Chemistry: The Central Science (13th Edition)

Published by Prentice Hall
ISBN 10: 0321910419
ISBN 13: 978-0-32191-041-7

Chapter 3 - Chemical Reactions and Reaction Stoichiometry - Exercises: 3.47c

Answer

Na3AlF6

Work Step by Step

We have 32.79 % Na, 13.02% Al and the remainder (100 - 32.79 - 13.02 = 54.19%) F. We assume 100g basis so we have 32.79g Na, 13.02g Al and 54.19g F Then we convert to moles so Na is 22.99 g/mol then 32.79g is 1.43 mole Al is 26.98 g/mol so 13.02g is 0.483 mole F is 18.998 so 54.19g is 2.84 mole we divide each by the smallest number, 0.483 1.43/0.483 = 3 0.483/0.483 = 1 2.84/0.483 = 6 The ratio is 3:1:6 so we have Na3AlF6
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