Chapter 2.2, Problem 2.3CYU

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Neon has three stable isotopes, one with a small abundance. What are the abundances of the other two isotopes?20Ne, mass = 19.992435 u; percent abundance = ?21Ne mass = 20.993843 u; percent abundance = 027%22Ne mass = 21.991383 u: percent abundance = ?

Interpretation Introduction

Interpretation:

The relative abundance of two isotopes of Neon has to be determined if the isotope-20Ne has a mass of 19.992435u, the isotope -21Ne has a mass of 20.993843u and percent abundance of 0.27% and isotope-22Ne has a mass of 21.991383u.

Concept introduction:

The average weight (atomic weight) of Neon is 20.1797u.

Equation for average weight is,

average weight=(%abundanceisotop-1100)×(massofisotop-1) +(%abundanceisotop-2100)×(massofisotop-2)+etc

Explanation

There are three isotopes for neon atom.

The mass of isotope-1 (20Ne) is 19.992435u

The mass of isotope-2 is (21Ne) is 20.993843u.

The mass of isotope-3 is (22Ne) is 21.991383u.

The percent abundance of 21Ne is 0.27%.

Letâ€™s take the percent abundance of 20Ne as â€˜xâ€™ and 22Ne is â€™99.73-xâ€™ (since x+.27+99.73âˆ’x=100 ).

Therefore,

Equation for average weight of neon is,

20.1797â€‰â€‰=â€‰â€‰(x100)â€‰â€‰Ã—â€‰(19.992435)â€‰+â€‰(0.27100)Ã—(20.993843)+(99.73-x100)â€‰â€‰Ã—â€‰(21.991383)20.1797â€‰â€‰=â€‰â€‰(19.992435x100)â€‰â€‰+5

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