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 32E
Interpretation Introduction

(a)

Interpretation:

The number of formula units in 85.5g of beryllium 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
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Answer to Problem 32E

The number of formula units in 85.5g of beryllium nitrate is 3.87×1023formula units.

Explanation of Solution

The molecular formula for beryllium nitrate is Be(NO3)2.

The molar mass of nitrogen is 14.01gmol1.

The molar mass of oxygen is 16.00gmol1.

The molar mass of beryllium is 9.012gmol1.

The molar mass of Be(NO3)2 is calculated below.

Totalmolarmass=9.012+(2×14.01)+(6×16.00)=133.032gmol1

Therefore, the molar mass of Be(NO3)2 is 133.032gmol1.

Thus one mole of Be(NO3)2 is 133.032g.

The number of moles in 85.5g of Be(NO3)2 is given below.

MolesofBe(NO3)2=1moleofBe(NO3)2133.032gofBe(NO3)2×weightofBe(NO3)2ing=1moleofBe(NO3)2133.032gofBe(NO3)2×85.5gofBe(NO3)2=0.643molesofBe(NO3)2

The number of formula units in Be(NO3)2 is calculated as,

Formula unitsofBe(NO3)2=(6.022×1023formula unitsofBe(NO3)2×molesofBe(NO3)21moleofBe(NO3)2)

Substitute the number of moles of Be(NO3)2 in above equation,

Formula unitsofBe(NO3)2=(6.022×1023formula unitsofBe(NO3)2×0.643molesofBe(NO3)21moleofBe(NO3)2)=3.87×1023formula unitsofBe(NO3)2

Conclusion

The number of formula units in 85.5g of beryllium nitrate is 3.87×1023formula units3.87×1023.

Interpretation Introduction

(b)

Interpretation:

The number of formula units in 9.42g manganese 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 32E

The number of formula units in 9.42g manganese is 1.03×1023formula units.

Explanation of Solution

The molar mass of manganese atom is 54.94gmol1.

Thus, one mole of manganese is 54.94g.

The number of moles in 9.42g of manganese is given below.

Molesofmanganese=1moleofmanganese54.94gofmanganese×weightofmanganeseing=1moleofmanganese54.94gofmanganese×9.42gofmanganese=0.171molesofmanganese

The number of formula units in manganese is calculated below.

Formula unitsofmanganese=(6.022×1023formula unitsofmanganese×molesofmanganese1moleofmanganese)

Substitute the number of moles of manganese in the above equation.

Formula unitsofmanganese=(6.022×1023formula unitsofmanganese×0.171molesofmanganese1moleofmanganese)=1.03×1023formula unitsofmanganese

Conclusion

The number of formula units in 9.42g manganese is 1.03×1023formula units.

Interpretation Introduction

(c)

Interpretation:

The number of molecules in 0.0948gC3H7OH 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 32E

The number of molecules in 0.0948gC3H7OH is 9.49×1020molecules.

Explanation of Solution

The molecular formula is C3H7OH.

The molar mass of carbon is 12.01gmol1.

The molar mass of oxygen is 16.00gmol1.

The molar mass of hydrogen is 1.008gmol1.

The molar mass of C3H7OH is calculated below.

Totalmolarmass=3×12.01+(8×1.008)+16.00=60.094gmol1

Therefore, the molar mass of C3H7OH is 60.094.

Thus one mole of C3H7OH is 60.0094.

The number of moles in 0.0948gC3H7OH is given below.

MolesofC3H7OH=1moleofC3H7OH60.094gofC3H7OH×weightofC3H7OHing=1moleofC3H7OH360.094ggofC3H7OH×0.0948gofC3H7OH=0.001577molesofC3H7OH

The number of molecules in C3H7OH is calculated below.

MoleculesofC3H7OH=(6.022×1023moleculesofC3H7OH×molesofC3H7OH1moleofC3H7OH)

Substitute the number of moles of C3H7OH in above equation.

MoleculesofC3H7OH=(6.022×1023moleculesofC3H7OH×0.001577molesofC3H7OH1moleofC3H7OH)=9.49×1020moleculesofC3H7OH

Conclusion

The number of molecules in 0.0948gC3H7OH is 9.49×1020molecules.

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Chapter 7 Solutions

Introductory Chemistry: An Active Learning Approach

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