Calculate the percentage of (i) Fluorine (ii) Sodium (iii) Aluminium in sodium aluminium fluoride [Na₃AlF₆] to the nearest whole number

[Na = 23, Al = 27, F = 19]

(i) Molecular weight of sodium aluminium fluoride (Na₃AlF₆)
= 3(23) + 27 + 6(19)
= 69 + 27 + 114
= 210 g

210 g of sodium aluminium fluoride contains 114 g of fluorine

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{114}{210}$ x 100 = 54.28% = 54%

(ii) 210 g of sodium aluminium fluoride contains 69 g of sodium

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{69}{210}$ x 100 = 32.85% = 33%

(iii) 210 g of sodium aluminium fluoride contains 27 g of aluminium

∴ 100 g of sodium aluminium fluoride will contain = $\dfrac{27}{210}$ x 100 = 12.85% = 13%

Hence, percentage of F = 54%, Na = 33%, Al = 13%