## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 2 Atoms, Molecules, and Ions - Study Questions - Page 95i: 133

#### Answer

$C_9H_{7}O_3Mn$

#### Work Step by Step

Atomic weights (g/mol): $C: 12.011, \ H: 1.008,\ O: 15.999,\ Mn: 54.938$ In 100 g : C: $49.5\ g \div 12.011 \ g/mol=4.12\ mol$ H: $3.2\ g\div1.008\ g/mol=3.17\ mol$ O: $22.0\ g\div15.999\ g/mol=1.38\ mol$ Mn: $25.2\div54.938\ g/mol=0.46\ mol$ Dividing all these number of moles by the smallest value (0.46) leads to the empirical formula: C: $4.12\div0.46=9.0$ H: $3.17\div0.46=7.0$ O: $1.38\div0.46=3.0$ $C_9H_{7}O_3Mn$

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