Summer Mini-Session 3:
Nomenclature and Chemical Reactions
Answer Key
Nomenclature (Ionic Compounds)
Give the appropriate chemical formulas for the following compounds. Be sure to include the
proper subscripts for each atom.
1. Iron(III) sulfide
Fe2S3
2. Copper(I) oxide
Cu2O
3. Magnesium chloride
MgCl2
4. Potassium bromide
KBr
5. Lithium carbonate
Li2CO3
6. Beryllium chloride
BeCl2
7. Magnesium nitride
Mg3N2
8. Lead(IV) phosphate
Pb3(PO4)4
9. Ammonium chloride
NH4Cl
10. Calcium fluoride
CaF2
List the name of each of the compounds represented by the following chemical formulas:
11. NH4NO3
ammonium nitrate
12. Ca(OH)2
calcium hydroxide
13. CaCO3
calcium carbonate
14. CuSO4
copper (II) sulfate
15. Fe2O3
iron (III) oxide
16. CrS
chromium (II) sulfide
17. BeBr2
beryllium bromide
18. NaF
sodium fluoride
19. Li2O
lithium oxide
20. KCl
potassium chloride
Nomenclature (Acids/Covalent Compounds)
Give the appropriate chemical formulas for the following compounds. Be sure to include the
proper subscripts for each atom.
1. Hydrobromic acid
HBr
2. Carbon tetrachloride
CCl4
3. Nitrous acid
HNO2
4. Sulfur dioxide
SO2
5. Nitric acid
HNO3
6. Silicon dioxide
SiO2
7. Hydrofluoric acid
HF
8. Chlorine tetrafluoride
ClF4
9. Dinitrogen tetroxide
N2O4
10. Sulfur tetrabromide
SBr4
List the name of each of the compounds represented by the following chemical formulas:
11. ClF3
chlorine trifluoride
12. HI
hydroiodic acid
13. SI6
sulfur hexaiodide
14. H2SO3
sulfurous acid
15. CO
carbon monoxide
16. H3PO4
phosphoric acid
17. PF5
phosphorus pentafluoride
18. H2Se
hydroselenic acid
19. SiSe2
silicon diselenide
20. P2S5
diphosphorus pentasulfide
Balancing Equations
Balance the following chemical equations by using the appropriate coefficients for the various
compounds. Note: not every compound will need a coefficient. Also, for each reaction, indicate
the type: synthesis, decomposition, single displacement, double displacement, or combustion.
1. ____ BaCl2 + ____ H2SO4 –> ____ BaSO4 + _2_ HCl
Type: double displacement
2. _4_ P + _5_ O2 –> ____ P4O10
Type: synthesis
3. ____ C3H8 + _5_ O2 –> _3_ CO2 + _4_ H2O
Type: combustion
4. _2_ KClO3 –> _2_ KCl + _3_ O2
Type: decomposition
5. ____ Cu + _2_ AgNO3 –> ____ Cu(NO3)2 + _2_ Ag
Type: single displacement
6. ____ NH4NO3 –> ____ N2O + _2_ H2O
Type: decomposition
7. ____ BF3 + _3_ H2O –> ____ B(OH)3 + _3_ HF
Type: double displacement
8. ____ P4O10 + _6_ H2O –> _4_ H3PO4
Type: synthesis
9. _4_ NH3 + _3_ O2 –> _2_ N2 + _6_ H2O
Type: single displacement
10. _2_ Ag + ____ H2S –> ____ Ag2S + ____ H2
Type: single displacement
Balancing Equations 2
Balance the following chemical equations by using the appropriate coefficients for the various
compounds. Note: not every compound will need a coefficient. Also, for each reaction, indicate
the type: synthesis, decomposition, single displacement, double displacement, or combustion.
