HCCS CHEM 1405
PRACTICE EXAM II (Ch. 3-4) 2nd edition of Blei and Odian
Multiple Choice: Choose the best answer. The correct answer is shown in bold character.
1. How many valence electrons total in water?
(A) Two
(B) four
(C) six
(D) eight
(E) ten
Note: p. 77, example 3.9.
Valence electron number is same as the group number. Water is H 2O indicating two hydrogen and
one oxygen in there. So the total valence electron number is 1x2 + 6x1= 8
2. How many lone pairs in the Lewis structure for water, H 2O?
(A) One
(B) two
(C) three
(D) four
(E) five
Note: p.p. 78, example & problem 3.9. This is the most important question.
To draw the Lewis structure, you must count the total valence electrons first and then follow the
rules stated in the textbook to draw the Lewis structure.
The total valence electron number is 1x2 + 6x1= 8.
The central atom is O and the surrounding atoms are H. So use a dash line to connect each H to O
as following:
H-O-H
Note that each dash line represents two electrons, and thus there are 2x2=4 electrons are used and
there are 8-4=4 left over electrons. The left over electrons must go to the surrounding atom H, but
here is a big problem because each H can have two electrons only, and thus the left over 4
electrons are forced to go to the central atom O and we must put two electrons at one time, see the
following:
..
H-O-H
..
So this is the Lewis structure of water. Note that lone pair electrons are always shown as dots, and
here we see four lone pair electrons. But the question asks for lone pair electrons, and each pair
means two electrons, and thus four lone pair electrons means two pairs of lone pair electrons.
3.
Which of the following formulas represents an ionic compound (p.p. 67-73)?
(A) NO
(B) HCl
(C) AsBr3
(D) MgO
(E) CO
Note: p.p. 65-73. This is the definition. Ionic compound, composing cation(s) and anion(s),
contains one or more metal and nonmetal atoms. While molecular compounds contain all
nonmetal atoms.
4.
Which of the following is a polar covalent compound?
(A) KCl
(B) HF
(C) CCl4
(D) BF3
(E) CO2
Note: p.p. 74-80. This is the definition. Polar covalent compounds are molecular compounds.
Among which the nonmetal atoms do not share electrons evenly. You must use molecular
geometry (i.e. VSEPR model or VSEPR rule) to determine it. Rule of thumb, if a molecular
compound does not possess symmetry, then it will be polar.
5.
Which formula represents manganese (II) phosphate?
(A) Mn PO4
(B) Mn2PO4
(C) Mn3PO42
(D) Mn2 (PO4)3
(E) Mn3 (PO4)2
Note: p.p. 70-72, example & problem 3.5. This is the definition and is very important. You first
need to know the formula for each ion and then put them excluding the signs together by
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exchanging the charges if the charges are different. If the charges are the same except the signs,
then simply put them together.
Like manganese (II) oxide: manganese has +2 charges and oxide has –2 charges as Mn2+ and O2-.
Since the charges are the same except the signs, we simply put them together as MnO
Like manganese (II) nitride: manganese has +2 charges and nitride has –3 charges as Mn2+ and N3. Since the charges are different, and thus we simply exchange the charges excluding the signs and
put them as subscripts as Mn3N2
Like manganese (II) phosphate: manganese has +2 charges and phosphate has –3 charges as Mn2+
and PO43-. Since the charges are different, and thus we simply exchange the charges excluding the
signs and put them as subscripts; be careful that since phosphate has subscript in oxygen atom and
to avoid confusion, we must write is as Mn3(PO4)2 instead of Mn3PO42. Because the former
indicates there are two P atoms and 2x4=8 oxygen atoms, while the latter indicates there is one P
atom and there are 42 oxygen atoms.
6. Which of the following represents a carbonate ion?
(A) NH4+
(A) CO32(B) C2H3O2(C) NO3(D) OHNote: p. 71, Table 3.2. This is the definition of naming and should be memorized.
7.
What is the correct name for BaHPO4?
(A) barium phosphate
(B) barium hydrogen phosphate
(C) barium dihydrogen phosphate
(D) monobarium hydrogen phosphate
(E) barium phosphoric acid
Note: p.p. 70-72, example & problem 3.5. This is the definition of naming and should be
memorized.
8. What is the correct name for N2O5?
(A) chromium (III) oxide
(B) dinitrogen pentoxide.
