Question

How can I find the percent compositions of ${\text{Mg}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$?

Answer 1

Hint: We can find the percent compositions of each atom i.e. magnesium (${\text{Mg}}$), nitrogen (${\text{N}}$) and oxygen (${\text{O}}$) present in given compound in the terms of mass percent composition of ${\text{Mg}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$

Complete answer:
Formula which is used to calculate mass percent composition of each atom is as follow:
\[\% {\text{ }}composition{\text{ }}of{\text{ }}atom{\text{ }} = \dfrac{{Mass{\text{ }}of{\text{ }}atom}}{{Mass{\text{ }}of{\text{ }}molecule}} \times 100\]
Mass of the given compound i.e. Magnesium nitrate (${\text{Mg}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$) is calculate by adding masses of all the atoms i.e. magnesium (${\text{Mg}}$), nitrogen (${\text{N}}$) and oxygen (${\text{O}}$) present in ${\text{Mg}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ and for this first we have to list the atomic mass of all atoms that constituent the ${\text{Mg}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$ compound.
-Atomic mass of Magnesium (${\text{Mg}}$) atom = ${\text{24}}{\text{.31g/mol}}$
-Atomic mass of two Nitrogen (${\text{N}}$) atoms = ${\text{2}} \times {\text{14}}{\text{.01 = 28}}{\text{.02g/mol}}$
-Atomic mass of six Oxygen (${\text{O}}$) atoms = $6 \times {\text{15}}{\text{.99 = 95}}{\text{.94g/mol}}$
-So molecular mass of Magnesium nitrate (${\text{Mg}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$) = ${\text{24}}{\text{.31 + 28}}{\text{.02 + 95}}{\text{.94 = 148}}{\text{.27g/mol}}$
Now % composition of each atom present in the ${\text{Mg}}{\left( {{\text{N}}{{\text{O}}_{\text{3}}}} \right)_{\text{2}}}$compound is calculated as follow:
-% composition of Magnesium atom (${\text{Mg}}$) = $\dfrac{{24.31}}{{148.27}} \times 100 = 16.39\% $
-% composition of Nitrogen atom (${\text{N}}$) = $\dfrac{{28.02}}{{148.27}} \times 100 = 18.9\% $
-% composition of Oxygen atom (${\text{O}}$) = $\dfrac{{95.94}}{{148.27}} \times 100 = 64.7\% $

Note:
Here some of you may do wrong calculation if you only put the mass of only one nitrogen and one oxygen in place of two nitrogen atoms (${\text{N}}$) as well as for six oxygen atoms (${\text{O}}$) also. So, always keep in mind while calculating the atomic mass of each constituent first we have to count the number of that atom in the given compound and then multiply that number to the atomic mass of that atom.