South Pasadena • Honors Chemistry
Name
2 • Elements and Compounds
Period
2.5
1. Write the balanced chemical equations for the
decomposition of the following compounds:
(a) Solid iron(III) oxide
2 Fe2O3 (s) → 4 Fe (s) + 3 O2 (g)
(b) Liquid phosphorus pentachloride
PROBLEMS
(d) Write the balanced equation for this reaction.
Pb (s) + O2 (g) → PbO2 (s)
4. A 4.00 g sample of mercury chloride is heated.
After heating, a 3.43 g product remains.
(a) What are the possible formulas for mercury
chloride?
2 PCl5 (ℓ) → 2 P (s) + 5 Cl2 (g)
Hg2Cl2
(c) Solid aluminum bromide
2 AlBr3 (s) → 2 Al (s) + 3 Br2 (ℓ)
2. Write the balanced chemical equations for the
formation of the following compounds:
(a) Solid manganese(IV) oxide
HgCl2
mercury(I) chloride or
mercurous chloride
mercury(II) chloride or
mercuric chloride
(b) Find the percent composition of the mercury
chloride.
mHg = 3.43 g
mCl = 4.00 g HgxCly – 3.43 g Hg = 0.57 g Cl
3.43 g
%Hg =
× 100% = 85.8%
4.00 g
0.57 g
%Cl =
× 100% = 14%
4.00 g
Mn (s) + O2 (g) → MnO2 (s)
(b) Liquid benzene, C6H6
6 C (s) + 3 H2 (g) → C6H6 (ℓ)
(c) Water (H2O)
2 H2 (g) + O2 (g) → 2 H2O (ℓ)
3. A 2.74 g sample of lead is burned in oxygen. After
the reaction, 3.12 g of the product is formed.
(a) What are formulas and names for possible
compounds that can form?
PbO lead(II) oxide or plumbus oxide
PbO2 lead(IV) oxide or plumbic oxide
(b) Find the percent composition of lead and
oxygen in the product.
mPb = 2.74 g
mO = 3.12 g PbxOy – 2.74 g Pb = 0.38 g O
2.74 g
%Pb =
× 100% = 87.8%
3.12 g
0.38 g
%O =
× 100% = 12%
3.12 g
(c) Find the empirical formula for the compound.
nPb = 2.74 g Pb 
1 mol Pb 
207.2 g Pb = 0.0132 mol Pb
1 mol O 
nO = 0.38 g O 
16.00 g O = 0.0238 mol O
Empirical Formula = Pb0.0132 O0.0238 = PbO2
0.0132
CLASS
Date
0.0132
(c) Find the empirical and molecular formulas for
the compound.
nHg = 3.43 g Hg 
1 mol Hg 
200.6 g Hg
= 0.0171 mol Hg
1 mol Cl 
nCl = 0.57 g Cl 
35.45 g Cl = 0.0161 mol Cl
Empirical Formula = Hg0.0171 Cl0.0161 = HgCl
0.0161
0.0161
Molecular Formula = Hg2Cl2
(d) Write the balanced equation for the
decomposition reaction.
Hg2Cl2 (s) → 2 Hg (ℓ) + O2 (g)
(e) How many atoms of mercury are in the
product?
1 mol Hg 
3.43 g Hg 
200.6 g Hg
23
6.022 × 10 atoms Hg
1 mol Hg


22
= 1.03 × 10 atoms Hg
5. A 6.95 g sample of a manganese oxide is heated.
After heating, 4.80 g of a product remains.
(a) Find the percent composition of all of the
elements the manganese oxide.
mMn = 4.80 g
mO = 6.95 g MnxOy – 4.80 g Mn = 2.15 g O
4.80 g
%Mn =
× 100% = 69.1
6.95 g
2.15 g
%O =
× 100% = 30.9%
6.95 g
(b) Find the empirical formula for the compound.
nMn = 4.80 g Mn 
1 mol Mn 
54.9 g Mn
= 0.0874 mol Mn
1 mol O 
nO = 2.15 g O 
16.00 g O = 0.134 mol O
Empirical Formula = Mn0.0874 O 0.134
0.134
0.0874
= MnO1.5 = Mn2O3
(c) Write the name of the compound.