1. _2_ KOH + ____ H2SO4 –> ____ K2SO4 + _2_ H2O
Type: double displacement
2. ____ Na2O + ____ H2O –> _2_ NaOH
Type: synthesis
3. _2_ C2H6 + _7_ O2 –> _4_ CO2 + _6_ H2O
Type: combustion
4. _2_ PbO2 –> _2_ PbO + ____ O2
Type: decomposition
5. _3_ Fe + _4_ H2O –> ____ Fe3O4 + _4_ H2
Type: single displacement
6. _2_ H2O –> _2_ H2 + ____ O2
Type: decomposition
7. _2_ Al(NO3)3 + _3_ H2SO4 –> ____ Al2(SO4)3 + _6_ HNO3
Type: double displacement
8. _3_ CaO + ____ P2O5 –> ____ Ca3(PO4)2
Type: synthesis
9. _2_ C2H2 + _5_ O2 –> _4_ CO2 + _2_ H2O
Type: combustion
10. _2_ Al + _3_ CuCl2 –> _2_ AlCl3 + _3_ Cu
Type: single diplacement
Chpt. 6, Sect. 1 #1-5
1. The main distinction between an ionic and a covalent bond is the degree to which the
electrons are attracted to one or the other of the two atoms in the bond. This is
determined by the difference in electronegativities between the two atoms.
2. If the difference in the electronegativities between the two elements forming the
bond is greater than 1.7, then the bond is said to be ionic in character. Differences
of less than 1.7 are said to be more covalent in character. Even though that
arbitrary point is used, it’s important to understand that the character of the bonds
constitutes a continuum rather than a discreet two groups or types of bonds.
3. a. Li and F would form an ionic bond.
b. Cu and S would also form an ionic bond.
c. I and Br form a covalent bond.
4. The pairs in increasing ionic character would be: I-Br, Cu-S, Li-F
5. a. A Cu-Cl bond would have more ionic character than a I-Cl bond.
b. Cl would have a greater negative charge in the Cu-Cl bond.
Chpt. 6, Sect. 2 #1-5
1. a. Bond length is the average distance between two atoms sharing a bond.
b. Bond energy is the amount of energy required to pull two atoms apart from one
another (the amount of energy required to “break” the bond).
2. The octet rule says that atoms will form bonds with other atoms by sharing, losing,
or gaining electrons in order to fill up their outermost energy level with eight
electrons.
3. a. Atoms with a single bond share one pair of electrons.
b. Double bonds involve two pairs of electrons.
c. A triple bond has three pairs of electrons in it.
4.
Chpt. 6, Sect. 2 #1-5
(cont.)
5. HNNH has a stronger N-N bond than H2NNH2 because the N-N bond in HNNH is a
triple bond whereas the N-N bond in H2NNH2 is a double bond. Triple bonds tend to
be stronger than double bonds which tend to be stronger than single bonds.
Chpt. 6, Sect. 3 #1-5
1. Two examples of ionic compounds are sodium chloride and magnesium nitrate. (There
are many other examples).
2.
3. The basic unit of an ionic compound is a “formula unit”. A “molecule” is the basic unit
in molecular compounds.
4. Compound B is the molecular compound and compound A is ionic. The intermolecular
forces in covalent molecules (molecular compounds) tend to be much weaker than the
attractive forces in ionic compounds. This means it’s harder to pull apart the ions in
an ionic lattice, so the melting and boiling points are higher than those of the
molecular compound.
5. Listed in order of increasing lattice energy: Rb2S, K2S, Li2S. The higher the melting
point, the higher the lattice energy.
Chpt. 6, Sect. 5 #5-6
5. The high boiling point of water is due in part to the hydrogen bonds between the
water molecules. There is a large difference in electronegativity between the
oxygen and hydrogen atoms in the water molecule. Since the oxygen atom is much
more electronegative, the outer electrons spend more time on average near the
oxygen atom than either of the hydrogen atoms. This gives the oxygen atom a slight
or partial negative charge compared to the hydrogen atom creating an charge dipole.
The slightly negative oxygen atoms of one water molecule are attracted to the
slightly positive hydrogen atoms of the surrounding water molecules. This extra
intermolecular attraction makes it tougher to pull water molecules apart giving water
a higher melting and boiling point than many molecules of similar molecular mass.
Chpt. 6, Sect. 5 #5-6 (cont.)
