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: $C_{10}H_8$
