage,

$$\begin{gathered} 1\mathrm{mole} \mathrm{of} F{e}_2{\mathrm{SiO}}_4(203.77 \mathrm{g}) \mathrm{contains} 111.7 \mathrm{g} \mathrm{of} Fe \\ \mathrm{Then} 5 \mathrm{g} \mathrm{of} F{e}_2{\mathrm{SiO}}_4\mathrm{will} \mathrm{contain} \frac{111.7 \mathrm{g}}{203.77 \mathrm{g}}\times 5 \mathrm{g}=2.74 \mathrm{g} \\ 1\mathrm{mole} \mathrm{of} F{e}_2{\mathrm{SiO}}_4(203.77\mathrm{g}) \mathrm{contains} 28 \mathrm{g} \mathrm{of} \mathrm{Si} \\ \mathrm{Then} 5 \mathrm{g} \mathrm{of} F{e}_2{\mathrm{SiO}}_{4 }\mathrm{will} \mathrm{contain} \frac{28 \mathrm{g}}{203.77 \mathrm{g} }\times 5 \mathrm{g}=0.687 \mathrm{g} \\ 1\mathrm{mole} \mathrm{of} F{e}_2{\mathrm{SiO}}_4(203.77 \mathrm{g}) \mathrm{contains} 64 \mathrm{g} \mathrm{of} \mathrm{O} \\ \mathrm{Then} 5 \mathrm{g} \mathrm{of} F{e}_2{\mathrm{SiO}}_4\mathrm{will} \mathrm{contain}\frac{64 \mathrm{g}}{203.77 \mathrm{g}}\times 5 \mathrm{g}=1.57 \mathrm{g} \end{gathered}$$

Step 4: Finding the total percentage

On adding all the percentages,

 

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