(C) nitrogen oxide
(D) nitric acid
(E) Nitrogen (V) oxide
Note: p. 74, example & problem 3.6 & Table 3.4. This is the definition of naming and should be
memorized.
9.
What is the formula mass in gram of Ca(NO3)2 ?
A) 164.1
B) 106.4
C) 52
D) 98.5
Note: p. 93: example & problem 4.1. In the formula, Ca(NO3)2, it indicates there are 1 Ca, 1x2 = 2
N and 3x2 = 6 O. Go to the periodic table and locate the atomic mass for each atom. Thus the
formula mass of Ca(NO3)2 = 1x40 + 2x14 + 6x16 = 164
10. A compound has a formula mass of 602 g. What number is m in the chemical formula Bam (PO4)2?
A) 2
B) 3
C) 4
D) 5
Note: p. 93: example & problem 4.1. Also see Q9. Here mx137.3 + 2x31 + 8x16 = 602, and thus
m = 3.
11. How many moles of NaHCO3 in 508 g of this substance?
A) 4.06
B) 8.54
C) 6.05
2
D) 21.9
Note: p. 96: example & problem 4.3. This is the definition and is very important. Memorize the
formula: mole = mass (g) / molar mass (g/mole) = 508/{23x1+1x1+12x1+16x3} = 6.05
12. The formula mass of Ca(NO3)2 contains which of the following?
A) 1 mole of Ca atom, 2 moles of N atoms and 6 moles of O atoms.
B) 6.022 x 1023 Ca atoms, 1.204 x 1024 N atoms and 3.613 x 1024 O atoms.
C) 40 grams of Ca atoms, 28 grams of N atoms and 96 grams of O atoms.
D) All of the above
Note: Comprehensive: p.p. 93-97, Table 4.1 & p. 96: example & problem 4.5. This is the
definition and is very important. The formula, Ca(NO3)2, indicates there is one mole Ca atoms,
1x2=2 moles N atoms, and 3x2=6 moles O atoms.
Note that 1 mole of atoms contains 6.022 x 1023 atoms; that is to say, to use the number 6.022 x
1023 to replace the word mole. So 1 mole Ca atoms contains 1 x 6.022 x 1023 Ca atoms, 2 moles N
atoms contain 2 x 6.022 x 1023 = 1.204 x 1024 N atoms and 6 moles O atoms contain 6 x 6.022 x
1023 = 3.613 x 1024 O atoms.
Note that 1 mole Ca atoms weighs 40 grams, 2 moles N atoms weigh 2x14 = 28 grams, and 6
moles O atoms weigh 6x16 = 96 grams.
13. The mass of silver (atomic mass unit 107.9) in 3.14 moles is determined by which of the following
calculation?
A)107.9 x 3.14
B) 47 x 3.14
C) 107.9 x 47
D) 107.9 x 3.14 x 47
Note: p. 96: example & problem 4.4. This is a very important question. Apply the definition or
say formula: Since mole = mass (g) / molar mass (g/mole), and thus mass (g) = molar mass
(g/mole) x mole = 107.9 x 3.14
14. The number of atoms of sodium (atomic mass unit 23) in 4.7 moles is determined by using which
of the following calculation?
A) 4.7 x 6.022 x 1023
B) 23 x 6.022 x 1023
C) 4.7x 23
D) None of the above
Note: p.p. 95-98: example & problem 4.5 & p. 117: exercise 4.10. Also see Q12. This is a very
important question. Apply the definition or say formula: number of atoms = moles x 6.022 x 1023
15. What is the empirical formula for methyl benzoate, a compound used in the manufacture of
perfumes, contains 70.57% carbon, 5.94% hydrogen, and 23.49% oxygen? Note: subscripts must
be integers.
A) C4H4O
B) C2H2O0.5
C) C8H8O2
D) CHO
Note: p. p. 98-101: example & problem 4.8, 4.9 & 4.10. This is a very, very important question.
The empirical formula is the simplest integral ratio of moles among each atom. Here, there are
three different kinds of atoms, C, H and O. Thus mole of C = 70.57/12 = 5.88; mole of H = 5.94/1
= 5.94; mole of O = 23.49/16 = 1.47.
Note that as long as one of the moles is not an integer, we have to divide the smallest value among
them: here the smallest value is 1.47. So C : H : O = 5.88/1.47 : 5.94/1.47 : 1.47/1.47 = 4 : 4.04 :
1. Since 4.04 is very close to 4.00 and thus we can round it to 4.00. So C : H : O = 4 : 4 : 1, which
indicates that the empirical formula contains four C, four H and one O. Thus, the empirical
formula is written as C4H4O as 1 is usually not written in the formula.
