## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 10 Gases and Their Properties - Study Questions - Page 403h: 104

#### Answer

See the answer below.

#### Work Step by Step

Number of moles of water: $0.09912\ g\div 18.015\ g/mol=5.502×10^{-3}\ mol$ $0.011\ mol\ H\rightarrow 0.0111\ g$ Number of moles of NaOH: $0.3283\ M×28.81/1000=0.00946\ mol$ From stoichiometry: $0.00473\ mol\ C\rightarrow 0.0568\ g$ Number of moles of nitrogen: $n=(65.12/760×1\ atm)×0.225\ L/((0.082)(298\ K))$ $n=0.000788\ mol\ N_2$ $0.00158\ mol\ N\rightarrow 0.0221\ g$ Number of moles of oxygen: $(0.1152-0.0111-0.0568-0.0221)\ g/(15.999\ g/mol)=0.00158\ mol$ H/O ratio: $0.0110/0.00158=7.0$ C/O ratio: $0.00473/0.00158=3.0$ N/O ratio: $0.00158/0.00158=1.0$ Empirical formula: $C_4H_7NO$ Molecular weight of the empirical formula: $85.11\ g/mol$ Molecular formula: $C_8H_{14}O_2N_2$

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