6. The boiling point of a compound correlates with the strength of the intermolecular
forces in various compounds. The stronger the attractive forces, the higher the
boiling point. However, there are other factors which are important (such as the
mass of the molecules), so the boiling point can be affected by factors other than
intermolecular attractions.
Chpt. 6 Review
#6-7, 15, 19-21, 45-47, 63-64, 68-69, 71
6.
a.
b.
c.
d.
e.
f.
g.
Electroneg. Diff
H-I
0.4
S-O
1.0
K-Br
2.0
Si-Cl
1.2
K-Cl
2.2
Se-S
0.1
C-H
0.4
Bond type
polar covalent
polar covalent
ionic
polar covalent
ionic
(non)polar covalent
polar covalent
More electronegative atom
I
O
Br
Cl
Cl
S
C
7. Bonding pairs from prob. #6 listed by increasing covalent character:
KCl, KBr, SiCl, SO, HI, CH, SeS
15. a. H: 1 valence electron, b. F: 7, c. Mg: 2, d. O: 6, e. Al: 3, f. N: 5, g. C: 4
19.
20.
21.
45. In each of these four cases the resulting dipole would point away from the hydrogen
atom and toward the halogen atom.
Chpt. 6 Review (cont.)
#6-7, 15, 19-21, 45-47, 63-64, 68-69, 71
46. a. nonpolar
47. a. polar
63.
b. polar
b. nonpolar
Electroneg. Diff
a. Zn-O
1.9
b. Br-I
0.3
c. S-Cl
0.5
c. polar
d. nonpolar
c. nonpolar
Bond type
ionic
(non)polar covalent
polar covalent
d. polar
e. polar
f. polar
e. nonpolar
More electronegative atom
O
Br
Cl
64.
68. The C2H6 molecule will have the longest C-C bond as it is only a single bond. The C2H4
molecule will have a shorter bond, and the C2H2 molecule will have the shortest C-C
bond of the three.
69. Fluorine is the most electronegative element on the periodic table, so it will have the
greatest difference in electronegativity between it and the atom it’s bonding with
compared to any other element on the table. The greater the electronegativity
difference between the two atoms, the greater the polarity of the bond.
71. Even though methane and ammonia are both covalent compounds and have similar
molecular masses, ammonia is a polar molecule and methane is a nonpolar molecule.
The polarity of ammonia causes intermolecular attractions between ammonia
molecules (in this case, hydrogen bonds). The intermolecular attraction between the
ammonia molecules requires greater energy to overcome their attraction and pull
them apart than a non polar molecule giving the ammonia a higher boiling point.
Chpt. 7, Sect. 1 #1-4
1. The chemical formula gives you the relative numbers of the different elements
found in a single molecule (for a molecular compound) or formula unit (for an ionic
compound).
2. a. AlBr3 b. Na2O c. MgI2 d. PbO e. SnI2 f. Fe2S3 g. Cu(NO3)2 h. (NH4)2SO4
3. a. sodium iodide b. magnesium sulfide c. calcium oxide d. potassium sulfide
e. copper bromide f. iron (II) chloride
4. a. NaOH b. Pb(NO3)2 c. FeSO4 d. P2O3 e. CSe2 f. CH3COOH
g. HClO3 h. H2SO4
Chpt. 7 Review #1, 3-11, 14-18
1. a. A monoatomic ion is a single atom with a positive or negative charge on it.
b. Examples of monoatomic ions include: Na+, O2-, and Ca2+
3. a. K+
4. a. Na+
b. Ca2+
c. S2-
b. Al3+
c. Cl -
d. Cl -
e. Ba2+
d. N3-
f. Br -
e. Fe2+
f. Fe3+
5. a. potassium ion b. magnesium ion c. aluminum ion d. chloride ion
e. peroxide ion f. calcium ion
6. a. NaI
b. CaS
c. ZnCl2
d. BaF2
e. Li2O
7. a. potassium chloride b. calcium bromide c. lithium oxide d. magnesium chloride
8. a. CrF2 (chromium (II) fluoride) b. NiO (nickel (II) oxide)
c. Fe2O3 (iron (III) oxide)
9. When naming a binary molecular compound, the element from the lower group
number is named first. If both elements are from the same group, then the element
from the highest period is named first.
