## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 2 Atoms, Molecules, and Ions - Study Questions - Page 95i: 138

#### Answer

Empirical formula: $V_2S_3$

#### Work Step by Step

Atomic weights (g/mol): $V: 50.9415, \ S: 32.06$ Number of moles consumed: V: $2.04\ g\div50.9415=4.00\times10^{-2}\ mol\ of\ V$ S: $1.93\ g\div32.06=6.02\times10^{-2}\ mol\ of\ S$ Dividing by the smallest number of moles: S: $6.02\times10^{-2}\div4.00\times10^{-2}=1.50=3/2$ Empirical formula: $V_2S_3$

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