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

Explanation:

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

nuclear fission produces elements that are heavier than helium.

the elements that are used in nuclear fission and their products are much heavier than helium.