(a) Interpretation: The number of formula units in 29 .6 g 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 × 10 23 units of some quantity which can be atoms, molecules or ions. The quantity 6.022 × 10 23 is called the Avogadro’s number .

Question

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Answer 1

Question

Chapter 7, Problem 31E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $29.6 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium nitrate is $LiNO3$ .

The molar mass of nitrogen is $14.01 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of $LiNO3$ is calculated below.

$Total molar mass =6.94+14.01+(3×16.00)=68.95 g mol−1$

Therefore, the molar mass of $LiNO3$ is $68.95 g mol−1$ .

Thus, one mole of $LiNO3$ is $68.95 g$ .

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

$Moles of LiNO3=1 mole of LiNO368.95 g of LiNO3×weight of LiNO3 in g=1 mole of LiNO368.95 g of LiNO3×29.6 g of LiNO3 =0.429 moles of LiNO3$

The number of formula units in $LiNO3$ is calculated as,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×moles of LiNO31 mole of LiNO3)$

Substitute the number of moles of $LiNO3$ in above equation,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×0.429 moles of LiNO31 mole of LiNO3)=2.58×1023 formula units of LiNO3$

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $0.151 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium sulfide is $Li2S$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of $Li2S$ is calculated below.

$Total molar mass =(2×6.94)+32.06=45.94 g mol−1$

Therefore, the molar mass of $Li2S$ is $45.94 g mol−1$ .

Thus one mole of $Li2S$ is $45.94 g$ .

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

$Moles of Li2S=1 mole of Li2S45.94 g of Li2S×weight of Li2S in g=1 mole of Li2S45.94 g of Li2S×0.151 g of Li2S =0.00328 moles of Li2S$

The number of formula units in $Li2S$ is calculated as,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×moles of Li2S1 mole of Li2S)$

Substitute the number of moles of $Li2S$ in above equation,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×0.00328 moles of LiNO31 mole of Li2S)=1.98×1021 formula units of Li2S$

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $457 g$ 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

Answer to Problem 31E

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

Explanation of Solution

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

The molar mass of iron is $55.85 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of oxygen is $16.00 g mol−1$ .

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

$Total molar mass =(2×55.85) +(3×32.06)+(12×16)=399.88 g mol−1$

Therefore, the molar mass of $Fe2(SO4)3$ is $399.88 g mol−1$ .

Thus one mole of $Fe2(SO4)3$ is $399.88 g mol−1$ .

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

$Moles of Fe2(SO4)3=(1 mole of Fe2(SO4)3×weight of Fe2(SO4)3 in g399.88g of Fe2(SO4)3)=(1 mole of Fe2(SO4)3×457 g of Fe2(SO4)3399.88 g of Fe2(SO4)3)=1.1428 moles of Fe2(SO4)3$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×moles of Fe2(SO4)31 mole of Fe2(SO4)3)$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×1.1428 moles of Fe2(SO4)31 mole of Fe2(SO4)3)=6.88×1023 formula units of Fe2(SO4)3$

Conclusion

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

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Answer 2

Question

Chapter 7, Problem 31E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $29.6 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium nitrate is $LiNO3$ .

The molar mass of nitrogen is $14.01 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of $LiNO3$ is calculated below.

$Total molar mass =6.94+14.01+(3×16.00)=68.95 g mol−1$

Therefore, the molar mass of $LiNO3$ is $68.95 g mol−1$ .

Thus, one mole of $LiNO3$ is $68.95 g$ .

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

$Moles of LiNO3=1 mole of LiNO368.95 g of LiNO3×weight of LiNO3 in g=1 mole of LiNO368.95 g of LiNO3×29.6 g of LiNO3 =0.429 moles of LiNO3$

The number of formula units in $LiNO3$ is calculated as,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×moles of LiNO31 mole of LiNO3)$

Substitute the number of moles of $LiNO3$ in above equation,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×0.429 moles of LiNO31 mole of LiNO3)=2.58×1023 formula units of LiNO3$

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $0.151 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium sulfide is $Li2S$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of $Li2S$ is calculated below.

$Total molar mass =(2×6.94)+32.06=45.94 g mol−1$

Therefore, the molar mass of $Li2S$ is $45.94 g mol−1$ .

