Q3.`17P - Chemistry: Molecular Nature Of Matter And Change

Question

3.17 Calculate each of the following quantities:

(a) Mass in grams of 8.42 mol of chromium(III) sulfate decahydrate

(b) Mass in grams of molecules of dichlorineheptaoxide

(c) Number of moles and formula units in 6.2 g of lithium sulfate

(d) Number of lithium ions, sulfate ions, S atoms, and O atoms in the mass of compound in part (c)

Step-by-Step Solution

Answer

a) Mass in grams of 8.42 mol of chromium(III) sulfate decahydrate is $4819.69\text{ g }C{r}_2{\left(S{O}_4\right)}_3\cdot 10H_{2}O$ 

b) Mass in grams of  $1.83\times {10}^{24}$molecules of dichlorineheptaoxide is$4819.69\text{ g }C{r}_2{\left(S{O}_4\right)}_3\cdot 10H_{2}O$ .

c) The number of moles and formula units in 6.2 g of lithium sulfate is$3.40\times {10}^{22}FU{\text{ Li}}_2{\text{SO}}_4$ .

d) The number of lithium ions, sulfate ions, S atoms, and O atoms in the mass of the compound in part (c) is $3.40\times {10}^{22}{\text{ SO}}_4^{2-}\text{ ions}$,$3.40\times {10}^{22}\text{ S atoms}$, $1.36\times {10}^{23}\text{ O atoms}$

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 Chromium (III) sulfate decahydrate by the stated number of moles.

The molar mass of$C{r}_2{\left(S{O}_4\right)}_3\times 10H_{2}O$  is $572.41\frac{g}{mol}$.

$\begin{array}{c}Mass\text{ of }C{r}_2{\left(S{O}_4\right)}_3\cdot 10H_{2}O=8.42\text{ mol }C{r}_2{\left(S{O}_4\right)}_3\cdot 10H_{2}O\times \left(\frac{572.41\text{ g }C{r}_2{\left(S{O}_4\right)}_3\cdot 10H_{2}O}{\text{mol }C{r}_2{\left(S{O}_4\right)}_3\cdot 10H_{2}O}\right)\\ =\boxed{4819.69\text{ g }C{r}_2{\left(S{O}_4\right)}_3\cdot 10H_{2}O}\end{array}$

Step 3: Solution Explanation

b)

 Multiply the number of molecules of dichlorineheptaoxide, $C{l}_2{O}_7$, by the reciprocal of the number of molecules.

The Avogadro's number and the molar mass of ${\mathrm{Cl}}_2{O}_7$, respectively, are .$182.90\frac{g}{mol}$

$\begin{array}{c}Mass\text{ of }C{l}_2{O}_7=1.83\times {10}^{24}\text{ molecules }C{l}_2{O}_7\times \left(\frac{molC{l}_2{O}_7}{6.022\times {10}^{23}\text{ molecules }C{l}_2{O}_7}\right)\times \left(\frac{182.90\text{ g }C{l}_2{O}_7}{molC{l}_2{O}_7}\right)\\ =\boxed{555.81\text{ g }C{l}_2{O}_7}\end{array}$

Step 4: Solution Explanation

c)

 We multiply the given number of moles in 6.2 g of lithium sulfate, ${\text{Li}}_2{\text{SO}}_4$, by the reciprocal of its molar mass,$109.95\frac{g}{mol}$ .    $\begin{array}{c}Molesof{\text{ Li}}_2{\text{SO}}_4=6.2{\text{ g Li}}_2{\text{SO}}_4\times \left(\frac{mol{\text{ Li}}_2{\text{SO}}_4}{109.95{\text{ g Li}}_2{\text{SO}}_4}\right)\\ =5.64\times {10}^{-2}Molof{\text{ Li}}_2{\text{SO}}_4\end{array}$

We multiply the calculated number of moles of by ${\text{Li}}_2{\text{SO}}_4$ the Avogadro's number to get the formula units (FU).

$\begin{array}{c}Formula\text{ units }of{\text{ Li}}_2{\text{SO}}_4=5.64{\text{ mol Li}}_2{\text{SO}}_4\times \left(\frac{6.022\times {10}^{23}FU{\text{ Li}}_2{\text{SO}}_4}{mol{\text{ NLi}}_2{\text{SO}}_4}\right)\\ =\boxed{3.40\times {10}^{22}FU{\text{ Li}}_2{\text{SO}}_4}\end{array}$

Step 5: Solution Explanation

d)

One mole of ${\mathrm{Li}}_{2}S{O}_4$ contains two moles of ions ${\text{Li}}^{\text{+}}$. In section (c), the number of moles of  ions in the mass of the compound is as follows.

$\begin{array}{c}MolesofL{i}^{+}\text{ atoms}=5.64\times {10}^{-2}\text{ mol }L{i}_{2}S{O}_4\times \left(\frac{2molL{i}^{+}ions}{molL{i}_{2}S{O}_4}\right)\\ =0.113molL{i}^{+}\text{ ions}\end{array}$

The number of moles of ${\text{Li}}^{\text{+}}$ ions is multiplied by Avogadro's number.

$\begin{array}{c}MolesofL{i}^{+}\text{ atoms}=0.113\text{ mol }L{i}^{+}\text{ ions}\times \left(\frac{6.022\times {10}^{23}L{i}^{+}ions}{molL{i}^{+}\text{ ions}}\right)\\ =6.80\times {10}^{22}L{i}^{+}\text{ ions}\end{array}$

Because one mole of $S{O}_4^{2-}$ ions and one mole of S atoms make up one mole of ${\text{Li}}_2{\text{SO}}_4$, their mole numbers are numerically equal. As a result, the amount of $S{O}_4^{2-}$ ions and S atoms in  ${\text{Li}}_2{\text{SO}}_4$ equals the number of formula units.

$\begin{array}{l}No.{\text{ of SO}}_4^{2-}\text{ ions}=3.40\times {10}^{22}{\text{ SO}}_4^{2-}\text{ ions}\\ No.\text{ of S atoms}=3.40\times {10}^{22}\text{ S atoms}\end{array}$

One mole of ${\text{Li}}_2{\text{SO}}_4$ 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).

$\begin{array}{c}MolesofO\text{ atoms}=5.64\times {10}^{-2}{\text{ mol Li}}_2{\text{SO}}_4\times \left(\frac{4mol\text{ O }atoms}{mol{\text{ Li}}_2{\text{SO}}_4}\right)\\ =0.226mol\text{ O atoms}\end{array}$

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

$\begin{array}{c}No.ofO\text{ atoms}=0.226\text{ mol }O\text{ atoms}\times \left(\frac{6.022\times {10}^{23}\text{ O }atoms}{mol\text{ O atoms}}\right)\\ =\boxed{1.36\times {10}^{23}\text{ O atoms}}\end{array}$