A compound of lead has following percentage composition, Pb = 90.66%, O = 9.34%. Calculate empirical formula of the compound. [Pb = 207, O = 16]

Element	Percentage	Atomic mass	Relative number of moles	Simplest mole ratio	Whole number ratio
Pb	90.66	207	$\frac{90.66}{207}=0.44$	$\frac{0.44}{0.44}=1$	(1$\times$3) = 3
O	9.34	16	$\frac{9.34}{16}=0.58$	$\frac{0.58}{0.44}=1.32$	(1.32$\times$3) = 3.96$\approx$4

Therefore, the empirical formula of the compound is Pb₃O₄.