NumericalModerate1 mark
5(b)
In sodium aluminium fluoride [Na₃AlF₆], [Atomic Mass: Na = 23, Al= 27, F= 19] Calculate the percentage to the nearest whole number of:
Fluorine
Topic: Mole concept, gas volumes and stoichiometry
Previous-year question practice database
Answer
Exam answer
Na₃AlF₆ [Na = 23, Al = 27, F = 19]
Molar mass = (3 × 23) + 27 + (6 × 19) = 69 + 27 + 114 = 210 g mol⁻¹
Percentage of fluorine = (mass of F ÷ molar mass) × 100 = (114 ÷ 210) × 100 = 54.28 %
Percentage of fluorine ≈ 54 %
Explanation
Work out the molar mass once, carefully, and keep it for all three parts — it is the denominator every time.
The subscripts are where marks are lost: SIX fluorines, so 6 × 19 = 114, not 19. And three sodiums, 3 × 23 = 69. A useful check at the end: the three percentages must add up to 100, and 54 + 33 + 13 = 100.
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