General Chemistry (4th Edition)

Published by University Science Books
ISBN 10: 1891389602
ISBN 13: 978-1-89138-960-3

Chapter 2 Atoms and Molecules - Problems - Page 75: 55

Answer

(Isotopic Mass of N-14)(Abundance of N-14) + (Isotopic Mass of N-15) (Abundance of N-15) = Atomic Mass of Nitrogen Suppose x = Abundance of N-14 This implies that Abundance of N-15 = 1-x Therefore (14.0031)(x) + (15.0001)(1-x) = 14.0067 14.0031x + 15.0001 – 15.0001x = 14.0067 -0.997x = -0.9934 x = 0.996 1 – x = 1 – 0.996 = 0.0036 Abundance of N-14 = 99.6 % Abundance of N-15 = 0.4 %

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