Question

What is the percent composition of nitrogen in a compound of lead (Il) nitride?

Answer 1

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

$PercentComposition=\frac{AtomicMassElement}{MolarMassMolecule}\times100\%$

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:

$MolarMassPb_3N_2=207.2u\times3+14.0067\times2=649.6g/mol$

The percent composition of nitrogen in Pb3N2 will be:

$\begin{gathered} PercentCompositionN_2=\frac{14.0067\times2}{649.6}\times100\% \\ PercentCompos\imaginaryI t\imaginaryI onN_2=\frac{28.0134}{649.6}\times100\operatorname{\%} \\ PercentCompos\imaginaryI t\imaginaryI onN_2=4.31\% \end{gathered}$

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