# 9_ Lesson 5- Isotopes and average atomic mass by hedongchenchen

VIEWS: 7 PAGES: 9

• pg 1
Lesson 5
Isotopes
Elements that have the same atomic number, but
different atomic mass.

Many elements are present as natural isotopes
Isotopic abundance
Natural isotopes occur, but not at the same amounts
Natural isotopes of Neon

Neon-20             Neon-21             Neon-22
90.60%               0.26%               9.20%

The isotope abundance is used to calculate the
average mass
Average atomic mass
The average atomic mass can be calculated using weighted
averages

1. Multiply the atomic mass x % abundance for each
isotope

2. Add these values together to get the average atomic
mass
Average atomic mass

Neon-20       Neon-21     Neon-22
% abundance   90.60%        0.26%       9.20%
Atomic mass   19.992439     20.993845   21.991384
(amu)

= 18.11 + 0.055 + 2.02
= 20.19 amu
Try this one!!!

Find the average atomic mass of Argon if argon-36,
35.967546 amu, 0.34%, argon-38, 37.962732 amu, 0.063%,
argon-40, 39.962383 amu, 99.60%.

= 0.122 + 0.0239 + 39.802
= 39.948 amu
Calculating the Isotopic Abundance
Calculate the isotope abundance of chlorine if the average
mass is 35.45, and chlorine-35 has a 34.968853 and
chlorine-37 has a mass of 36.965903
Let isotope abundance for Cl-35 be x and for Cl-37 be 1 –x
Average mass      = x(atomic mass Cl-35) + (1-x)(atomic
mass Cl-37)
35.45      = x(34.968853) + (1-x)(36.965903)
35.45      = 34.968853x + 36.965903 – 36.965903x
35.45 - 36.965903 = -2.00x
- 1.515903 = -2.00x
0.758 = x

 Cl-35 is 75.8% and Cl-37 is 24.2%
Determine the isotope abundance of Nitrogen if N-14 is
14.0031, N-15 is 15.0001, and the average mass is 14.01
amu.

Atomic mass = x(a.m. N-14) + (1-x)(a.m. N-15)
14.01       = x(14.0031) + (1-x)(15.0001)
14.01       = 14.0031x + 15.0001 -15.0001x
14.01-15.0001= 14.0031x -15.0001x
-0.990      = -0.997x
0.9931      =x
 The N-14 is 99.31% and N-15 is 0.6921%
Homework: See worksheet

To top