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Chemistry & Chemical Reactivity

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

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BuyFindarrow_forward

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%.

Let’s take the percent abundance of 20Ne as ‘x’ and 22Ne is ’99.73-x’ (since x+.27+99.73x=100 ).

Therefore,

Equation for average weight of neon is,

20.1797=(x100)×(19.992435)+(0.27100)×(20.993843)+(99.73-x100)×(21.991383)20.1797=(19.992435x100)+5

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