Introduction to Chemistry
Introduction to Chemistry
4th Edition
ISBN: 9780073523002
Author: Rich Bauer, James Birk Professor Dr., Pamela S. Marks
Publisher: McGraw-Hill Education
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Chapter 4, Problem 55QP

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

a  H 2 b  Ca NO 3 2   c  N 2 O 2 d  K 2 SO 4

(a)

Expert Solution
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Interpretation Introduction

Interpretation:

The number of atoms (or ions) present in 140 g H2 is to be determined.

Explanation of Solution

For the number of formula units:

mgMMnNANumber of formula units

Here, m is the given mass, MM is the molar mass and n is the number of moles.

For the number of atoms or ions:

Number of formula unitsformula ratioAtoms or ions

The number of formula units present is determined as follows:

Number of formula units=mMM×NA ...(1)

Here, NA is Avogadro’s number with a value of 6.022×1023mol1 .

The number of atoms is determined as follows:

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

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

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

The molar mass of one mole of H2 is 2.016g . The weight of H2 is 140 g . In the H2 molecule, there are two hydrogen atoms. Substitute these values in equation (2) as follows.

Number of H atoms=140 g2.016g/mol×6.022×1023mol1×2 H atoms=8.364×1025 H atoms

Therefore, there are 8.364×1025 H atoms present in 140 g of H2 .

(b)

Expert Solution
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Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in 140 g CaNO32 is to be determined.

Explanation of Solution

The molar mass of CaNO32 is 164.10g/mol . The weight of CaNO32 is 140 g . Substitute these values in equation (1) to determine the number of formula units present in 140 g of CaNO32 as follows.

Formula unitsCaNO32=140 g164.1g/molg/mol×6.022×1023mol1=5.14×1023 formula unitsCaNO32

There is one Ca2+ in CaNO32 . By using equation (3), the number of Ca2+ ions in 140 g CaNO32 is determined as follows.

Number of atoms=5.14×1023 ×1 Ca2+=5.14×1023 Ca2+

There are two NO3 ions in CaNO32 . By using equation (3), the number of NO3 in 140 g CaNO32 is determined as follows.

Number of atoms=5.14×1023 ×2 NO3=1.028×1024 NO3

Therefore, there are 5.14×1023 Ca2+ and 1.028×1024 NO3 present in 140 g of CaNO32 .

(c)

Expert Solution
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Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in 140 g N2O2 is to be determined.

Explanation of Solution

The molar mass of N2O2 is 60.02g/mol . The weight of N2O2 is 140 g . Substitute these values in equation (1) to determine the number of formula units present in 140 g of N2O2 as follows.

Formula unitsN2O2=140 g60.02g/molg/mol×6.022×1023mol1=1.405×1024 formula unitsN2O2

There are two N in N2O2 . By using equation (3), the number of N atoms in 140 g N2O2 is determined as follows.

Number of atoms=1.405×1024 ×1 N=2.8×1024 Natoms

There are two O ions in N2O2 . By using equation (3), the number of O in 140 g N2O2 is determined as follows.

Number of atoms=1.405×1024 ×2 O=2.8×1024 Oatoms

Therefore, there are 2.8×1024 N and 2.8×1024 O atoms present in 140 g of N2O2 .

(d)

Expert Solution
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Interpretation Introduction

Interpretation:

The number of atoms (or ions) of each element present in 140 g K2SO4 is to be determined.

Explanation of Solution

The molar mass of K2SO4 is 174.26g/mol . The weight of K2SO4 is 140 g . Substitute these values in equation (1) to determine the number of formula units present in 140 g of K2SO4 as follows.

Formula unitsK2SO4=140 g174.26g/molg/mol×6.022×1023mol1=4.838×1023 formula unitsK2SO4

There are 2 K+ in K2SO4 . By using equation (3), the number of K+ ions in 140 g K2SO4 is determined as follows.

Number of atoms=4.838×1023 ×2 K+=9.676×1023 K+ions

There is one SO24 in K2SO4 . By using equation (3), the number of SO24 in 140 g K2SO4 is determined as follows.

Number of atoms=4.838×1023 ×1 SO24=4.838×1023 SO24

Therefore, there are 9.676×1023 K+ and 4.838×1023 SO24 present in 140 g of K2SO4 .

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