Answer
$M(vitamin B12)=1356.3\frac{g}{mol}$
Work Step by Step
$\omega (Co) = \frac{Ar(Co)}{M_{r}(vitamin B12)}\implies 0.0435=\frac{59}{M_{r}(vitamin B12)}\implies M(vitamin B12)\approx 1356.3\frac{g}{mol}$
Chemistry 10th Edition
by Whitten, Kenneth W.; Davis, Raymond E.; Peck, Larry; Stanley, George G.
Published by
Brooks/Cole Publishing Co.
ISBN 10:
1133610668
ISBN 13:
978-1-13361-066-3
Chapter 2 - Chemical Formulas and Composition Stoichiometrhy - Exercises - Building Your Knowledge - Page 79: 111
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