Question

How many nitrate ions($N{{O}_{3}}^{1-}$), and how many oxygen atoms are present in 1.00 microgram of magnesium nitrate, $Mg{{(N{{O}_{3}})}_{2}}$ ?

Answer 1

Hint: We will attempt this question by calculating total no. of magnesium nitrate molecules in the given mass and then moving further, we will calculate total number of nitrate ions and oxygen atoms present in it.
Formula used:
Total no. of atoms or molecules or ions = ${{N}_{A}}\times n$

Complete answer:
First of all, we need to convert a given mass unit into standard units for easy calculations.
Given mass= 1.00 microgram = $1.00\mu g\times \dfrac{1\text{ }g}{{{10}^{6}}\mu g}=1.00\times {{10}^{-6}}$
 Now, let’s calculate total no. of moles of $Mg{{(N{{O}_{3}})}_{2}}$=
\[\dfrac{1.00\times {{10}^{-6}}g}{148.3g}\times 1mole\text{ Mg(N}{{\text{O}}_{3}}{{\text{)}}_{2}}=6.743\times {{10}^{-9}}\text{moles Mg(N}{{\text{O}}_{3}}{{\text{)}}_{2}}\]
Total no. of molecules (formula units) of $Mg{{(N{{O}_{3}})}_{2}}$= ${{N}_{A}}\times n$
Where
${{N}_{A}}=6.022\times {{10}^{23}}$
n= no. of moles of
=$6.743\times {{10}^{-9}}moles\text{ }Mg{{(N{{O}_{3}})}_{2}}\times \dfrac{6.022\times {{10}^{23}}\text{ formula units of }Mg{{(N{{O}_{3}})}_{2}}}{1mole\text{ }Mg{{(N{{O}_{3}})}_{2}}}$
=$\text{4}\text{.061}\times \text{1}{{\text{0}}^{15}}\text{formula units of }Mg{{(N{{O}_{3}})}_{2}}$
We know that, 1 formula unit of $Mg{{(N{{O}_{3}})}_{2}}$ contain 1 ion of $M{{g}^{2+}}$and 2 ions of $N{{O}_{3}}^{1-}$ .
Therefore, no. of nitrate ions =
$\text{4}\text{.061}\times \text{1}{{\text{0}}^{15}}\text{formula units of }Mg{{(N{{O}_{3}})}_{2}}\times \dfrac{2\text{ N}{{\text{O}}_{3}}^{1-}}{\text{1 formula unit of }Mg{{(N{{O}_{3}})}_{2}}}$
 = $8.12\times {{10}^{15}}N{{O}_{3}}^{1-}$ ions
Further we see that, 1 $N{{O}_{3}}^{1-}$ion has 1 N atom and 3 O atoms.
Therefore, no. of oxygen atom =
 $8.12\times {{10}^{15}}N{{O}_{3}}^{1-}\times \dfrac{3\text{ O atoms}}{1N{{O}_{3}}^{1-}}$
 = $2.44\times {{10}^{16}}\text{ }atoms\text{ }of\text{ }O$
 Hence, there are $8.12\times {{10}^{15}}N{{O}_{3}}^{1-}$ions and $2.44\times {{10}^{16}}\text{ }atoms\text{ }of\text{ }O$ in 1 microgram of $Mg{{(N{{O}_{3}})}_{2}}$

Note:
If question does not asked about total no. of $N{{O}_{3}}^{1-}$ions and we have to directly calculate total no. of N or O atoms, then we must directly count total no. of N or O atoms in $Mg{{(N{{O}_{3}})}_{2}}$ molecule and multiply it with total formula units of$Mg{{(N{{O}_{3}})}_{2}}$.
For example:-
no. of oxygen atom =\[\text{4}\text{.061}\times \text{1}{{\text{0}}^{15}}\text{formula units of }Mg{{(N{{O}_{3}})}_{2}}\times \dfrac{6\text{ O atoms}}{\text{1 formula unit of }Mg{{(N{{O}_{3}})}_{2}}}=2.44\times {{10}^{16}}\text{atoms of O}\]
Similarly, it can be done for N atoms as well.