16. The molar mass (i.e. molecular weight) for methyl benzoate is 136.1 g /mol. What is its molecular
formula?
A) C4H4O
B) C2H2O0.5
C) C8H8O2
D) CHO
Note: p.p. 95-102: example & problem 4.1 & 4.2. Also see Q9 and Q10.
17. The empirical formula of ethylene glycol is CH3O. The weight of 1 mole of ethylene glycol is 62.1
g. What is its molecular formula? Note: subscripts must be integers.
A) C2H6O2
B) CHO
C) CH3O
D) C0.5H0.5O0.5
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Note: p. 101. Remember the definition that the molecular formula is an integral multiple of
empirical formula. That is, the molar mass (i.e. molecular weight) = empirical formula weight x
integer. So the empirical formula weight of CH3O = 12x1+1x3+16x1=31. So the integer = 62.1/31
= 2. Thus there are two empirical formulas in a molecular formula. Therefore, the molecular
formula is C2H6O2
18. If the equation C4H10 + O2  CO2 + H2O is balanced, which of the following quantities is correct?
Note: Coefficients must be integers.
A) 2 C4H10, 13 O2, 8 CO2, and 10 H2O
B) 13 C4H10, 2 O2, 10 CO2, and 8 H2O
C) C4H10, 6.5 O2, 4 CO2, and 5 H2O
D) C4H10, 2 O2, 5 CO2, and H2O
Note: p.p. 102-105: example & problem 4.11. This is a very, very important question. To
balance a chemical equation, you must make sure the number of atoms of each kind at both
sides of the arrow is identical. Also start examining the most bulky species, that is the one with
the most different kinds of atoms and number of atoms.
In this question, C4H10 is the most bulky one, we put 1 in front of it to remind us we have done
examining C4H10. Now the equation is updated to be 1 C4H10 + __ O2  __ CO2 + __ H2O
Since C4H10 contains 4 carbon atoms, so we need four carbon atoms at the right side, which leads
us to put 4 (called coefficient) in front of the CO2.
Now the equation is updated to be 1 C4H10 + __ O2  4 CO2 + __ H2O
As there are 10 hydrogen atoms in C4H10, thus we need to balance the hydrogen atoms, which lead
us to put 5 in front of the H2O.
Now the equation is updated to be 1 C4H10 + __ O2  4 CO2 + 5 H2O
Now we need to balance the oxygen atoms. Since there are 4x2+5x1 = 13 oxygen atoms at the
right side, and thus the left side must have the same number. That is to say, __x2 = 13. So __ = 6.5
According to the rule, all the coefficients must be whole numbers and you cannot round them, so
we must multiply 6.5 by a whole number, which must be 2 or greater, until it reaches an integer.
We find that 2 will do the job. So each coefficient must be multiply by 2 as well. That is,
2 x (1 C4H10 + 6.5 O2  4 CO2 + 5 H2O) which leads to 2 C4H10 + 13 O2  8 CO2 + 10 H2O
Thus, we have balanced this equation.
19. If the equation Pb(NO3)2 + K2CrO4  PbCrO4 + KNO3 is balanced, which of the following
quantity is correct?
A)Pb(NO3)2, K2CrO4, PbCrO4 and 2 KNO3
B) 2 Pb(NO3)2, K2CrO4, PbCrO4 and KNO3
C) ½Pb(NO3)2, ½K2CrO4, ½PbCrO4 and KNO3
D)2Pb(NO3)2, 2K2CrO4, 2PbCrO4 and KNO3
Note: p.p. 102-105: example & problem 4.11. Also see Q18.
20. Which of the following is not a correct unit-conversion factor for the reaction
Pb(NO3)2 + K2CrO4  PbCrO4 + 2KNO3 ?
A)Pb(NO3)2/K2CrO4 or Pb(NO3)2/PbCrO4
B)K2CrO4/PbCrO4 or K2CrO4/2KNO3
C)Pb(NO3)2/2KNO3 or 2Pb(NO3)2/2K2CrO4
D) 3K2CrO4/3PbCrO4 or 2Pb(NO3)2/4KNO3
E) All of the above are correct unit-conversion factors.
Note: p. p. 112-113: example & problem 4.16.
