Answer:
The answer to your question is: SnF₂
Explanation:
Empirical formula of Stannous fluoride
Mass percent composition:
F = 24.25%
Sn = 75.75%
Express percents as grams
F = 24.25 g AW F = 19 g/mol
Sn = 75.75g AW Sn = 119 g(mol
Rule of three
19 g of F ------------------ 1 mol
24.25 g ------------------ x
x = (24.25 x 1) / 19 = 1.28 mol
119 g of Sn ---------------- 1 mol
75.75 g of Sn -------------- x
x = (75.75 x 1) / 119 = 0.64 mol
Divide both results by the lowest number of moles
F = 1.28 / 0.64 = 2
Sn = 0.64 / 0.64 = 1
Then, the empirical formula is: Sn₁ F₂ = Sn F₂