# Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179h: 96

#### Work Step by Step

Number of moles of $TiO_2$: $1.598\ g\div 79.87\ g/mol= 0.0200\ mol$, From stoichiometry: $0.0200\ mol$ of Ti, $47.867\ g/mol\times0.0200\ mol=0.958\ g$ Number of moles of $O$: $(1.438-0.958)\ g\div 15.9994\ g/mol= 0.0300\ mol$ Ratio: $0.0300/0.0200=1.50\approx 3/2$ Empirical formula: $Ti_2O_3$

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