Question

How do you calculate the percentage composition of oxygen in \[Zn{\left( {N{O_3}} \right)_2}\]?

Answer 1

Hint: The percent composition is the percentage of the individual atoms which makes a compound. The percentage is obtained by the ratio of the mass of an atom and the total mass of the compound multiplied by \[100\].

Complete step by step answer:
A composition is defined as the mixture of several elements in a definite ratio or proportion. The moles of elements and the atomic masses of the elements are used for calculation. The percent composition is expressed by the relation
$\% {C_E} = \dfrac{{{m_E}}}{M} \times 100$ where \[{C_E}\] is the percentage composition of an element, \[{m_E}\] is the amount of the individual element and \[M\] is the total amount of the compound.
The given compound is zinc nitrate. It is composed of one zinc ion and two nitrate ions. The zinc ion is the cation and the nitrate ion is the anion. The element zinc is in \[ + 2\] oxidation state. The nitrate ion carries unit negative charge. So to satisfy the two positive charges of zinc two nitrate ions are required. The molecule is a neutral species.
So at first we need to evaluate the atomic mass of the elements involved in the compound. The elements which consist of zinc nitrate are zinc, nitrogen and oxygen. The chemical formula of zinc nitrate is \[Zn{\left( {N{O_3}} \right)_2}\].
The atomic mass of \[Zn\] is \[65.38u\], atomic mass of \[N\] is \[14.0067u\] and atomic mass of \[O\] is \[15.999u\]. The chemical formula is simply written as \[Zn{N_2}{O_6}\] for calculation.
Mass of \[Zn\] = \[1{\text{ }} \times \] atomic mass of \[Zn\] = \[1 \times 65.38u = 65.38u\].
Mass of \[N\] = \[2{\text{ }} \times \] atomic mass of \[N\] = \[2 \times 14.0067u = 28.0134u\].
Mass of \[O\] = \[6{\text{ }} \times \] atomic mass of \[O\] = \[6 \times \;15.999u\; = 95.994u\].
Thus the mass of \[Zn{\left( {N{O_3}} \right)_2}\] = \[65.38u + 28.0134u + 95.99u = 189.3834u\].

Hence the percentage composition of oxygen in \[Zn{\left( {N{O_3}} \right)_2}\] is determined as
\[\% {\text{ O}} = \dfrac{{95.994u}}{{189.3834u}}\; \times {\text{ }}100\% = 50.69\% \].

Note: Percent composition is the amount of each element by mass present in a compound. It helps in determining the stoichiometry of each element present in a chemical composition.