10. a. carbon dioxide b. carbon tetrachloride c. phosphorus pentachloride
d. selenium hexafluoride e. diarsenic pentoxide
11. a. CBr4
b. SiO2
c. P4O10
d. As2S3
Chpt. 7 Review #1, 3-11, 14-18 (cont.)
14. a. hydrofluoric acid b. hydrobromic acid c. nitric acid
d. sulfuric acid e. phosphoric acid
15. a. H2SO3 b. HClO3 c. HCl d. HClO e. HClO4 f. H2CO3 g. CH3COOH
16. a. NaF b. CaO c. K2S d. MgCl2 e. AlBr3 f. Li3N g. FeO
17. a. ammonium ion b. chlorate ion c. hydroxide ion d. sulfate ion
e. nitrate ion f. carbonate ion g. phosphate ion h. acetate ion
i. bicarbonate ion j. chromate ion
18. a. NH4+ b. CH3COO - c. OH - d. CO32- e. SO42- f. PO43- g. Cu2+ h. Sn2+
i. Fe3+ j. Cu+ k. Hg+ l. Hg2+
Chpt. 8, Sect. 1 #1-5
1. A “word equation” just tells you which chemicals are involved in the reaction. A
“formula equation” tells you the formula or makeup of each of those compounds, but
not the quantities of each compound involved in the reaction. A “chemical equation”
gives both the formulas of the compounds and the relative amounts of each one.
2. Word equation; sodium hydroxide + sulfuric acid -> sodium sulfate + water
Formula equation: NaOH (aq) + H2SO4(aq) -> Na2SO4(aq) + H2O(l)
3. a. Solid potassium reacts with water to form aqueous potassium hydroxide and
hydrogen gas.
b. Solid iron reacts with chlorine gas to form solid iron (III) chloride.
4. Word equation: hydrogen sulfide + oxygen -> sulfur dioxide + water
Formula equation: H2S(g) + O2(g) -> SO2(g) + H2O(g)
Chemical equation: 2H2S(g) + 3O2(g) -> 2SO2(g) + 2H2O(g)
5. 2VO + 3Fe2O3 -> V2O5 + 6FeO
Chpt. 8, Sect. 2 #1-2
1. Five ways to classify reactions include: single displacement, double displacement,
combustion, synthesis, and decomposition reactions.
2. a. synthesis
b. single displacement
c. decomposition
d. combustion
Chpt. 8, Sect. 3 #1-3
1. In the case of single displacement reactions, the activity series is a good tool for
predicting whether one species will actually replace another. This lets you predict
whether the reaction will proceed or not.
2. a. no reaction
b. reaction proceeds
c. no reaction
d. reaction proceeds
e. reaction proceeds
3. b. Br2 + 2KI -> I2 + 2KBr
d. Cd + 2HCl -> H2 + CdCl2
e. Mg + Co(NO3)2 -> Mg(NO3)2 + Co
Chpt. 8 Review #1, 7-8, 10-15, 22-27, 29, 34-35, 44
1. Some ways to tell if a chemical reaction is occurring include: a change in color, a
change in smell, a change in temperature, the formation of a precipitate, and the
formation of a gas (bubbles).
7. a. An aqueous solution is simply when a compound (the solute) is dissolved in water
(the solvent).
b. A catalyst is a substance that lowers the activation energy of a reaction without
providing any energy or being used up in the reaction. Catalysts also tend to be
specific for a specific reaction.
c. A reversible reaction is a reaction in which the products can go “backwards” and
reform the reactants.
8. a. KOH
b. Ca(NO3)2
c. Na2CO3
d. CCl4
e. MgBr2
10. a. 6N b. 2O, 4H c. 4H, 4N, 12O d. 2Ca, 4O, 4H e. 3Ba, 6Cl, 18O f. 5Fe, 10N, 30O
g. 4Mg, 8P, 32O h. 4N, 16H, 2S, 8O i. 12Al, 18Se, 72O j. 12C, 32H
Chpt. 8 Review #1, 7-8, 10-15, 22-27, 29, 34-35, 44 (cont.)
