What is the percent composition of muscovite mica if its chemical formula is (KF) 2 (Al2O3) 3 (SiO2) 6 (H2O)

Hello,

To find the percentage composition of muscovite mica, we'll have to first find the molecular mass of the compound.

Chemical formula = (KF) ₂ (Al₂O₃) ₃ (SiO₂) ₆ (H₂O)

(KF) ₂ = 58.097 * 2 = 116.194g/mol

(Al₂O₃) ₃ = 3 * 101.96 = 305.88g/mol

(SiO₂) ₆ = 6 * 60.08 = 360.48g/mol

H₂O = 18g/mol

(KF) ₂ (Al₂O₃) ₃ (SiO₂) ₆ (H₂O) = 116.194 + 305.88 + 360.48 + 18 = 800.554g/mol

Potassium = (78.18 / 800.554) * 100 = 9.765%

Fluorine = (38 / 800.554) * 100 = 4.75%

Aluminium = (162 / 800.554) * 100 = 20.23%

Silicon = (168.48/800.554) * 100 = 21.04%

Oxygen = (352/800.554) * 100 = 43.97%

Hydrogen = (2 / 800.554) * 100 = 0.24%

Muscovite mica is an aluminosilicate compound or a polysillicate compound found in rocks