Calculate the percentage composition of all the elements present in lead nitrate.
[ $P b = 207, N = 14, O = 16$ ]

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Solution

Molecular weight of $P b (N O_{3})_{2} = 331 g / m o l$
% $P b = \frac{207}{331} \times 100 = 62.5$ %
% $N = \frac{2 \times 14}{331} \times 100 = 8.45$ %
% $O = \frac{6 \times 16}{331} \times 100 = 29$ %

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Similar questions

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

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

P b = 90.66 %; O = 9.34 %

[P b = 207; O = 16]

2 P b (N O_{3})_{2} \to 2 P b O + 4 N O_{2} + O_{2}

(P b = 207; N = 14; O = 16)

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