Chapter 3, Problem 3.28QP

Interpretation Introduction

Interpretation:

The number of atoms has to be calculated given $7.14\times {10}^{3}g$ of dimethyl sulfoxide

Concept Introduction:

• Mass percentage is one way to represent the concentration of an element in a compound or a component in a mixture. Mass percent of elements of a compound is the ratio of mass of element to the mass of whole compound and multiplied with hundred.
• The molar mass of an element or compound is the mass in grams of 1 mole of that substance, and it is expressed in the unit of grams per mol (g/mol)
• The fundamental units may be molecules, formula units or atoms.  In chemistry mole is represented as mol.  It is a counting unit.  In a mole Avogadro’s number represent the number of fundamental entities.
• A mole is the number of atoms (or) molecules in $(1\text{g})$ of hydrogen or $12\text{g}$ of carbon. Avogadro’s number $6.023\times {10}^{23}$ is called one mole.

Explanation of Solution

Let’s calculate the number of carbon atoms in $7.14\times {10}^{3}g$ of dimethyl sulfoxide. First we convert grams of $\text{DMSO}$ to the number of molecules of $\text{DMSO}$.

The molecular formula for DMSO is ${{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}$

$\begin{array}{l}\text{Molar}\text{mass}\text{of}{{\text{(CH}}_{\text{3}}}_{\text{)}}\text{SO}=78.13\text{g/mol}\\ \text{No}\text{.}\text{of}\text{C}\text{atoms}=2(12.01g)=24.02g\\ \text{No}\text{.}\text{of}\text{O}\text{atoms}=1(16.00g)=16.00g\\ \text{No}\text{.}\text{of}\text{S}\text{atoms}=1(\mathrm{32.07.}g)=32.07g\\ \text{No}\text{.}\text{of}\text{H}\text{atoms}=6(1.008g)=6.048g\\ -----------------------\\ \text{Total}\text{mass}\text{value}\text{= 78}\text{.14}\text{g}\end{array}$

The conversion factor needed is,

$\frac{1\text{mol}\text{DMSO}}{78.14g\text{DMSO}}$

Avogadro’s number is $1\text{mol=}\text{6}\text{.022}\times {\text{10}}^{23}\text{particles}$

From this equality, we write the conversion factor.

$\frac{\text{6}\text{.022}\times {\text{10}}^{23}\text{DMSO}\text{molecule}}{1\text{mole}\text{DMSO}}$

Let’s complete the two conversions in single step

$\begin{array}{l}\text{Grams}\text{of}\text{DMSO}\to \text{Moles}\text{of}\text{DMSO}\to \text{molecules}\text{of}\text{DMSO}\to \text{Atoms}\text{of}\text{C}\\ \text{=}7.14\times {10}^3\cancel{\text{g}}{{\text{(CH}}_{\text{3}}}_{\text{)}}\text{SO}\times \frac{1\cancel{\text{mol}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}{78.14\cancel{g}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}\times \frac{\text{6}\text{.022}\times {\text{10}}^{23}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}\text{molecules}}{1\cancel{\text{mol}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}\times \frac{2\text{C atoms}}{\text{1}{\text{molecules (CH}}_{\text{3}}{\text{)}}_{\text{2}}\text{SO}}\\ =\underset{\_}{1.10\times {10}^{26}\text{carbon}\text{atoms}}\end{array}$

Therefore $7.14\times {10}^{3}g$ of $\text{DMSO}$ contains $1.10\times {10}^{26}$ atoms of carbons

Calculate the numbers of hydrogen are present in dimethyl sulfoxide,

$\begin{array}{l}\text{Grams}\text{of}\text{DMSO}\to \text{Moles}\text{of}\text{DMSO}\to \text{molecules}\text{of}\text{DMSO}\to \text{Atoms}\text{of}\text{H}\\ \text{=}7.14\times {10}^3\cancel{\text{g}}{{\text{(CH}}_{\text{3}}}_{\text{)}}\text{SO}\times \frac{1\cancel{\text{mol}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}{78.14\cancel{g}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}\times \frac{\text{6}\text{.022}\times {\text{10}}^{23}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}\cancel{\text{molecules}}}{1\cancel{\text{mol}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}\times \frac{\text{6}\text{H atoms}}{\text{1}\cancel{\text{molecules}}{\text{ (CH}}_{\text{3}}{\text{)}}_{\text{2}}\text{SO}}\\ =\underset{\_}{3.30\times {10}^{26}\text{hydrogem}\text{atoms}}\end{array}$

Thus, we have found that $7.14\times {10}^{3}g$ of $\text{DMSO}$ contains $3.30\times {10}^{26}$ atoms of hydrogen.

Calculate the number of oxygen is present in dimethyl sulfoxide,

$\begin{array}{l}\text{Grams}\text{of}\text{DMSO}\to \text{Moles}\text{of}\text{DMSO}\to \text{molecules}\text{of}\text{DMSO}\to \text{Atoms}\text{of}\text{O}\\ \text{=}7.14\times {10}^3\cancel{\text{g}}{{\text{(CH}}_{\text{3}}}_{\text{)}}\text{SO}\times \frac{1\cancel{\text{mol}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}{78.14\cancel{g}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}\times \frac{\text{6}\text{.022}\times {\text{10}}^{23}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}\cancel{\text{molecules}}}{1\cancel{\text{mol}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}\times \frac{\text{1}\text{O atoms}}{\text{1}\cancel{\text{molecules}}{\text{ (CH}}_{\text{3}}{\text{)}}_{\text{2}}\text{SO}}\\ =\underset{\_}{5.50\times {10}^{25}\text{oxygen}\text{atom}}\end{array}$

Calculate the number of sulfur is present in dimethyl sulfoxide,

$\begin{array}{l}\text{Grams}\text{of}\text{DMSO}\to \text{Moles}\text{of}\text{DMSO}\to \text{molecules}\text{of}\text{DMSO}\to \text{Atoms}\text{of}\text{S}\\ \text{=}7.14\times {10}^3\cancel{\text{g}}{{\text{(CH}}_{\text{3}}}_{\text{)}}\text{SO}\times \frac{1\cancel{\text{mol}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}{78.14\cancel{g}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}\times \frac{\text{6}\text{.022}\times {\text{10}}^{23}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}\cancel{\text{molecules}}}{1\cancel{\text{mol}}{{\text{(CH}}_{\text{3}}\text{)}}_{\text{2}}\text{SO}}\times \frac{\text{1}\text{S atoms}}{\text{1}\cancel{\text{molecules}}{\text{ (CH}}_{\text{3}}{\text{)}}_{\text{2}}\text{SO}}\\ =\underset{\_}{5.50\times {10}^{25}\text{sulfur}\text{atom}}\end{array}$

Therefore the number of $\text{C, S,}\text{H,}\text{O}$ are calculated in $7.14\times {10}^{3}g$ of $\text{DMSO}$ molecule.