EBK INTRODUCTION TO CHEMISTRY
EBK INTRODUCTION TO CHEMISTRY
5th Edition
ISBN: 9781260162165
Author: BAUER
Publisher: MCGRAW HILL BOOK COMPANY
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Chapter 4, Problem 56QP

How many atoms (or ions) of each element are in 140.0 g of the following substances?

a  BaSO 4 b  Mg 3 PO 4 2   c  O 2 d  KBr

(a)

Expert Solution
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Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in 140.0 g of BaSO4 is to be determined.

Explanation of Solution

The number of formula units present is determined as follows.

Number of formula units=mMM×NA ...(1)

Here, m is the given mass, MM is the molar mass, n is the number of moles, and NA is Avogadro’s number with a value of 6.022×1023mol1 .

The number of atoms is determined as follows.

Number of atoms/ions=mMM×NA×Present ions ...(2)

By combining equations (1) and (2), the number of atoms can be determined as follows.

Number of atoms/ions=Number of formula units×Present ions ...(3)

The molar mass of one mole of BaSO4 is 233.38 g/mol and its weight is 140.0 g . Substitute these values in equation (1) to determine the number of formula units present in 140.0 g of BaSO4 as follows.

Formula unitof BaSO4=140.0 g233.38g/mol×6.022×1023mol1=3.613×1022 Formula unit BaSO4

In the BaSO4 molecule, there is one barium ion Ba2+ and one sulfate ion SO42 . By using equation (3), the number of barium ions Ba2+ in 140.0 g of BaSO4 is determined as follows.

Number of  Ba2+ions=3.613×1023 ×1 Ba2+=3.613×1023  Ba2+ ions

Similarly, the number of sulfate ions is calculated as follows.

Number of  SO42 ions=3.613×1023 ×1 SO42=3.613×1023  SO42 ions

Therefore, there are 3.613×1022  Ba2+ ions and 5.14×1023  SO42 ions present in 140.0 g of BaSO4 .

The total number of barium and sulfate ions in BaSO4 is calculated as follows.

Total ions of  BaSO42=3.613×1023  Ba2+ ions+3.613×1023  SO42 ions=7.225×1023 ions KBr

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in 140.0 g of Mg3PO42 is to be determined.

Explanation of Solution

The molar mass of one mole of Mg3PO42 is 233.38 g/mol and its weight is 140.0 g . Substitute these values in equation (1) to determine the number of formula units present in 140.0 g of Mg3PO42 as follows.

Formula unitof  Mg3PO42=140.0 g265.85g/mol×6.022×1023mol1=3.2×1023 Formula unit Mg3PO42

In the Mg3PO42 molecule, there are three magnesium ions 3 Mg2+ and two phosphate ions 2 PO43- . By using equation (3), the number of magnesium ions 3 Mg2+ in 140.0 g of Mg3PO42 is determined as follows.

Number of  Mg2+ions=3.2×1023 ×3 Mg2+=9.6×1023  Mg2+ ions

Similarly, the number of phosphate ions is calculated as follows.

Number of  PO43- ions=3.2×1023 ×2 PO43-=6.4×1023 PO43- ions

Therefore, there are 9.6×1023  Mg2+ ions and 6.4×1023 PO43- ions present in 140.0 g of Mg3PO42 .

The total number of potassium and bromide ions in Mg3PO42 is calculated as follows.

Total ions of Mg3PO42=9.6×1023  Mg2+ ions+6.4×1023 PO43- ions=1.6×1024 ions KBr

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The number of atoms of each element present in 140.0 g of O2 is to be determined.

Explanation of Solution

The molar mass of two moles of O2 is 32 g/mol and its weight is 140.0 g . In the O2 molecule, there are two oxygen atoms. Substitute these values in equation (2) as follows.

Number of  O2 atoms=140.0 g32g/mol×6.022×1023mol1×2 O2 atoms=5.269×1024 O2 atoms

Therefore, there are 5.269×1024 O2 atoms present in 140.0 g of O2 .

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The number of ions of each element present in 140.0 g of KBr is to be determined.

Explanation of Solution

The molar mass of one mole of KBr is 119.002 g/mol and its weight is 140.0 g . Substitute these values in equation (1) to determine the number of formula units present in 140.0 g of KBr as follows.

Formula unitof KBr=140.0 g119.002g/mol×6.022×1023mol1=7.085×1023 Formula unit KBr

In the KBr molecule, there is one potassium ion K+ and one barium ion Br . By using equation (3), the number of potassium ions K+ in 140.0 g of KBr is determined as follows.

Number of  K+ions=7.085×1023 ×1 K+=7.085×1023  K+ ions

Similarly, the number of sulfate ions is calculated as follows.

Number of  Br ions=7.085×1023 ×1 Br=7.085×1023  Br ions

Therefore, there are 5.14×1023  Ba2+ ions and 5.14×1023  SO42 ions present in 140.0 g of BaSO4 .

The total number of potassium and bromide ions in KBr is calculated as follows.

Total ions of  KBr= Total ions of KBr=7.085×1023  K+ ions+7.085×1023  Br ions=1.417×1024 ions KBr

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Chapter 4 Solutions

EBK INTRODUCTION TO CHEMISTRY

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