## Chemistry and Chemical Reactivity (9th Edition)

Published by Cengage Learning

# Chapter 2 Atoms, Molecules, and Ions - Study Questions - Page 95i: 129

#### Answer

a) Empirical and molecular formula: $CO_2F_2$ b) Empirical formula: $C_{5}H_4$ Molecular formula: $C_{10}H_8$

#### Work Step by Step

Atomic weights (g/mol): $C: 12.011,\ O: 15.999, \ H: 1.008,\ F: 18.998$ a) In 100 g : C: $14.6\ g \div 12.011 \ g/mol=1.22\ mol$ O: $39.0\ g\div15.999\ g/mol=2.44\ mol$ F: $46.3\ g\div18.998\ g/mol=2.44\ mol$ Dividing all these number of moles by the smallest value (1.22) leads to the empirical formula: $CO_2F_2$ Molar mass of the empirical formula: $82.0\ g/mol$ Molar mass of the compound: $82.0\ g/mol$ Ratio of the two: 1.00 Molecular formula: $CO_2F_2$ b) Since this is a hydrocarbon, the remaining fraction is of H. In 100 g : C: $93.71\ g \div 12.011 \ g/mol=7.80\ mol$ H: $6.29\ g\div 1.008\ g/mol=6.24\ mol$ Dividing all these number of moles by the smallest value (6.24) leads to the empirical formula: $C_5H_4$ ($7.80\div6.24=1.25=5/4$) Molar mass of the empirical formula: $64.09\ g/mol$ Molar mass of the compound: $128.16\ g/mol$ Ratio of the two: 2.00 Molecular formula: $C_{10}H_8$

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