Q3.16P - Chemistry: Molecular Nature Of Matter And Change

Question

3.16 Calculate each of the following quantities:

(a) Mass in grams of 8.35 mol of copper(I) carbonate

(b) Mass in grams of${\text{4.04×10}}^{\text{20}}$ molecules of dinitrogen pentaoxide

(c) Number of moles and formula units in 78.9 g of sodium perchlorate

(d) Number of sodium ions, perchlorate ions, Cl atoms, and O atoms in the mass of the compound in part (c).

Step-by-Step Solution

Answer

Answer

a). Mass in grams of 8.35 mol of copper(I) carbonate is$1562.37\text{g }C{u}_{2}C{O}_3$ 

b). Mass in grams of $4.04\times {10}^{20}$ molecules of dinitrogen pentaoxide is.$7.25\times {10}^{-2}{\text{g N}}_2{O}_5$

c). The number of moles and formula units in 78.9g of sodium perchlorate is$3.88\times {10}^{23}FU{\text{ NaClO}}_4$ .

d). The number of sodium ions, perchlorate ions, Cl atoms, and O atoms in the mass of the compound in part (c) is ,$3.88\times {10}^{23}{\text{ Na}}^{+}\text{ ions}$ $3.88\times {10}^{23}{\text{ ClO}}_4^{-}\text{ ions}$,$1.55\times {10}^{24}\text{ O atoms}$.

 

Step 1: Introduction to the Concept

The mass of a substance made up of an equal number of fundamental units is defined as a mole.

Step 2: Solution Explanation

a)

Multiply the molar mass of copper (I) carbonate,,$C{u}_{2}C{O}_3$ by the stated number of moles.

The molar mass of $C{u}_{2}C{O}_3$ is $187.11\frac{g}{mol}$.

$\begin{array}{c}Mass\text{ of }C{u}_{2}C{O}_3=8.35\text{ mol }C{u}_{2}C{O}_3\left(\frac{187.11{\text{ g Cu}}_{2}C{O}_3}{{\text{mol Cu}}_{2}C{O}_3}\right)\\ =\boxed{1562.37\text{ g }C{u}_{2}C{O}_3}\end{array}$

Step 3: Solution Explanation

b)

 Multiply the number of molecules of dinitrogen pentaoxide,$N_2{O}_5$ , by the reciprocal of the number of molecules.

The Avogadro's number and the molar mass of$N_2{O}_5$ , respectively, are $108.02\frac{g}{mol}$

$\begin{array}{c}Mass{\text{ of N}}_2{O}_5=4.04\times {10}^{20}{\text{ molecules N}}_2{O}_5\times \left(\frac{mol{\text{ N}}_2{O}_5}{6.022\times {10}^{23}{\text{ molecules N}}_2{O}_5}\right)\times \left(\frac{108.02{\text{ g N}}_2{O}_5}{mol{\text{ N}}_2{O}_5}\right)\\ =\boxed{7.25\times {10}^{-2}{\text{ g N}}_2{O}_5}\end{array}$

.

Step 4: Solution Explanation

c)

We multiply the given number of moles in 78.9 g of sodium perchlorate,$NaCl{O}_4$, by the reciprocal of its molar mass$122.44\frac{g}{mol}$,
$\begin{array}{c}Molesof{\text{ NaClO}}_4=78.9{\text{ g NaClO}}_4\times \left(\frac{mol{\text{ NaClO}}_4}{122.44{\text{ g NaClO}}_4}\right)\\ =0.644Molof{\text{ NaClO}}_4\end{array}$ .

We multiply the calculated number of moles of $NaCl{O}_4$by the Avogadro's number to get the formula units (FU).

$\begin{array}{c}Formula\text{ units }of{\text{ NaClO}}_4=0.644{\text{ mol NaClO}}_4\times \left(\frac{6.022\times {10}^{23}FU{\text{ NaClO}}_4}{mol{\text{ NaClO}}_4}\right)\\ =\boxed{3.88\times {10}^{23}FU{\text{ NaClO}}_4}\end{array}$

Step 5: Solution Explanation

Because one mole of${\text{Na}}^{\text{+}}$ ions, one mole of$Cl{O}_4^{-}$ ions, and one mole of Cl atoms make up one mole of $NaCl{O}_4$, they have the same number of moles$NaCl{O}_4$ . As a result, the number of$NaCl{O}_4$ formula units is equal to the number of ${\text{Na}}^{\text{+}}$ions, $Cl{O}_4^{-}$ions, and Cl atoms.

$\begin{array}{c}No.{\text{ of Na}}^{+}\text{ ions}=3.88\times {10}^{23}{\text{ Na}}^{+}\text{ ions}\\ No.{\text{ of ClO}}_4^{-}\text{ ions}=3.88\times {10}^{23}{\text{ ClO}}_4^{-}\text{ ions}\\ No.\text{ of Cl atoms}=3.88\times {10}^{23}\text{ Cl atoms}\end{array}$

One mole of$NaCl{O}_4$ contains four moles of O atoms. The following is the number of moles of O atoms in the mass of the compound in section(c).

$\begin{array}{c}MolesofO\text{ atoms}=0.644{\text{ mol NaClO}}_4\times \left(\frac{4mol\text{ O }atoms}{mol{\text{ NaClO}}_4}\right)\\ =2.576mol\text{ O atoms}\end{array}$

The number of moles of O atoms is multiplied by Avogadro's number.

$\begin{array}{c}No.ofO\text{ atoms}=2.576\text{ mol }O\text{ atoms}\times \left(\frac{6.022\times {10}^{23}\text{ O }atoms}{mol\text{ O atoms}}\right)\\ =\boxed{1.55\times {10}^{24}\text{ O atoms}}\end{array}$