Chemistry: A Molecular Approach (3rd Edition)

Published by Prentice Hall
ISBN 10: 0321809246
ISBN 13: 978-0-32180-924-7

Chapter 3 - Sections 3.1-3.12 - Exercises - Problems by Topic - Page 133: 86a



Work Step by Step

Convert the mass of each constituent element into moles by dividing by its molar mass: $1.245g\ Ni\times\frac{1mol}{58.693g}=0.02121mol\ Ni$ $5.381g\ I\times\frac{1mol}{126.90g}=0.04240mol\ I$ Formulas have whole number subscripts so divide all the amounts by the smallest and round if the number is really close. Then multiply by whatever factor makes them all a whole number or very close to a whole number. $\frac{0.02121mol\ Ni}{0.02121}=1$ $\frac{0.04240mol\ I}{0.02121}=2$ So the empirical formula for this compound is: $NiI_{2}$
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