11. a. 2ZnS(s) + 3O2(g) -> 2ZnO(s) + 2SO2(g)
b. 2HCl(aq) + Ba(OH)2(aq) -> BaCl2 + 2H2O(l)
c. HNO3(aq) + Ca(OH)2(aq) -> Ca(NO3)2(aq) + H2O(l)
12. a. Solid zinc sulfide reacts with oxygen gas to form solid zinc oxide and sulfur
dioxide gas.
b. Solid calcium hydride reacts with water to produce aqueous calcium hydroxide and
hydrogen gas.
c. Aqueous solutions of silver nitrate and potassium iodide react to form a
precipitate of silver iodide and an aqueous solution of potassium nitrate.
13. a. H2 + Cl2 -> 2HCl
b. 2Al + Fe2O3 -> Al2O3 + 2Fe
c. Pb(CH3COO)2 + H2S -> PbS + 2CH3COOH
14. a. 4Li + O2 -> 2Li2O
b. H2 + Cl2 -> 2HCl
c. MgCO3 -> MgO + CO2
d. 2NaI + Cl2 -> 2NaCl + I2
15. a. 4Al + 3O2 -> 2Al2O3
b. P4O10 + 6H2O -> 4H3PO4
c. Fe2O3 + 3CO -> 2Fe + 3CO2
22. a. Sodium reacts with oxygen to form sodium oxide.
4Na + O2 -> 2Na2O
b. Magnesium reacts with fluorine to produce magnesium fluoride.
Mg + F2 -> MgF2
23. a. 2HgO -> 2Hg + O2
b. 2H2O -> 2H2 + O2
c. 2Ag2O -> 4Ag + O2
d. CuCl2 -> Cu + Cl2
24. a. Zn + Pb(NO3)2 -> Pb + Zn(NO3)2
b. 2Al + 3Hg(CH3COO)2 -> 2Al(CH3COO)3 + 3Hg
c. 2Al + 3NiSO4 -> 3Ni + Al2(SO4)3
d. 2Na + H2O -> H2 + Na2O
Chpt. 8 Review #1, 7-8, 10-15, 22-27, 29, 34-35, 44 (cont.)
25. a. AgNO3 + NaCl -> NaNO3 + AgCl
b. Mg(NO3)2 + 2KOH -> 2KNO3 + Mg(OH)2
c. 3LiOH + Fe(NO3)3 -> Fe(OH)3 + 3LiNO3
26. a. CH4 + 2O2 -> CO2 + 2H2O
b. 2C3H6 + 9O2 -> 6CO2 + 6H2O
c. C5H12 + 8O2 -> 5CO2 + 6H2O
27. a. I2 + H2 -> 2HI (synthesis)
b. 2Li + 2HCl -> 2LiCl + H2 (single displacement)
c. Na2CO3 -> Na2O + CO2 (decomposition)
d. 2HgO -> 2Hg + O2 (decomposition)
e. Mg(OH)2 -> MgO + H2O (decomposition)
29. a. C3H8 + 5O2 -> 3CO2 + 4H2O
b. C5H12 + 8O2 -> 5CO2 + 6H2O
c. C2H5OH + 3O2 -> 2CO2 + 3H2O
34. a. K replaces Na b. Al replaces Ni c. Cr replaces Bi d. F replaces Cl
e. Ag replaces Au f. Cl replaces I g. Sr replaces Fe h. F replaces I
35. a. Ni + CuCl2 -> Cu + NiCl2
b. Zn + Pb(NO3)2 -> Zn(NO3)2 + Pb
c. Cl2 + 2KI -> I2 + 2KCl
d. No reaction
e. Ba + H2O -> BaO + H2
44. a. Cu + Cl2 -> CuCl2 (synthesis)
b. Ca(ClO3)2 -> CaCl2 + 3O2 (decomposition)
c. 2Li + 2H2O -> 2LiOH + H2 (single displacement)
d. PbCO3 -> PbO + CO2 (decomposition)