Question

How can I find the percent composition of \[{\left( {N{H_4}} \right)_2}S\]?

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 diammonium sulphide. It is composed of one ammonium ion and one sulphide ion. The ammonium ion is the cation and the sulphide ion is the anion. The molecule is a neutral species.
So at first we need to known the atomic mass of the elements involved in the compound. The elements which consists diammonium sulfide are nitrogen, hydrogen and sulphur. The chemical formula of diammonium sulphide is \[{\left( {N{H_4}} \right)_2}S\].
The atomic mass of \[N\] is \[14.01u\], atomic mass of \[H\] is \[1.008u\] and atomic mass of \[S\] is \[32.06u\]. The chemical formula is simply written as \[{N_2}{H_8}S\] for calculation.
Mass of \[N\] = \[2{\text{ }} \times \] atomic mass of \[N\] = \[2 \times 14.01u = 28.02u\].
Mass of \[H\] = \[8{\text{ }} \times \] atomic mass of \[H\] = \[8 \times 1.008u = 8.064u\].
Mass of \[S\] = \[1{\text{ }} \times \] atomic mass of \[S\] = \[1 \times \;32.06u\; = 32.06{\text{ }}u\].
Thus the mass of \[{\left( {N{H_4}} \right)_2}S\]= \[28.02u + 8.064u + 32.06u = 68.144u\].
Hence the percentage composition of the \[{\left( {N{H_4}} \right)_2}S\] is determined as
\[\% {\text{ }}S = \dfrac{{28.02u}}{{68.144u}}\; \times {\text{ }}100\% = 41.12\% \].
\[\% {\text{ }}H = \dfrac{{8.064u}}{{68.144u}}\; \times 100\% = 11.83\% \].
\[\% N = \dfrac{{32.06u}}{{68.144u}}\; \times 100\% = 47.05\% \].

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.