Introductory Chemistry: An Active Learning Approach
Introductory Chemistry: An Active Learning Approach
6th Edition
ISBN: 9781305079250
Author: Mark S. Cracolice, Ed Peters
Publisher: Cengage Learning
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Chapter 7, Problem 31E
Interpretation Introduction

(a)

Interpretation:

The number of formula units in 29.6g of lithium nitrate is to be calculated.

Concept introduction:

A mole is a basic unit that is used in the International system of units (SI). It is abbreviated as mol. It can be defined as 6.022×1023 units of some quantity which can be atoms, molecules or ions. The quantity 6.022×1023 is called the Avogadro’s number.

Expert Solution
Check Mark

Answer to Problem 31E

The number of formula units in 29.6g of lithium nitrate is 2.58×1023formulaunits.

Explanation of Solution

The molecular formula for lithium nitrate is LiNO3.

The molar mass of nitrogen is 14.01gmol1.

The molar mass of oxygen is 16.00gmol1.

The molar mass of lithium is 6.94gmol1.

The molar mass of LiNO3 is calculated below.

Totalmolarmass=6.94+14.01+(3×16.00)=68.95gmol1

Therefore, the molar mass of LiNO3 is 68.95gmol1.

Thus, one mole of LiNO3 is 68.95g.

The number of moles in 29.6g of LiNO3 is given below.

MolesofLiNO3=1moleofLiNO368.95gofLiNO3×weightofLiNO3ing=1moleofLiNO368.95gofLiNO3×29.6gofLiNO3=0.429molesofLiNO3

The number of formula units in LiNO3 is calculated as,

Formula unitsofLiNO3=(6.022×1023formula unitsofLiNO3×molesofLiNO31moleofLiNO3)

Substitute the number of moles of LiNO3 in above equation,

Formula unitsofLiNO3=(6.022×1023formula unitsofLiNO3×0.429molesofLiNO31moleofLiNO3)=2.58×1023formula unitsofLiNO3

Conclusion

The number of formula units in 29.6g of lithium nitrate is 2.58×1023formulaunits.

Interpretation Introduction

(b)

Interpretation:

The number of formula units in 0.151g of lithium sulfide is to be calculated.

Concept introduction:

A mole is a basic unit that is used in the International system of units (SI). It is abbreviated as mol It can be defined as 6.022×1023 units of some quantity which can be atoms, molecules or ions. The quantity 6.022×1023 is called the Avogadro’s number.

Expert Solution
Check Mark

Answer to Problem 31E

The number of formula units in 0.151g of lithium sulfide is 1.98×1021formulaunits.

Explanation of Solution

The molecular formula for lithium sulfide is Li2S.

The molar mass of lithium is 6.94gmol1.

The molar mass of sulfur is 32.06gmol1

The molar mass of Li2S is calculated below.

Totalmolarmass=(2×6.94)+32.06=45.94gmol1

Therefore, the molar mass of Li2S is 45.94gmol1.

Thus one mole of Li2S is 45.94g.

The number of moles in 0.151g of Li2S is given below.

MolesofLi2S=1moleofLi2S45.94gofLi2S×weightofLi2Sing=1moleofLi2S45.94gofLi2S×0.151gofLi2S=0.00328molesofLi2S

The number of formula units in Li2S is calculated as,

Formula unitsofLi2S=(6.022×1023formula unitsofLi2S×molesofLi2S1moleofLi2S)

Substitute the number of moles of Li2S in above equation,

Formula unitsofLi2S=(6.022×1023formula unitsofLi2S×0.00328molesofLiNO31moleofLi2S)=1.98×1021formula unitsofLi2S

Conclusion

The number of formula units in 0.151g of lithium sulfide is 1.98×1021formulaunits.

Interpretation Introduction

(c)

Interpretation:

The number of formula units in 457g of iron (III) sulfate is to be calculated.

Concept introduction:

A mole is a basic unit that is used in the International system of units (SI). It is abbreviated as mol. It can be defined as 6.022×1023 units of some quantity which can be atoms, molecules or ions. The quantity 6.022×1023 is called the Avogadro’s number.

Expert Solution
Check Mark

Answer to Problem 31E

The number of formula units in 457g of iron (III) sulphate is 6.88×1023formulaunits.

Explanation of Solution

The molecular formula for iron (III) sulfate is Fe2(SO4)3.

The molar mass of iron is 55.85gmol1.

The molar mass of sulfur is 32.06gmol1

The molar mass of oxygen is 16.00gmol1.

The molar mass of Fe2(SO4)3 is calculated below.

Totalmolarmass=(2×55.85)+(3×32.06)+(12×16)=399.88gmol1

Therefore, the molar mass of Fe2(SO4)3 is 399.88gmol1.

Thus one mole of Fe2(SO4)3 is 399.88gmol1.

The number of moles in 457g of Fe2(SO4)3 is given below.

MolesofFe2(SO4)3=(1moleofFe2(SO4)3×weightofFe2(SO4)3ing399.88gofFe2(SO4)3)=(1moleofFe2(SO4)3×457gofFe2(SO4)3399.88gofFe2(SO4)3)=1.1428molesofFe2(SO4)3

The number of formula units in Fe2(SO4)3 is calculated as,

Formula unitsofFe2(SO4)3=(6.022×1023formula unitsofFe2(SO4)3×molesofFe2(SO4)31moleofFe2(SO4)3)

Substitute the number of moles of Fe2(SO4)3 in above equation,

Formula unitsofFe2(SO4)3=(6.022×1023formula unitsofFe2(SO4)3×1.1428molesofFe2(SO4)31moleofFe2(SO4)3)=6.88×1023formula unitsofFe2(SO4)3

Conclusion

The number of formula units in 457g of iron (III) sulphate is 6.88×1023formulaunits.

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Chapter 7 Solutions

Introductory Chemistry: An Active Learning Approach

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