Question

What is the percent composition of muscovite mica. Its chemical formula is

(KF)2(Al2O3)3(SiO2)6(H2O)

Answer 1

`\boxed{\text{0.25 \% H, 43.96 \% O, 4.75 \% F, 20.22 \% Al, 21.05 \% Si, 9.77 \% K}}`

Step-by-step explanation:

The oxide formula is (KF)₂(Al₂O₃)₂(SiO₂)(H₂O).

Rewrite it as a molecular formula — H₂O₂₂F₂Al₆Si₆K₂

The formula for the mass percent of an element is

`\text{Mass \%} = \frac{\text{mass of element}}{\text{mass of compound}} * 100 \, \%`

We can set up a table to calculate the percent of each element.

`\begin{array}{rrrrr}\textbf{Atom} & \textbf{No.} &\textbf{MM/u} & \textbf{Mass/u} & \mathbf{\%} \\\text{H} & 2& 1.01 & 2.02& 0.25 \\\text{O} &22 & 16.00 & 351.98 & 43.96 \\\text{F} & 2 & 19.00 & 38.00 & 4.75 \\\text{Al} & 6 & 26.98 & 161.89 & 20.22 \\\text{Si} & 6 & 28.08 & 168.51 & 21.05 \\\text{K} & 2 & 39.10 & 78.20 & 9.77 \\& & \text{TOTAL =} & \mathbf{800.60} & \mathbf{100.00} \\\end{array}\\`

`\text{The percent composition of muscovite mica is}\\ \boxed{\textbf{0.25 \% H, 43.96 \% O, 4.75 \% F, 20.22 \% Al, 21.05 \% Si, 9.77 \% K}}`