Thus one mole of $Li2S$ is $45.94 g$ .

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

$Moles of Li2S=1 mole of Li2S45.94 g of Li2S×weight of Li2S in g=1 mole of Li2S45.94 g of Li2S×0.151 g of Li2S =0.00328 moles of Li2S$

The number of formula units in $Li2S$ is calculated as,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×moles of Li2S1 mole of Li2S)$

Substitute the number of moles of $Li2S$ in above equation,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×0.00328 moles of LiNO31 mole of Li2S)=1.98×1021 formula units of Li2S$

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $457 g$ 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

Answer to Problem 31E

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

Explanation of Solution

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

The molar mass of iron is $55.85 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of oxygen is $16.00 g mol−1$ .

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

$Total molar mass =(2×55.85) +(3×32.06)+(12×16)=399.88 g mol−1$

Therefore, the molar mass of $Fe2(SO4)3$ is $399.88 g mol−1$ .

Thus one mole of $Fe2(SO4)3$ is $399.88 g mol−1$ .

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

$Moles of Fe2(SO4)3=(1 mole of Fe2(SO4)3×weight of Fe2(SO4)3 in g399.88g of Fe2(SO4)3)=(1 mole of Fe2(SO4)3×457 g of Fe2(SO4)3399.88 g of Fe2(SO4)3)=1.1428 moles of Fe2(SO4)3$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×moles of Fe2(SO4)31 mole of Fe2(SO4)3)$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×1.1428 moles of Fe2(SO4)31 mole of Fe2(SO4)3)=6.88×1023 formula units of Fe2(SO4)3$

Conclusion

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

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Answer 3

Question

Chapter 7, Problem 31E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $29.6 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium nitrate is $LiNO3$ .

The molar mass of nitrogen is $14.01 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of $LiNO3$ is calculated below.

$Total molar mass =6.94+14.01+(3×16.00)=68.95 g mol−1$

Therefore, the molar mass of $LiNO3$ is $68.95 g mol−1$ .

Thus, one mole of $LiNO3$ is $68.95 g$ .

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

$Moles of LiNO3=1 mole of LiNO368.95 g of LiNO3×weight of LiNO3 in g=1 mole of LiNO368.95 g of LiNO3×29.6 g of LiNO3 =0.429 moles of LiNO3$

The number of formula units in $LiNO3$ is calculated as,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×moles of LiNO31 mole of LiNO3)$

Substitute the number of moles of $LiNO3$ in above equation,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×0.429 moles of LiNO31 mole of LiNO3)=2.58×1023 formula units of LiNO3$

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $0.151 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium sulfide is $Li2S$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of $Li2S$ is calculated below.

$Total molar mass =(2×6.94)+32.06=45.94 g mol−1$

Therefore, the molar mass of $Li2S$ is $45.94 g mol−1$ .

Thus one mole of $Li2S$ is $45.94 g$ .

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

$Moles of Li2S=1 mole of Li2S45.94 g of Li2S×weight of Li2S in g=1 mole of Li2S45.94 g of Li2S×0.151 g of Li2S =0.00328 moles of Li2S$

The number of formula units in $Li2S$ is calculated as,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×moles of Li2S1 mole of Li2S)$

Substitute the number of moles of $Li2S$ in above equation,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×0.00328 moles of LiNO31 mole of Li2S)=1.98×1021 formula units of Li2S$

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $457 g$ 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

Answer to Problem 31E

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

Explanation of Solution

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

The molar mass of iron is $55.85 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of oxygen is $16.00 g mol−1$ .

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

$Total molar mass =(2×55.85) +(3×32.06)+(12×16)=399.88 g mol−1$

Therefore, the molar mass of $Fe2(SO4)3$ is $399.88 g mol−1$ .

Thus one mole of $Fe2(SO4)3$ is $399.88 g mol−1$ .

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

$Moles of Fe2(SO4)3=(1 mole of Fe2(SO4)3×weight of Fe2(SO4)3 in g399.88g of Fe2(SO4)3)=(1 mole of Fe2(SO4)3×457 g of Fe2(SO4)3399.88 g of Fe2(SO4)3)=1.1428 moles of Fe2(SO4)3$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×moles of Fe2(SO4)31 mole of Fe2(SO4)3)$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×1.1428 moles of Fe2(SO4)31 mole of Fe2(SO4)3)=6.88×1023 formula units of Fe2(SO4)3$

Conclusion

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

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Answer 4

Question

Chapter 7, Problem 31E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $29.6 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium nitrate is $LiNO3$ .

