21) Vanadium has two naturally occurring isotopes, 50V with an atomic mass of 49.9472 amu and 51v with an atomic mass of 50.9440. The atomic weight of vanadium is 50.9415. The percent abundances of the vanadium isotopes are % 51v % 50V and

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter2: Atoms, Molecules, And Ions
Section: Chapter Questions
Problem 21QAP: Bromine has two occuring isotopes: 79Br with atomic mass 78.9183 and 81Br with atomic mass 80.9163....
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I'm confused on how to get the atomic mass average.
It's not as straight forward as adding the two given and dividing by two is it?

The question is asking for the % abundance if that isn't clear.

21) Vanadium has two naturally occurring isotopes, 50V with an atomic mass of 49.9472 amu and 51v with an atomic
mass of 50.9440. The atomic weight of vanadium is 50.9415. The percent abundances of the vanadium isotopes are
% 51v
% 50V and
Transcribed Image Text:21) Vanadium has two naturally occurring isotopes, 50V with an atomic mass of 49.9472 amu and 51v with an atomic mass of 50.9440. The atomic weight of vanadium is 50.9415. The percent abundances of the vanadium isotopes are % 51v % 50V and
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Why does the x go from:
-0.9968x = -0.0025

To:
x= 0.0025x/0.9968

And not:
x= 0.9968/0.0025

Wouldn't the negatives simply be ignored and the x would remain with:
0.9968x and then you divide then 0.0025 without the coefficient x to get the value of x by cancelling out the 0.9968?

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