Chapter 9, Problem 88QRT

Interpretation Introduction

Interpretation:

The mass percent of each type of atom present has to be determined.  The empirical formula of the material has to be determined.

Concept Introduction:

Mass percentage:

It represents the concentration of an element in a compound or a mixture.  Mathematically, it can be represented as follows,

  Mass percentage $=\frac{m}{M}\times 100$

Where, $m=$ Mass of element in one mole of compound and $M=$ Mass of one mole of compound

Answer to Problem 88QRT

The mass percent of each type of atom present are $46\%Ca$, $9\%Si$, $3\%Al$, $4\%Fe$, $2\%Mg$ and $36\%O.$  The empirical formula of the material is $C{a}_{16}S{i}_{4}AlFeMg{O}_{30}.$

Explanation of Solution

Given that, typical cement contains $65\%CaO,20\%Si{O}_2,5\%A{l}_2{O}_3,6\%F{e}_2{O}_3,and4\%MgO.$  It implies that in $100g$ of cement $65gCaO,20gSi{O}_2,5gA{l}_2{O}_3,6gF{e}_2{O}_3,and4gMgO$ are present.

From the empirical formula of $CaO$, it can be concluded that one mole of $Ca$ and one mole of oxygen are present.

The mass percent of $Ca$ in $CaO$ can be calculated as follows,

Molecular mass of $CaO$ and $Ca$ are $56.078g/mol$ and $40.078g/mol$ respectively.

  $\frac{65gCaO}{100gcement}\times \frac{1molCaO}{56.078gCaO}\times \frac{1molCa}{1molCaO}\times \frac{40.078gCa}{1molCa}\times 100=46\%Ca.$

Similarly, for others the same calculation can be done.

	Molecular mass $\left(g/mol\right)$
$Si{O}_2$	$60.0854$
$Si$	$28.0855$
$F{e}_2{O}_3$	$159.6882$
$Fe$	$55.845$
$MgO$	$40.305$
$Mg$	$24.305$
$A{l}_2{O}_3$	$101.96$
$Al$	$26.98$
$O$	$15.9994$

Calculation of mass percentage:

  $\frac{20gSi{O}_2}{100gcement}\times \frac{1molSi{O}_2}{60.0854gSi{O}_2}\times \frac{1molSi}{1molSi{O}_2}\times \frac{28.0855gSi}{1molSi}\times 100=9\%Si.$

  $\frac{5gA{l}_2{O}_3}{100gcement}\times \frac{1molA{l}_2{O}_3}{101.96gA{l}_2{O}_3}\times \frac{2molAl}{1molA{l}_2{O}_3}\times \frac{26.98gAl}{1molAl}\times 100=3\%Al.$

  $\frac{6gF{e}_2{O}_3}{100gcement}\times \frac{1molF{e}_2{O}_3}{159.6882gF{e}_2{O}_3}\times \frac{2molFe}{1molF{e}_2{O}_3}\times \frac{55.845gFe}{1molFe}\times 100=4\%Fe.$

  $\frac{4gMgO}{100gcement}\times \frac{1molMgO}{40.305gMgO}\times \frac{1molMg}{1molMgO}\times \frac{24.305gMg}{1molMg}\times 100=2\%Mg.$

Mass percentage of oxygen:

  $100\%-46\%Ca-9\%Si-3\%Al-4\%Fe-2\%Mg=36\%O.$

A $100g$ sample will have $46gCa$, $9gSi$, $3gAl$, $4gFe$, $2gMg$ and $36gO.$

Calculation of moles:

  $n\left(Ca\right)=\frac{46g}{40.078g/mol}=1.1mol.$

  $n\left(Si\right)=\frac{9g}{28.0855g/mol}=0.3mol.$

  $n\left(Al\right)=\frac{3g}{26.9815g/mol}=0.1mol.$

  $n\left(Fe\right)=\frac{4g}{55.845g/mol}=0.07mol.$

  $n\left(Mg\right)=\frac{2g}{24.305g/mol}=0.08mol.$

  $n\left(O\right)=\frac{36g}{15.9994g/mol}=2.3mol.$

Mole ratio:

  $1.1molCa:0.3molSi:0.1molAl:0.07molFe:0.08molMg:2.3molO$

Divide by the lowest value , here $0.07mol$:

  $16Ca:4Si:1Al:1Fe:1Mg:30O$

The empirical formula: $C{a}_{16}S{i}_{4}AlFeMg{O}_{30}.$