The molar mass of nitrogen is $14.01 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of $LiNO3$ is calculated below.

$Total molar mass =6.94+14.01+(3×16.00)=68.95 g mol−1$

Therefore, the molar mass of $LiNO3$ is $68.95 g mol−1$ .

Thus, one mole of $LiNO3$ is $68.95 g$ .

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

$Moles of LiNO3=1 mole of LiNO368.95 g of LiNO3×weight of LiNO3 in g=1 mole of LiNO368.95 g of LiNO3×29.6 g of LiNO3 =0.429 moles of LiNO3$

The number of formula units in $LiNO3$ is calculated as,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×moles of LiNO31 mole of LiNO3)$

Substitute the number of moles of $LiNO3$ in above equation,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×0.429 moles of LiNO31 mole of LiNO3)=2.58×1023 formula units of LiNO3$

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $0.151 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium sulfide is $Li2S$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of $Li2S$ is calculated below.

$Total molar mass =(2×6.94)+32.06=45.94 g mol−1$

Therefore, the molar mass of $Li2S$ is $45.94 g mol−1$ .

Thus one mole of $Li2S$ is $45.94 g$ .

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

$Moles of Li2S=1 mole of Li2S45.94 g of Li2S×weight of Li2S in g=1 mole of Li2S45.94 g of Li2S×0.151 g of Li2S =0.00328 moles of Li2S$

The number of formula units in $Li2S$ is calculated as,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×moles of Li2S1 mole of Li2S)$

Substitute the number of moles of $Li2S$ in above equation,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×0.00328 moles of LiNO31 mole of Li2S)=1.98×1021 formula units of Li2S$

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $457 g$ 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

Answer to Problem 31E

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

Explanation of Solution

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

The molar mass of iron is $55.85 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of oxygen is $16.00 g mol−1$ .

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

$Total molar mass =(2×55.85) +(3×32.06)+(12×16)=399.88 g mol−1$

Therefore, the molar mass of $Fe2(SO4)3$ is $399.88 g mol−1$ .

Thus one mole of $Fe2(SO4)3$ is $399.88 g mol−1$ .

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

$Moles of Fe2(SO4)3=(1 mole of Fe2(SO4)3×weight of Fe2(SO4)3 in g399.88g of Fe2(SO4)3)=(1 mole of Fe2(SO4)3×457 g of Fe2(SO4)3399.88 g of Fe2(SO4)3)=1.1428 moles of Fe2(SO4)3$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×moles of Fe2(SO4)31 mole of Fe2(SO4)3)$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×1.1428 moles of Fe2(SO4)31 mole of Fe2(SO4)3)=6.88×1023 formula units of Fe2(SO4)3$

Conclusion

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

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Answer 5

Chapter 7, Problem 31E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $29.6 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium nitrate is $LiNO3$ .

The molar mass of nitrogen is $14.01 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of $LiNO3$ is calculated below.

$Total molar mass =6.94+14.01+(3×16.00)=68.95 g mol−1$

Therefore, the molar mass of $LiNO3$ is $68.95 g mol−1$ .

Thus, one mole of $LiNO3$ is $68.95 g$ .

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

$Moles of LiNO3=1 mole of LiNO368.95 g of LiNO3×weight of LiNO3 in g=1 mole of LiNO368.95 g of LiNO3×29.6 g of LiNO3 =0.429 moles of LiNO3$

The number of formula units in $LiNO3$ is calculated as,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×moles of LiNO31 mole of LiNO3)$

Substitute the number of moles of $LiNO3$ in above equation,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×0.429 moles of LiNO31 mole of LiNO3)=2.58×1023 formula units of LiNO3$

Conclusion

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

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $0.151 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium sulfide is $Li2S$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of $Li2S$ is calculated below.

$Total molar mass =(2×6.94)+32.06=45.94 g mol−1$

Therefore, the molar mass of $Li2S$ is $45.94 g mol−1$ .