Manganese(III) oxide
(d) Write the balanced equation for the
decomposition reaction.
2 Mn2O3 (s) → 4 Mn (s) + 3 O2 (g)
(e) What is the volume of the oxygen gas released
at STP?
1 mol O2  22.4 L O2
2.15 g O2 
32.00 g O2  1 mol O2 
= 1.51 L O2
6. A 18.4 g liquid sample of an unknown organic
compound containing only carbon, hydrogen, and
oxygen is burned in air (undergoes combustion)
releasing 32.7 g CO2 and 13.4 g H2O. The equation
for the reaction is:
... CxHyOz (ℓ) + … O2 (g) → … CO2 (g) + … H2O (ℓ)
(a) Find the mass of each element in this
compound.
mC = 32.7 g CO2 
1 mol CO2   1 mol C 
44.01 g CO2 1 mol CO2
12.01 g C
= 0.743 mol C 
 1 mol C  = 8.92 g C
mH = 13.4 g H2O 
1 mol H2O   2 mol H 
18.02 g H2O 1 mol H2O
1.008 g H
= 1.49 mol H 
 1 mol H  = 1.50 g H
mO = 18.4 g CxHyOz – (8.92 g C + 1.50 g H)
= 7.98 g O
(b) Find the empirical formula for the compound.
nC = 0.743 mol C
nH = 1.49 mol H
1 mol O 
nO = 7.98 g O
16.00 g O = 0.499 mol O
Empirical Formula = C0.743 H 1.49 O0.499
0.499
0.499
0.499
= C1.5H3O = C3H6O2
(c) If the molar mass of the compound is 74 g/mol,
find the molecular formula of the compound.
Molar Mass
Empirical Formula mass
74 g/mol
=
3(12.01) + 6(1.008) + 2(16.00)
74 g/mol
1
=
=
74.08 g/mol 1
Molecular Formula
= 1 × (Empirical Formula)
= 1 × (C3H6O2) = C3H6O2
(d) What is the volume of the CO2 released at
STP?
32.7 g CO2 
1 mol CO2  22.4 L CO2
44.01 g CO2  1 mol CO2 
= 16.6 L CO2
(e) What is the volume occupied by the sample of
this compound? The density is 0.99 g/cm3.
18.4 g 
1 cm3 
3
0.99 g = 18.6 cm
7. A sample of lysine contains 0.865 g carbon, 0.170 g
hydrogen, 0.335 g nitrogen, and 0.385 g oxygen.
The molar mass of lysine is 146 g/mol.
(a) Find the empirical formula for lysine.
1 mol C
nC = 0.865 g C ×
= 0.0720 mol C
12.01 g C
1 mol H
nH = 0.170 g H ×
= 0.169 mol H
1.008 g H
1 mol N
nN = 0.335 g N ×
= 0.0239 mol N
14.01 g N
1 mol O
nO = 0.385 g O ×
= 0.0240 mol O
16.00 g O
Empirical Formula =
C0.0720 H 0.169 N0.0239 O0.0240 = C3H7NO
0.0239
0.0239
0.0239
0.0239
(b) Find the molecular formula for lysine.
Molar Mass
=
Empirical Formula mass
146 g/mol
=
3(12.01) + 7(1.008) + 1(14.01) + 1(16.00)
146 g/mol 2
=
73.10 g/mol 1
Molecular Formula
= 2 × (Empirical Formula)
= 2 × (C3H7NO) = C6H14N2O2
(c) How many molecules are in this sample of
lysine?
