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Lemmerich · Answer 1 · 2023-03-26T13:35:43+0000

The percent composition of an element in a compound can be found by the following formula:

$PercentComposition=(AtomicMassElement)/(MolarMassMolecule)*100\%$

Therefore, we must first find the weight of the lead(II) nitride molecule. The molecular formula of lead (Il) nitride is Pb3N2, so the molar mass of the element will be:

Molar Mass Pb3N2= Atomic Mass Pb x 3 + Atomic Mass N x 2

the values of the atomic mass of the elements are found in the periodic table.

Atomic Mass Pb=207.2u

Atomic Mass N=14.0067u

So, the molar mass of Pb3N2 will be:

MolarMassPb_3N_2=207.2u*3+14.0067*2=649.6g/mol

The percent composition of nitrogen in Pb3N2 will be:

$\begin{gathered} PercentCompositionN_2=(14.0067*2)/(649.6)*100\% \\ PercentCompos\imaginaryI t\imaginaryI onN_2=(28.0134)/(649.6)*100\operatorname{\%} \\ PercentCompos\imaginaryI t\imaginaryI onN_2=4.31\% \end{gathered}$

Answer: The percent composition of nitrogen in a compound of lead (Il) nitride is 4.31%

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