Chemistry
Chemistry
4th Edition
ISBN: 9780393919370
Author: Thomas R. Gilbert
Publisher: NORTON
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Chapter 2, Problem 2.120AP

(a)

Interpretation Introduction

Interpretation: The masses of the two isotopes of bromine are given. The possible masses of the bromine molecule formed from these isotopes are to be calculated and the natural abundance of each of the molecules of the predicted masses is to be calculated.

Concept introduction: The average mass is calculated by the formula,

Averagemass=i=1n(Naturalabundanceofisotope)i×(Massofisotope)i

To determine: The possible masses of the of the individual molecules of Br2 .

(a)

Expert Solution
Check Mark

Answer to Problem 2.120AP

Solution:

The possible masses of Br2 is 157.8366amu (79Br79Br) , 159.8346amu (81Br79Br) and 161.8326amu (81Br81Br) .

Explanation of Solution

Given

The mass of isotope of bromine, 79Br is 78.9183amu .

The mass of isotope of bromine, 81Br is 80.9163amu .

The possibilities of formation of molecules of bromine are,

  • 79Br and 79Br
  • 81Br and 79Br
  • 81Br and 81Br

Therefore, the mass of 79Br79Br is,

78.9183amu×2=157.8366amu

The mass of 81Br79Br is,

78.9183amu+80.9163amu=159.8346amu

The mass of 81Br81Br is,

80.9163amu×2=161.8326amu

Therefore, the possible masses of Br2 is 157.8366amu , 159.8346amu and 161.8326amu .

(b)

Interpretation Introduction

To determine: The natural abundance of each of the molecules of the predicted masses of Br2 .

(b)

Expert Solution
Check Mark

Answer to Problem 2.120AP

Solution:

The natural abundance possible masses of Br2 is 25% (79Br79Br) , 50% (81Br79Br) and 25% (81Br81Br) .

Explanation of Solution

The possible masses of Br2 is 157.8366amu (79Br79Br) , 159.8346amu (81Br79Br) and 161.8326amu (81Br81Br) .

Therefore, the average mass of Br2 is,

157.8366amu+159.8346amu+161.8326amu3=159.8346amu

The average mass of Br2 is same as that of 81Br79Br , therefore, the natural abundance of 81Br79Br is 50% and that of the others is 50% .

The natural abundance of 81Br81Br is assumed to be x100 .

Therefore, the natural abundance of 79Br79Br is 50x100 .

The average mass is calculated by the formula,

Averagemass=i=1n(Naturalabundanceofisotope)i×(Massofisotope)i

Therefore, the average mass of bromine molecule is calculated by the formula,

Averagemass=(Abundanceof79Br79Br×Massof79Br-79Br+Abundanceof81Br79Br×Massof81Br79Br+Abundanceof81Br81Br×Massof81Br81Br)

Substitute the natural abundance and mass of 79Br79Br , 81Br79Br and 81Br81Br in the above equation.

159.8346amu=((50x100)×157.8366amu+50100×159.8346amu+x100×161.8326amu)159.8346×100=7891.83157.8366x+7991.73+161.8326x15983.467891.837991.73=3.996xx=25%

Therefore, the abundance of both 79Br79Br and 81Br81Br is 25% .

Conclusion:

  1. a. The possible masses of Br2 is 157.8366amu (79Br79Br) , 159.8346amu (81Br79Br) and 161.8326amu (81Br81Br) .
  2. b. The natural abundance possible masses of Br2 is 25% (79Br79Br) , 50% (81Br79Br) and 25% (81Br81Br)

