Question

calculate the %composition of Fluorine in sodium aluminium fluoride​

Answer 1

Answer: 54.3% Fluoride

Explanation: Percent Composition=($\frac{mass}{total mass}$)(100%)

molar mass of $Na_{3} AlF_{6}$: (3(22.99g)+26.98g+6(19.00g))/mol=209.95g/mol

mass of fluoride per mole of compound= 6(19.00g)=114.00g

Plug into percent composition formula:

($\frac{114.00g}{209.95g}$)(100%)=54.3%