   # A fictitious element has two naturally occurring isotopes and has an atomic mass of 29 .5 amu . a. If the natural abundance of isotope 1 is 33 .7%, what is the natural abundance of isotope 2? b. If the mass of isotope 2 is 30 .0 amu, what is the mass of isotope 1? ### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692

#### Solutions

Chapter
Section ### Chemistry In Focus

7th Edition
Tro + 1 other
Publisher: Cengage Learning,
ISBN: 9781337399692
Chapter 3, Problem 41E
Textbook Problem
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## A fictitious element has two naturally occurring isotopes and has an atomic mass of 29 .5 amu . a. If the natural abundance of isotope 1 is 33 .7%, what is the natural abundance of isotope 2?b. If the mass of isotope 2 is 30 .0 amu, what is the mass of isotope 1?

Interpretation Introduction

Interpretation:

The natural abundance of isotope 2 and the mass of isotope 1 of fictitious element are to be calculated.

Concept Introduction:

 Atomic mass is defined as the mass of an atom. It is a characteristic of an element. It can be a whole number or fractional. Its unit is amu.

 An element can have more than one form of atoms. These forms differ in the number of neutrons contained in the nucleus, while the number of protons is same. These are termed as isotopes.

 As these isotopes are the atoms of the same element, the actual atomic mass depends on the natural abundance of the isotopes of that element. The total natural abundance of all the isotopes of an element is 100.

### Explanation of Solution

a) The natural abundance of isotope 2.

The natural abundance of isotope 2 can be calculated as:

The total abundance of isotope 1 and 2 of the fictitious element is 100.0 %.

The natural abundance of isotope 1 is 33.7 %.

Hence, the natural abundance of isotope 2 is the difference of both the above values.

Natural abundance of isotope 2=100 % - 33.7 %=66.3 %

b) The mass of isotope 1.

The mass of isotope 1 is calculated as:

Atomic mass of isotope 1 of the element is given by the expression given below:

Atomic mass =Mass of isotope×Natural abundance (%)100

Substitute the values of mass of isotope and its natural abundance in the above equation.

Atomic mass of fictitious element=(Atomic mass of isotope 1×33.7)+(30.0 amu×66.3)100                                     29.5 amu=(Atomic mass of isotope 1×33

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