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: 86b

Answer

$BaBr_{2}$

Work Step by Step

Convert the mass of each constituent element into moles by dividing by its molar mass: $2.677g\ Ba\times\frac{1mol}{137.33g}=0.01949mol\ Ba$ $3.115g\ Br\times\frac{1mol}{79.904g}=0.03898mol\ Br$ 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. $\frac{0.01949mol\ Ba}{0.01949}=1$ $\frac{0.03898mol\ Br}{0.01949}=2$ So the empirical formula for this compound is: $BaBr_{2}$
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