Note that if the equation is not balanced, you must balance it first. After that look at the
corresponding coefficient for each compound, usually the unit conversion factors will have the
constants same as the coefficients; if different, then they must be in an integral ratio. Thus, there
are numerous unit-conversion factors.
That is, when writing the unit-conversion factor, you must include the numerical part and the
chemical formula part. For instance, the unit-conversion factor for PbCrO4 and KNO3. Here, we
know that the coefficient in front of PbCrO4 is 1 and that in front of KNO3 is 2. Since there are
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two species, PbCrO4 and KNO3 involved, thus there are two unit-conversion factors can be
written: one is PbCrO4/2KNO3 and the other is 2KNO3/PbCrO4.
21. Which of the following is not a correct unit-conversion factor for the reaction
Zn(s) + 2 AgNO3(aq) 2 Ag(s) + Zn(NO3)2(aq)?
A) Zn/2AgNO3 or 2Ag/Zn(NO3)2
B)Zn/2Ag(s) or 2AgNO3/2 Ag(s)
C) 2AgNO3/Zn or Zn(NO3)2/2Ag
D)3Zn/6Ag(s) or AgNO3/ Ag(s)
E) None of the above is wrong unit-conversion factors.
Note: p. p. 112-113: example & problem 4.16. Also see Q 20.
22. How many lone pairs in the Lewis structure for ammonia, NH 3?
(A) one
(B) two
(C) three
(D) four
(E) five
Note: p. 78, example & problem 3.8 & 3.9. Also see Q2.
23. How many valence electrons total in CH4?
(A) 4
(B) 6
(C) 8
(D) 10
(E) 12
Note: p. 77 example 3.9. Also see Q1.
24. Which of the following formulas represents a molecular (or say covalent) compound?
(A) HCl
(B) MnBr2
(C) Al2O3
(D) CaO
(E) K2O
Note: p.p. 74-80. Also see Q3.
25. Which of the following is a non-polar covalent compound?
(A) KCl
(B) HF
(C) H2O
(D) BClF2
(E) CO2
Note: p.p. 74-80. Also see Q4.
26. Which formula represents magnesium floride?
(A) Mg F
(B) Mg2F
(C) MgF2
(D) Mg2 (PO4)3
(E) MnSO4
Note: p.p. 70-72 example & problem 3.5. Also see Q5.
27. Which of the following represents a dihydrogen phosphate ion?
(A) NH4+
(B) CO32(C) C2H3O2(D) H2PO4-
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(E) HSO4Note: p. 71, Table 3.2. Also see Q6.
28.What is the correct name for CoCl2?
(A) cobalt chloride
(B) cobalt (II) chloride
(C) carbon oxygen chloride
(D) carbon oxygen dichloride
(E) cobalt dichloride
Note: p. 74, example & problem 3.6 & Table 3.4. Also see Q7.
28. What is the correct name for S2Cl2?
(A) disulfur dichloride
(B) sulfur chloride.
(C) disulfurous chloride
(D) sulfurous dichloride
(E) All of the above
Note: p. 74.
29. How many single bonds, shown as a line, in the Lewis structure of C3H8?
(A) 7
(B) 8
(C) 9
(D) 10
(E) 11
Note: One line means one bond, that is, 2 electrons.
p.p. 76-79: Also see Q 2.
30. How many double bond(s) in the Lewis structure of the rocket fuel hydrazine, N 2H4?
A) 1
B) 2
C) 3
D) 4
(E) 0
Note: p.p. 79-80: examples 3.11 & 3.12. Also see Q 2.
Note: Two lines mean double (i.e. two) bonds, that is, 4 electrons.
31. What is the three-dimensional structure of ammonia, NH3 (p. 87), by using VSEPR theory?
(A) linear
(B) trigonal planar
(C) tetrahedral
(D) trigonal pyramidal
Note: p.p. 84-88: examples 3.14 & 3.15.
32. According to the VSEPR theory, what is the three-dimensional structure or say molecular
geometry of C2H2?
(A) linear
(B) trigonal planar
(C) tetrahedral
(D) trigonal pyramidal
Note: p.p. 84-88, examples 3.14 & 3.15. Also see Q 31.
33. How many moles of magnesium ions, Mg2+, are there in 1 mole of Mg3N2?
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Note: p. 96: example 4.5. Also see Q 12.
34. How many moles of sulfur in 57.8 g of sulfur?
(A) 1.35
(B) 1.80
(C) 3.06
Note: p. 96 example 4.3. Also see Q 11.
