Calculate the percentage composition of oxygen in lead nitrate [Pb(NO₃)₂].
(Pb = 207, N = 14, O = 16)

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Solution

Molecular mass of [Pb(NO₃)₂] = [207 + 2 $\times$ {14 + (3 $\times$ 16)}] g = 331 g
Mass of oxygen in [Pb(NO₃)₂] = 96 g
$Percentage of O = \frac{Mass of O in compound}{Molecular mass of Pb {({NO}_{3})}_{2}} \times 100 = \frac{96}{331} \times 100 = 29 %$
Hence, percentage composition of oxygen in [Pb(NO₃)₂] is 29 %.

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