Question

Calculate the percentage of:

1. Fluorine
2. Sodium and
3. Aluminum

in sodium aluminum fluoride [Na₃AlF), to the nearest whole number.

[Atomic Mass: Na = 23, Al = 27, F = 19]

Answer 1

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 = `114/210 xx 100`

= 54.28%

= 54%

ii. 210 g of sodium aluminium fluoride contains 69 g of sodium:

∴ 100 g of sodium aluminium fluoride will contain = `69/210 xx 100`

= 32.85%

= 33%

iii. 210 g of sodium aluminium fluoride contains 27 g of aluminium:

∴ 100 g of sodium aluminium fluoride will contain = `27/210 xx 100`

= 12.85%

= 13%