Calculate the percentage of:
(i) Fluorine
(ii) Sodium and
(iii) Aluminium in sodium aluminium fluoride [Na3AlF6], to the nearest whole number.
[Atomic Mass: Na = 23, Al= 27, F= 19]
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(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 = $\frac{114}{210}$ 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 = $\frac{69}{210}$ 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 = $\frac{27}{210}$ x 100 = 12.85% = 13%
Hence, percentage of F = 54%, Na = 33%, Al = 13%