Answer
Empirical and molecular formula: $C_5H_{14}N_2$
Work Step by Step
Atomic weights (g/mol):
$C: 12.011, \ H: 1.008,\ N: 14.007$
In 100 g :
C: $58.77\ g \div 12.011 \ g/mol=4.89\ mol$
H: $13.81\ g\div1.008\ g/mol=13.70\ mol$
N: $27.40\ g\div14.007\ g/mol=1.96\ mol$
Dividing all these number of moles by the smallest value (1.96) leads to the empirical formula:
C: $4.89\div1.96=2.50$
H: $13.70\div1.96=7.00$
$C_5H_{14}N_2$
Molar mass of the empirical formula: $102.18\ g/mol$
Molar mass of the compound: $102.2\ g/mol$
Ratio of the two: 1.00
Molecular formula: $C_5H_{14}N_2$
