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Solutions for Introduction and Review
H C
H
H
C
H
H
NO2
empirical formula = empirical weight = 75.05
molecular weight = 75, same as the empirical weight
N C
O
O
empirical formula = molecular formula =
some possible structures:
C
H
H
H
H
H= 2.66 moles O ÷ 1.34 moles = 1.99 ≈ 2 O16.0 g/mole
42.6 g O
12.0 g/mole = 2.67 moles C ÷ 1.34 moles = 1.99 ≈ 2 C
6.67 g H
1.01 g/mole = 6.60 moles H ÷ 1.34 moles = 4.93 ≈ 5 H
18.7 g N
14.0 g/mole = 1.34 moles N ÷ 1.34 moles = 1 N
32.0 g C(b)
12 continued
C2H5NO2
C2H5NO2
MANY other structures 
are possible.
empirical formula = molecular formula =
molecular weight = 93, same as the empirical weight
empirical weight = 93.49empirical formula =
= 1.07 moles Cl ÷ 1.07 moles = 1 Cl35.45 g/mole
37.9 g Cl
C2H4ClNO
C2H3Cl
C4H6Cl2
C C
Cl
C
H
H
H
Cl
C
H
H
H
C
C C
C
38.4 g C
12.0 g/mole = 3.20 moles C ÷ 1.60 moles = 2 C
4.80 g H
1.01 g/mole = 4.75 moles H ÷ 1.60 moles = 2.97 ≈ 3 H
56.8 g Cl
35.45 g/mole = 1.60 moles Cl ÷ 1.60 moles = 1 Cl
empirical formula = empirical weight = 62.45
molecular weight = 125, twice the empirical weight
twice the empirical formula = molecular formula =
some possible structures:(d)
MANY other structures 
are possible.
12.0 g/mole = 2.13 moles C ÷ 1.07 moles = 1.99 ≈ 2 C
4.32 g H
1.01 g/mole = 4.28 moles H ÷ 1.07 moles = 4 H
15.0 g N
14.0 g/mole = 1.07 moles N ÷ 1.07 moles = 1 N
25.6 g C(c)
C2H4ClNO
= 1.07 moles O ÷ 1.07 moles = 1 O16.0 g/mole
17.2 g O
some possible structures:
MANY other structures 
are possible.
C
C
N
Cl
O
H
H
H
H
O
C C
N
H
N
C
C
O
H
Cl
H
H
H
H
H
Cl
H
Cl Cl
HH
H
HH
H
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