Chapter 4 Stoichiometry: Quantitative Information about Chemical Reactions - Study Questions - Page 179h: 88

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Number of moles of the $B_2O_3$: $0.422\ g\div 69.62\ g/mol=6.06\ mmol$ From stoichiometry: $12.12\ mmol$ of B, $0.01212\ mol\times10.81\ g/mol=0.131\ g$ Number of moles of hydrogen: $(0.148-0.131)\ g\div 1.008\ g/mol=16.86\ mmol$ Ratio: $16.86/12.12=1.40= 7/5$ Empirical formula: $B_5H_7$