Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning
ISBN 10: 1133949649
ISBN 13: 978-1-13394-964-0

Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179c: 25

Answer

$91.95\%$

Work Step by Step

Number of moles of water removed: $(1.245\ g - 0.832\ g)/18.015\ g/mol=0.0229\ mol$ Number of moles of the hydrated salt: $0.0229\ mol/5=0.0046\ mol$ Mass fraction of the hydrated salt: $0.0046\ mol\times 249.685\ g/mol=1.145\ g$ $1.145/1.245\times100\%=91.95\%$
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