(D) 4.3
(E) 7.61
35. Determine the empirical formula for freon-114 which contains 69.58% Cl, 11.78% C, and 18.64%
F?
(A) CClF
(B) CCl2F
(C) CCl4F2
(D) CCl3F
(E) C2Cl3F2
Note: p.p. 98-101: examples 4.8, 4.9 & 4.10. Also see Q 15.
36. Which of the following contains the greatest number of moles of Cl?
A) 1.5 mole NaCl
B) 1.7 mole CH2Cl2
C) 2.2 mole Mg(ClO4)2
Note: p. 96: example 4.5.
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D) 3.5 mole HCl
Note that the shortcut of calculating the number of moles of Cl is that mole of Cl = given mole
number x subscript of Cl in the chemical formula. For instance, the mole of Cl in 1.5 mole NaCl is
1.5 mole x 1 = 1.5 mole; the mole of Cl in 1.7 mole CH2Cl2 is 1.7 mole x 2 = 3.4 mole; the mole
of Cl in 2.2 mole Mg(ClO4)2 = 2.2 mole x 1 x 2 = 4.4 mole.
37. If the equation Al(s) + H2SO4(aq)  Al2(SO4)3(aq) + H2(g) is balanced, which of the following
quantities is correct? Note: Coefficients must be in simplest integers.
A) 2 Al, 1 H2SO4, 1 Al2(SO4)3, and 2 H2
B) 2 Al, 3 H2SO4, 1 Al2(SO4)3, and 3 H2
C) 1 Al, 1.5 H2SO4, 0.5 Al2(SO4)3, and 1.5 H2
D) 4 Al, 6 H2SO4, 2 Al2(SO4)3, and 6 H2
Note: p.p. 102-105: example & problem 4.11. Also see Q 18.
38. Give the following equation, 3 BaCl2 + 2 Na3PO4  Ba3(PO4)2 + 6 NaCl, how many moles of
Na3PO4 will react with 0.45 mole of BaCl2?
(A) 0.3
(B) 13.3
(C) 3.5
(D) 4.6
(E) 2.4
Note: p.p. 112-115: examples 4.16, 4.17 & 4.18. From the equation, the involving species, 3 BaCl2
and 2 Na3PO4 with coefficients 3 and 2 respectively, tell us that for 3 moles of BaCl2 it requires 2
moles of Na3PO4. According to this proportion or ratio, 0.45 mole of BaCl 2 requires 0.45 x 2/3 =
0.3 mole of Na3PO4.
39. Give the following equation, 3 BaCl2 + 2 Na3PO4  Ba3(PO4)2 + 6 NaCl, how many moles of
Ba3(PO4)2 can be produced from 33.3 g of BaCl2?
(A) 1.36
(B) 0.35
(C) 0.067
(D) 0.053
(E) 2.4
Note: p. p. 112-115: examples 4.16, 4.17 & 4.18. Also see Q 38. Apply the road map or say the
dimensional analysis. The road map is grams of substance A  moles of substance A moles
of substance B  grams of substance B.
Note: (1) substance A is the one with given (or known) information of mass (or mole);
substance B is the one needed to be calculated.
(2) grams of substance A  molar mass of substance A = moles of substance A.
(3) moles of substance A  coefficient of substance B  coefficient of substance A =
moles of substance B.
(4) grams of substance B = moles of substance B  molar mass of substance B.
40. If the equation Na2CO3(aq) + CuCl2(aq)  NaCl(aq) + CuCO3(aq) is balanced, which of the
following quantities is correct? Note: Coefficients must be in simplest integers.
(A) 1 Na2CO3(aq), 1CuCl2(aq), 2NaCl(aq) and 1CuCO3(aq)
(B) 2Na2CO3(aq), 1CuCl2(aq), 2NaCl(aq) and 2CuCO3(aq)
(C) 1Na2CO3(aq), 3CuCl2(aq), 1NaCl(aq) and 3CuCO3(aq)
(D) 2Na2CO3(aq), 2CuCl2(aq), 2NaCl(aq) and 4CuCO3(aq)
(E) 0.5Na2CO3(aq), 0.5CuCl2(aq), 0.5NaCl(aq) and 1CuCO3(aq)
Note: p.p. 102-105 & 108-111: example & problem 4.14 & 4.15. Also see Q 18 & Q 37.
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