Thus one mole of $Li2S$ is $45.94 g$ .

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

$Moles of Li2S=1 mole of Li2S45.94 g of Li2S×weight of Li2S in g=1 mole of Li2S45.94 g of Li2S×0.151 g of Li2S =0.00328 moles of Li2S$

The number of formula units in $Li2S$ is calculated as,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×moles of Li2S1 mole of Li2S)$

Substitute the number of moles of $Li2S$ in above equation,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×0.00328 moles of LiNO31 mole of Li2S)=1.98×1021 formula units of Li2S$

Conclusion

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

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $457 g$ 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

Answer to Problem 31E

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

Explanation of Solution

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

The molar mass of iron is $55.85 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of oxygen is $16.00 g mol−1$ .

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

$Total molar mass =(2×55.85) +(3×32.06)+(12×16)=399.88 g mol−1$

Therefore, the molar mass of $Fe2(SO4)3$ is $399.88 g mol−1$ .

Thus one mole of $Fe2(SO4)3$ is $399.88 g mol−1$ .

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

$Moles of Fe2(SO4)3=(1 mole of Fe2(SO4)3×weight of Fe2(SO4)3 in g399.88g of Fe2(SO4)3)=(1 mole of Fe2(SO4)3×457 g of Fe2(SO4)3399.88 g of Fe2(SO4)3)=1.1428 moles of Fe2(SO4)3$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×moles of Fe2(SO4)31 mole of Fe2(SO4)3)$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×1.1428 moles of Fe2(SO4)31 mole of Fe2(SO4)3)=6.88×1023 formula units of Fe2(SO4)3$

Conclusion

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

Answer 6

Chapter 7, Problem 31E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $29.6 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium nitrate is $LiNO3$ .

The molar mass of nitrogen is $14.01 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of $LiNO3$ is calculated below.

$Total molar mass =6.94+14.01+(3×16.00)=68.95 g mol−1$

Therefore, the molar mass of $LiNO3$ is $68.95 g mol−1$ .

Thus, one mole of $LiNO3$ is $68.95 g$ .

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

$Moles of LiNO3=1 mole of LiNO368.95 g of LiNO3×weight of LiNO3 in g=1 mole of LiNO368.95 g of LiNO3×29.6 g of LiNO3 =0.429 moles of LiNO3$

The number of formula units in $LiNO3$ is calculated as,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×moles of LiNO31 mole of LiNO3)$

Substitute the number of moles of $LiNO3$ in above equation,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×0.429 moles of LiNO31 mole of LiNO3)=2.58×1023 formula units of LiNO3$

Conclusion

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

Answer 7

Chapter 7, Problem 31E

Interpretation Introduction

(a)

Interpretation:

The number of formula units in $29.6 g$ 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.

Answer 8

Expert Solution

Answer to Problem 31E

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

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The molecular formula for lithium nitrate is $LiNO3$ .

The molar mass of nitrogen is $14.01 g mol−1$ .

The molar mass of oxygen is $16.00 g mol−1$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of $LiNO3$ is calculated below.

$Total molar mass =6.94+14.01+(3×16.00)=68.95 g mol−1$

Therefore, the molar mass of $LiNO3$ is $68.95 g mol−1$ .

Thus, one mole of $LiNO3$ is $68.95 g$ .

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

$Moles of LiNO3=1 mole of LiNO368.95 g of LiNO3×weight of LiNO3 in g=1 mole of LiNO368.95 g of LiNO3×29.6 g of LiNO3 =0.429 moles of LiNO3$

The number of formula units in $LiNO3$ is calculated as,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×moles of LiNO31 mole of LiNO3)$

Substitute the number of moles of $LiNO3$ in above equation,

$Formula units of LiNO3=(6.022×1023 formula units of LiNO3×0.429 moles of LiNO31 mole of LiNO3)=2.58×1023 formula units of LiNO3$

Answer 13

Conclusion

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

Answer 14

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $0.151 g$ 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

Answer to Problem 31E

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

Explanation of Solution

The molecular formula for lithium sulfide is $Li2S$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of $Li2S$ is calculated below.

