Question

How would you calculate the number of moles of $Cl$ atoms in $3.61 \times {10^{24}}$ formula units of magnesium chloride, $MgC{l_2}$ ?

Answer 1

We are actually finding the number of moles of $MgC{l_2}$ by using Avogadro’s number. We have to know about formula units as it is mentioned in question . Formula units are those which are the empirical formula of any covalent or ionic network which is used as an independent entity to undergo stoichiometric calculations. It can also consider as a whole number ratio of ions which are represented in an ionic compound. Also, we have to understand that each formula unit of $MgC{l_2}$ possess two chlorine atoms. So, we have to proceed further by multiplying the number of formula units by conversion factor.

Formula used: $n = \dfrac{N}{{{N_A}}}$
Where $n =$ number of moles of particular substance
$N =$ number of formula units
${N_A} = 6.022 \times {10^{23}}mo{l^{ - 1}}$ ( Avogadro’s constant)

Complete step-by-step answer:
Actually the fact we use to deal with this question is that the ratio between Avogadro’s constant and number of formula units is equal to the number of moles which is already mentioned in the formula used above but in equation form.
As we know the formula units of $MgC{l_2}$ is given as follows,
${N_{MgC{l_2}}} = 3.61 \times {10^{24}}$
Now, let us find the number of moles of $MgC{l_2}$ using Avogadro’s constant which is as follows,
${n_{MgC{l_2}}} = \dfrac{{{N_{MgC{l_2}}}}}{{{N_A}}}$
Next, let us have a look on to the stoichiometry of the equation where the number of moles of chloride anions are twice as amount of moles of $MgC{l_2}$ , which will be as follows,
${n_{Cl}} = 2{n_{MgC{l_2}}} = 2\dfrac{{{N_{MgC{l_2}}}}}{{{N_A}}}$
Now, let us substitute the values in the above equation. Then it will be as follows,
${n_{Cl}} = 2\dfrac{{3.61 \times {{10}^{24}}}}{{6.022 \times {{10}^{23}}mo{l^{ - 1}}}}$
$= 12.0mol$
Hence, there are $12.0mol$ of $Cl$ atoms in $3.61 \times {10^{24}}$ formula units of magnesium chloride, $MgC{l_2}$ .

Note: Actually the formula $MgC{l_2}$ makes us understand that in every unit of $MgC{l_2}$ , there are two units of $Cl$. We have to know that we can simply divide the total number of formula units of a given compound by Avogadro number. In this case, we know that here approximately six moles of $MgC{l_2}$ are present but we should not divide it by Avogadro number because we should note that if there are six moles of $MgC{l_2}$ , then there needs to be $2 \times 6 = 12moles$ of $Cl$ as we know there are two moles of $Cl$ ion per mole of $MgC{l_2}$.

So, we should be aware while dealing with this question as there is a chance of doing wrong calculation as we may have a tendency of not calculating the chlorine ions per mole of $MgC{l_2}$. Same steps and formula can be used to deal with any type of these kind questions.