A contains 38.5 g of tin for each 12.3 g of fluorine:
<span>mole ratio: </span>
<span>(38.5 g)/(118.71 g/mol):(12.3 g)/(18.998 g/mol) = 0.324:0.647 = 1:2 ⇒ SnF₂ </span>
<span>B contains 56.5 g of tin for each 36.2 g of fluorine: </span>
<span>mole ratio: </span>
<span>(56.5 g)/(118.71 g/mol):(36.2 g)/(18.998 g/mol) = 0.476:1.905 = 1:4 ⇒ SnF₄
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Answer:
A3B3
Explanation:
Molecular formula = n x empirical formula
(AB) n = 90
MM of AB = 30 g/mol
30n = 90
Divide both side by the coefficient n i.e 30
n = 90/30 = 3
Molecular formula = n x empirical formula
Molecular formula = n x (AB)
Molecular formula = 3(AB) = A3B3
