answer is: empirical formula is tlâ‚‚o. if we use 100 grams of compound, tha:
m(tl) = 96.2% · 100 g ÷ 100%.
m(tl) = 96.2 g.
n(tl) = m(tl) ÷ m(tl). n(tl) = 96.2 g ÷ 204.4 g/mol.
n(tl) = 0.47 mol. m(o) = 3.77 g. n(o) = m(o) ÷ m(o). n(o) = 3.77 g ÷ 16 g/mol. n(ag) = 0.235 mol. n(tl) . n(o) = 0.47 mol : 0.235 mol /0.235. n(tl) . n(o) = 2 : 1.
m(tl₂o) = 204.4 g/mol · 2 + 16 g/mol = 424.8 g/mol.
m(compound) = 456. 8 g/mol - 424.8 g/mol = 32 g/mol ( 2 oxygens); tl₂o₃.