$Total molar mass =(2×6.94)+32.06=45.94 g mol−1$

Therefore, the molar mass of $Li2S$ is $45.94 g mol−1$ .

Thus one mole of $Li2S$ is $45.94 g$ .

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

$Moles of Li2S=1 mole of Li2S45.94 g of Li2S×weight of Li2S in g=1 mole of Li2S45.94 g of Li2S×0.151 g of Li2S =0.00328 moles of Li2S$

The number of formula units in $Li2S$ is calculated as,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×moles of Li2S1 mole of Li2S)$

Substitute the number of moles of $Li2S$ in above equation,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×0.00328 moles of LiNO31 mole of Li2S)=1.98×1021 formula units of Li2S$

Conclusion

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

Answer 15

Interpretation Introduction

(b)

Interpretation:

The number of formula units in $0.151 g$ 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.

Answer 16

Expert Solution

Answer to Problem 31E

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

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The molecular formula for lithium sulfide is $Li2S$ .

The molar mass of lithium is $6.94 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of $Li2S$ is calculated below.

$Total molar mass =(2×6.94)+32.06=45.94 g mol−1$

Therefore, the molar mass of $Li2S$ is $45.94 g mol−1$ .

Thus one mole of $Li2S$ is $45.94 g$ .

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

$Moles of Li2S=1 mole of Li2S45.94 g of Li2S×weight of Li2S in g=1 mole of Li2S45.94 g of Li2S×0.151 g of Li2S =0.00328 moles of Li2S$

The number of formula units in $Li2S$ is calculated as,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×moles of Li2S1 mole of Li2S)$

Substitute the number of moles of $Li2S$ in above equation,

$Formula units of Li2S=(6.022×1023 formula units of Li2S×0.00328 moles of LiNO31 mole of Li2S)=1.98×1021 formula units of Li2S$

Answer 21

Conclusion

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

Answer 22

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $457 g$ 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

Answer to Problem 31E

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

Explanation of Solution

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

The molar mass of iron is $55.85 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of oxygen is $16.00 g mol−1$ .

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

$Total molar mass =(2×55.85) +(3×32.06)+(12×16)=399.88 g mol−1$

Therefore, the molar mass of $Fe2(SO4)3$ is $399.88 g mol−1$ .

Thus one mole of $Fe2(SO4)3$ is $399.88 g mol−1$ .

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

$Moles of Fe2(SO4)3=(1 mole of Fe2(SO4)3×weight of Fe2(SO4)3 in g399.88g of Fe2(SO4)3)=(1 mole of Fe2(SO4)3×457 g of Fe2(SO4)3399.88 g of Fe2(SO4)3)=1.1428 moles of Fe2(SO4)3$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×moles of Fe2(SO4)31 mole of Fe2(SO4)3)$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×1.1428 moles of Fe2(SO4)31 mole of Fe2(SO4)3)=6.88×1023 formula units of Fe2(SO4)3$

Conclusion

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

Answer 23

Interpretation Introduction

(c)

Interpretation:

The number of formula units in $457 g$ 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.

Answer 24

Expert Solution

Answer to Problem 31E

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

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

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

The molar mass of iron is $55.85 g mol−1$ .

The molar mass of sulfur is $32.06 g mol−1$

The molar mass of oxygen is $16.00 g mol−1$ .

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

$Total molar mass =(2×55.85) +(3×32.06)+(12×16)=399.88 g mol−1$

Therefore, the molar mass of $Fe2(SO4)3$ is $399.88 g mol−1$ .

Thus one mole of $Fe2(SO4)3$ is $399.88 g mol−1$ .

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

$Moles of Fe2(SO4)3=(1 mole of Fe2(SO4)3×weight of Fe2(SO4)3 in g399.88g of Fe2(SO4)3)=(1 mole of Fe2(SO4)3×457 g of Fe2(SO4)3399.88 g of Fe2(SO4)3)=1.1428 moles of Fe2(SO4)3$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×moles of Fe2(SO4)31 mole of Fe2(SO4)3)$

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

$Formula units of Fe2(SO4)3=(6.022×1023 formula units of Fe2(SO4)3×1.1428 moles of Fe2(SO4)31 mole of Fe2(SO4)3)=6.88×1023 formula units of Fe2(SO4)3$