Answer
Mass percentage of Si$O_{2}$= 52.4 %.
Work Step by Step
The strategy: 1- find the mass of CaC$O_{3}$that produce 3.95 mg of C$O_{2}$.
2- Find the mass of Si$O_{2}$ in the sandstone
2- Find mass percentage of Si$O_{2}$ in the sandstone.
CaC$O_{3}$ (s) ${\longrightarrow} $ CaO (s) + C$O_{2}$ (g)
Step 1: to find the mass of CaC$O_{3}$ that produce 3.95 mg of C$O_{2}$ :
g C$O_{2}$ ${\rightarrow}$ mol C$O_{2}$ ${\rightarrow}$ mol CaC$O_{3}$${\rightarrow}$ g CaC$O_{3}$
The conversion factors are:
$\frac{1 mol (CO_{2})}{Molar mass (CO_{2})}$ to convert to moles C$O_{2}$, where Molar mass C$O_{2}$:
Molar mass C$O_{2}$= 1 x 12 + 2 x 16 = 44 g
$\frac{1 mol (CaCO_{3})}{1 mol (CO_{2})}$, to convert the moles of C$O_{2}$ to mol of CaC$O_{3}$.
$\frac{Molar mass(CaCO_{3})}{1 mol(CaCO_{3})}$ to convert to g of CaC$O_{3}$, where the molar mass of CaC$O_{3}$ is:
Molar mass CaC$O_{3}$= 1 x 40 + 1 x 12 + 3 x 16= 100.0 g
Now we use the conversion factors to find the mas of CaC$O_{3}$:
3.95 x $10^{-3}$g C$O_{2}$ x $\frac{1 mol (CO_{2})}{Molar mass (CO_{2})}$ x $\frac{1 mol (CaCO_{3})}{1 mol (CO_{2})}$ x $\frac{Molar mass(CaCO_{3})}{1 mol(CaCO_{3})}$ =
3.95 x $10^{-3}$g C$O_{2}$ x $\frac{1 mol (CO_{2})}{44 g}$ x $\frac{1 mol (CaCO_{3})}{1 mol (CO_{2})}$ x $\frac{100 g}{1 mol(CaCO_{3})}$ = 8.9 x $10^{-3}$g CaC$O_{3}$ or 8.9 mg CaC$O_{3}$
Step 2: So the mass of Si$O_{2}$ in the sandstone is
mass of sandstone - mass of CaC$O_{3}$ in the sandstone
18.7 mg - 8.9 mg = 9.8 mg Si$O_{2}$
Step 3 - Find the mass percentage of Si$O_{2}$ in sandstone.
mass percentage of Si$O_{2}$= $\frac{mass _(SiO_{2})}{mass_(sandstone)}$ x 100%
mass percentage of Si$O_{2}$= $\frac{9.8 mg}{18.7}$ x 100%
mass percentage of Si$O_{2}$= 52.4 %
