Question

How many formula units of sodium nitrate are present in 33.9 g of the substance?

Answer 1

Hint: A mole is the entity that denotes $6.022\times {{10}^{23}}$ particles. Formula units are considered in ionic compounds, as here the ions or the molecules are termed as formula units. 1 mole of any compound will have formula unit mass equal to its molecular mass and the units equal to Avogadro number that is $6.022\times {{10}^{23}}$.

Complete answer:
We have been given an ionic compound sodium nitrate $NaN{{O}_{3}}$, the molecules of this compound will be termed as formula units and its molecular mass as formula unit mass. Formula units are also the simplest whole number ratio of the formula of any compound, but it is used for ionic compounds basically.
We will take out the number of moles in given 33.9 g of the substance, and as 1 mole of the substance has $6.022\times {{10}^{23}}$ formula units, so we will take out the formula units in the calculated moles that are in 33.9 g substance.
The formula unit mass of $NaN{{O}_{3}}$ is the mass of [Na + N + 3(O)] = [23 + 14 + 3(16)] = 85 u.
As we know, number of moles = $\dfrac{given\,mass}{molar\,mass}$ , for sodium nitrate we have, 33.9 g that has moles as:
Number of moles of sodium nitrate = $\dfrac{33.9}{85}$= 0.39 moles
As we know, 1 mole of $NaN{{O}_{3}}$ has $6.022\times {{10}^{23}}$ particles, so in 0.39 moles we have,
$0.39\,moles\times \dfrac{6.022\times {{10}^{23}}F.U.}{1\,mole}=2.35\times {{10}^{23}}FU$
So, formula units of sodium nitrate in 33.9 g of substance are $2.35\times {{10}^{23}}FU$.

Note:
The molar mass of any substance is expressed in gram per mole as it is the mass in 1 mole of that substance. It can also be referred to as ‘u’ as formula unit mass, which is the mass in $6.022\times {{10}^{23}}$ formula units or 1 mole of that molecule. The number of particles is the same in 1 mole of any compound that was discovered in Avogadro’s law.