Total mass = 0.865 g + 0.170 g + 0.335 g +
0.385 g = 1.755 g lysine
1 mol lysine
1.755 g lysine 
 146 g lysine 
23
6.022 × 10 molecules lysine
1 mol lysine


= 7.24 × 1021 molecules lysine
8. Glutamine is 41.1% carbon, 6.9% hydrogen, 19.2%
nitrogen, and 32.8% oxygen by mass. The molar
mass of glutamine is 146 g/mol.
(a) Find the empirical formula for glutamine.
Assume a 100 g sample.
1 mol C
nC = 41.1 g C ×
= 3.42 mol C
12.01 g C
1 mol H
nH = 6.9 g H ×
= 6.8 mol H
1.008 g H
1 mol N
nN = 19.2 g N ×
= 1.37 mol N
14.01 g N
1 mol O
nO = 32.8 g O ×
= 2.05 mol O
16.00 g O
Empirical Formula = C3.42 H 6.8 N1.37 O2.05
1.37
1.37
1.37
1.37
= C2.5H5NO1.5 = C5H10N2O3
(b) Find the molecular formula for glutamine.
Molar Mass
Empirical Formula mass
146 g/mol
=
5(12.01) + 10(1.008) + 2(14.01) + 3(16.00)
146 g/mol
1
=
=
146.15 g/mol 1
Molecular Formula = 1 × (Empirical
Formula) = 1 × (C5H10N2O3) = C5H10N2O3
(c) What is the number of oxygen atoms in a 53.0
g sample of glutamine?
1 mol glutamine 
53.0 g glutamine 
146.15 g glutamine
23
 10 mol H  6.022 × 10 atoms H
1 mol H
1 mol glutamine 

22
= 2.18 × 10 atoms H
9. A 0.0850 mol sample of arginine contains 6.13 g
carbon, 1.20 g hydrogen, 4.75 g nitrogen, and 2.72
g oxygen.
(a) Find the molar mass of arginine.
m = 6.13 g + 1.20 g + 4.75 g + 2.72 g = 14.80 g
14.80 g
m
molar mass = =
= 174 g/mol
n 0.0850 mol
(b) Find the empirical formula for arginine.
1 mol C
= 0.510 mol C
12.01 g C
1 mol H
nH = 1.20 g H ×
= 1.19 mol H
1.008 g H
1 mol N
nN = 4.75 g N ×
= 0.339 mol N
14.01 g N
1 mol O
nO = 2.72 g O ×
= 0.170 mol O
16.00 g O
Empirical Formula = C0.510 H 1.19 N0.339 O0.170
nC = 6.13 g C ×
0.170
= C3H7N2O
0.170
0.170
0.170
(c) Find the molecular formula for arginine.
Molar Mass
Empirical Formula mass
174 g/mol
=
3(12.01) + 7(1.008) + 2(14.01) + 1(16.00)
174 g/mol 2
=
=
87.11 g/mol 1
Molecular Formula = 2 × (Empirical
Formula) = 2 × (C3H7N2O) = C6H14N4O2
10. A 36.4 g sample of a gaseous compound containing
only nitrogen and oxygen is 30.5% nitrogen by
mass occupies 8.85 L at STP.
(a) Find the molar mass of this compound.
1 mol 
n = 8.85 L 
22.4 L = 0.395 mol
36.4 g
m
molar mass = =
= 92.1 g/mol
n 0.395 mol
(b) Find the empirical formula for this compound.
Assume 100 g sample.
1 mol N
nN = 30.5 g N ×
= 2.17 mol N
14.01 g N
1 mol O
nO = 69.5 g O ×
= 4.34 mol O
16.00 g O
Empirical Formula = N2.17 O4.34 = NO2
2.17
2.17
(c) Find the molecular formula for this compound.
Molar Mass
Empirical Formula mass
92.1 g/mol
92.1 g/mol 2
=
=
=
1(14.01) + 2(16.00) 46.01 g/mol 1
Molecular Formula = 2 × (Empirical
Formula) = 2 × (NO2) = N2O4