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Chapter 2 Solutions

Chemistry

Ch. 2 - Prob. 2.3VPCh. 2 - Prob. 2.4VPCh. 2 - Prob. 2.5VPCh. 2 - Prob. 2.6VPCh. 2 - Prob. 2.7VPCh. 2 - Prob. 2.8VPCh. 2 - Prob. 2.9QPCh. 2 - Prob. 2.10QPCh. 2 - Prob. 2.11QPCh. 2 - Prob. 2.12QPCh. 2 - Prob. 2.13QPCh. 2 - Prob. 2.14QPCh. 2 - Prob. 2.15QPCh. 2 - Prob. 2.16QPCh. 2 - Prob. 2.17QPCh. 2 - Prob. 2.18QPCh. 2 - Prob. 2.19QPCh. 2 - Prob. 2.20QPCh. 2 - Prob. 2.21QPCh. 2 - Prob. 2.22QPCh. 2 - Prob. 2.23QPCh. 2 - Prob. 2.24QPCh. 2 - Prob. 2.25QPCh. 2 - Prob. 2.26QPCh. 2 - Prob. 2.27QPCh. 2 - Prob. 2.28QPCh. 2 - Prob. 2.29QPCh. 2 - Prob. 2.30QPCh. 2 - Prob. 2.31QPCh. 2 - Prob. 2.32QPCh. 2 - Prob. 2.33QPCh. 2 - Prob. 2.34QPCh. 2 - Prob. 2.35QPCh. 2 - Prob. 2.36QPCh. 2 - Prob. 2.38QPCh. 2 - Prob. 2.39QPCh. 2 - Prob. 2.40QPCh. 2 - Prob. 2.41QPCh. 2 - Prob. 2.42QPCh. 2 - Prob. 2.43QPCh. 2 - Prob. 2.44QPCh. 2 - Prob. 2.45QPCh. 2 - Prob. 2.46QPCh. 2 - Prob. 2.47QPCh. 2 - Prob. 2.48QPCh. 2 - Prob. 2.49QPCh. 2 - Prob. 2.50QPCh. 2 - Prob. 2.51QPCh. 2 - Prob. 2.52QPCh. 2 - Prob. 2.53QPCh. 2 - Prob. 2.54QPCh. 2 - Prob. 2.55QPCh. 2 - Prob. 2.56QPCh. 2 - Prob. 2.57QPCh. 2 - Prob. 2.58QPCh. 2 - Prob. 2.59QPCh. 2 - Prob. 2.60QPCh. 2 - Prob. 2.61QPCh. 2 - Prob. 2.62QPCh. 2 - Prob. 2.63QPCh. 2 - Prob. 2.64QPCh. 2 - Prob. 2.65QPCh. 2 - Prob. 2.66QPCh. 2 - Prob. 2.67QPCh. 2 - Prob. 2.68QPCh. 2 - Prob. 2.69QPCh. 2 - Prob. 2.70QPCh. 2 - Prob. 2.71QPCh. 2 - Prob. 2.72QPCh. 2 - Prob. 2.73QPCh. 2 - Prob. 2.74QPCh. 2 - Prob. 2.75QPCh. 2 - Prob. 2.76QPCh. 2 - Prob. 2.77QPCh. 2 - Prob. 2.78QPCh. 2 - Prob. 2.79QPCh. 2 - Prob. 2.80QPCh. 2 - Prob. 2.81QPCh. 2 - Prob. 2.82QPCh. 2 - Prob. 2.83QPCh. 2 - Prob. 2.84QPCh. 2 - Prob. 2.85QPCh. 2 - Prob. 2.86QPCh. 2 - Prob. 2.87QPCh. 2 - Prob. 2.88QPCh. 2 - Prob. 2.89QPCh. 2 - Prob. 2.90QPCh. 2 - Prob. 2.91QPCh. 2 - Prob. 2.92QPCh. 2 - Prob. 2.93QPCh. 2 - Prob. 2.94QPCh. 2 - Prob. 2.95QPCh. 2 - Prob. 2.96QPCh. 2 - Prob. 2.97QPCh. 2 - Prob. 2.98QPCh. 2 - Prob. 2.99QPCh. 2 - Prob. 2.100QPCh. 2 - Prob. 2.101QPCh. 2 - Prob. 2.102QPCh. 2 - Prob. 2.103QPCh. 2 - Prob. 2.104APCh. 2 - Prob. 2.105APCh. 2 - Prob. 2.106APCh. 2 - Prob. 2.107APCh. 2 - Prob. 2.108APCh. 2 - Prob. 2.109APCh. 2 - Prob. 2.110APCh. 2 - Prob. 2.111APCh. 2 - Prob. 2.112APCh. 2 - Prob. 2.113APCh. 2 - Prob. 2.114APCh. 2 - Prob. 2.115APCh. 2 - Prob. 2.116APCh. 2 - Prob. 2.117APCh. 2 - Prob. 2.118APCh. 2 - Prob. 2.119APCh. 2 - Prob. 2.120APCh. 2 - Prob. 2.121APCh. 2 - Prob. 2.122APCh. 2 - Prob. 2.123APCh. 2 - Prob. 2.125APCh. 2 - Prob. 2.126APCh. 2 - Prob. 2.127AP
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