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

= 114\over210[\katex] × 100

= 54.28% = 54%

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

∴ 100 g of sodium aluminium fluoride will contain

= 69\over210[\katex] × 100

= 32.85% = 33%

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

∴ 100 g of sodium aluminium fluoride will contain

= 17\over210[\katex] × 100

= 12.85% = 13%

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