Chemistry: The Molecular Science
Chemistry: The Molecular Science
5th Edition
ISBN: 9781285199047
Author: John W. Moore, Conrad L. Stanitski
Publisher: Cengage Learning
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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 =mM×100

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

Expert Solution & Answer
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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 Ca16Si4AlFeMgO30.

Explanation of Solution

Given that, typical cement contains 65%CaO,20%SiO2,5%Al2O3,6%Fe2O3,and4%MgO.  It implies that in 100g of cement 65gCaO,20gSiO2,5gAl2O3,6gFe2O3,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.

  65gCaO100gcement×1molCaO56.078gCaO×1molCa1molCaO×40.078gCa1molCa×100=46%Ca.

Similarly, for others the same calculation can be done.

 Molecular mass (g/mol)
SiO260.0854
Si28.0855
Fe2O3159.6882
Fe55.845
MgO40.305
Mg24.305
Al2O3101.96
Al26.98
O15.9994

Calculation of mass percentage:

  20gSiO2100gcement×1molSiO260.0854gSiO2×1molSi1molSiO2×28.0855gSi1molSi×100=9%Si.

  5gAl2O3100gcement×1molAl2O3101.96gAl2O3×2molAl1molAl2O3×26.98gAl1molAl×100=3%Al.

  6gFe2O3100gcement×1molFe2O3159.6882gFe2O3×2molFe1molFe2O3×55.845gFe1molFe×100=4%Fe.

  4gMgO100gcement×1molMgO40.305gMgO×1molMg1molMgO×24.305gMg1molMg×100=2%Mg.

Mass percentage of oxygen:

  100%46%Ca9%Si3%Al4%Fe2%Mg=36%O.

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

Calculation of moles:

  n(Ca)=46g40.078g/mol=1.1mol.

  n(Si)=9g28.0855g/mol=0.3mol.

  n(Al)=3g26.9815g/mol=0.1mol.

  n(Fe)=4g55.845g/mol=0.07mol.

  n(Mg)=2g24.305g/mol=0.08mol.

  n(O)=36g15.9994g/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: Ca16Si4AlFeMgO30.

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Chapter 9 Solutions

Chemistry: The Molecular Science

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