CHEM 1411
PRACTICE EXAM I (Chapters 1, 2, 3): 25 questions.
Q1-7: Chapter 1; Q8-13: Chapter 2; Q14-25: Chapter 3.
Multiple Choice: Select one best answer.
1. Which of the following is a physical change?
(a) color of carpet faded (by sun light)
(b) water evaporates
(c) zinc strip dissolves in vinegar (to produce hydrogen gas)
(d) wine turns sour
(e) scramble an egg
Hint: p.p. 14-15. Physical change is the change using the physical properties, which are the ones you
can get them back by changing the temperature or pressure. The procedure used by adapting physical
properties is called the physical process. Chemical properties are the ones you cannot get them back.
2. Which of the following is a heterogeneous mixture?
(a) benzene and hexane
(b) salt, water, and sugar (c) oil and alcohol
(d) 14-K gold ring
(e) air
Hint: p. 11. Heterogeneous mixture indicates there are two or more phases in the mixture/solution. If
the mixture is made by solid and liquid, seeing the solid indicates it’s the heterogeneous mixture. If the
mixture is made by liquid and liquid, seeing drops (by shaking it) or layer indicating it’s the
heterogeneous mixture.
3. Which of the following is inaccurate?
(a) silicon, Si
(b) mercury, Hg (c) silver, Ag
Hint: Check with periodic table and memorize them
(d) chromium, Cr
(e) iron, Ir
4. What is the equivalent temperature for 98.6 Fahrenheit in Kelvin?
(a) 37
(b) 310
(c) 471.6
(d) 594.8
(e) -273.1
Hint: p.p. 19-20. Example 1.3. There is no direct conversion between the K and oF. So in this question,
you must convert oF to oC first and then convert oC to K by applying two formulas: one is oC = (5/9) x
(oF – 32) and the other is K = 273.15 + oC.
5. An unknown sample has a mass of 13.9 g and a volume of 17.4 mL. What is its density (g/mL)?
(a) 0.798
(b) 1.04
(c) 3.16
(d) 4.62
(e) 5.07
Hint: p. 18. Example 1.1. Density = mass/volume
6. How many significant figures are there in the measurement of 6.124 x 10-5?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Hint: p.p. 23-26. Memorize the rules. Example 1.4.
7.
How many significant figures are appropriate to show in the result after carrying out the operation
below?
(223.7 + 0.27) ÷ 4.21 =?
(a) 1
(b) 2
(c) 3
(d) 4
(e) 5
Hint: p. 25. Example 1.5. You need to memorize all the rules concerning the significant figures and the
rules related to addition or subtraction and multiplication or division. Many students do not know the
differences between addition/subtraction and multiplication/division. See two examples below:
Remember that the answers obtained from calculators are always WRONG when considering the
significant figures. So when the question combining addition/subtraction and multiplication/division
together, each follows its own rule.
Thus, 223.7 + 0.27 = 223.97 from calculator, which must be corrected with the least decimal points of
the components, and thus it is 224.0. So the question turns to 224.0 ÷ 4.21 = 53.20665083 from
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calculator, which must be corrected with the lease digit of the significant figures of the components,
and thus it is 53.2.
Since this question only asks how many significant figures in the result, thus from the components,
224.0 and 4.21, we know one has four significant figures and one has three significant figures. Thus
the number of significant figures of the result should follow the one with the least significant figures,
that is, three significant figures.
8. How many protons, electrons, and neutrons are there in 72Br- ?
(a) 35p, 35e, 72n
(b) 34p, 35e, 37n
(c) 35p, 36e, 72n
(d) 36p, 34e, 72n
(e) 35p, 36e, 37n
Hint: p. 53. Positive charge indicates there are more protons than electrons; while the negative charge
indicates there are more electrons than protons. This is because neutron does not carry charge, a proton
carries one positive charge, and an electron carries one negative charge.
The ion in this question carries one negative charge indicating that the electron is one more than the
protons. From periodic table, the atomic number (that is, the proton number) of Br is 35, and thus there
are 35-(-1) = 36 electrons.
9. Which of the following pairs would have similar chemical and physical properties?
(a) Ni, Mg
(b) H, Na
(c) C, Si
(d) Cu, Ca
(e) Cr, Cl
Hint: p.p. 50-51. The elements from the same group (i.e. the vertical column) will have similar
physical and chemical properties.
10. Which of the following is a molecular compound?
(a) KCl
(b) CsF
(c) HCN
(d) AlBr3
(e) NaOH
Hint: p. 62. Molecular compound is composed by nonmetal reacts with nonmetal atoms.
11. Which of the following is the empirical formula?
(a) C8H18
(b) CH3COOH
(c) C4H10O2
(d) NH4NO3
(e) B2H6
Hint: p.p. 55-57. Example 2.3. Empirical formula is the simplest integral ratio among each atom.
12. Which of the following is not a correct match?
(a) AlCl3, aluminum chloride
(b) H2O dihydrogen monoxide
(c) CrF3, chromium (III) fluoride
(d) HNO3, nitrous acid
(e) CuSO4.5H2O, copper (II) sulfate pentahydrate
Hint: p.p. 59-68. Follow the rules of naming the compounds. Be able to differentiate the rules between
ionic and molecular compounds.
13. Which of the following pairs is a correct match?
(a) Ba3(PO4)2, barium (II) phosphate
(b) (NH4)2SO4, diammonium sulfate
(c) Na2O2, sodium oxide
(d) Ca(NO3)2, calcium nitrate
(e) CH3COOH, carbonic acid
Hint: p.p. 59-68. Follow the rules of naming the compounds. Be able to differentiate the rules between
ionic and molecular compounds.
14. The atomic masses of 10B and 11B are 10.0129 amu (natural abundance 19.78%) and 11.0093 amu
(natural abundance 80.22%), respectively. What is the average atomic mass of B?
(a) 9.467
(b) 9.966
(c) 10.042
(c) 10.504
(e) 10.810
Hint: p.p. 78-79. Example 3.1. You must convert the % into decimals first and then apply the formula
and memorize it: M = M1X1 + M2X2 = 10.0129x0.1978 + 11.0093x0.8022 = 10.810. Note that the
average atomic mass (with decimal) can be found in the periodic table, while the mass number (a
whole number) can not be found.
15. How many moles of S are there in 64.2 g of S?
(a) 1.325
(b) 1.764
(c) 1.968
(d) 2.003
(e) 2.475
Hint: p. 81. Examples 3.2 and 3.6. Mole = mass (g) / molar mass (g/mole) = 64.2/32 = 2.003
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16. What is the molar mass for calcium nitrate, Ca(NO3)2?
(a) 44
(b) 56
(c) 87
(d) 93
(e) 164
Hint: p.p. 83-84. Example 3.5. 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
17. How many hydrogen atoms are there in 48.0 g of CH4?
(a) 1.81x1023
(b) 7.22x1024
(c) 6.02x1023
(d) 1.20x1025
(e) 4.70x1025
Hint: p.p. 85. Example 3.7. According to the chemical formula, one mole of CH 4 contains 1 mole of C
atoms and 4 moles of hydrogen atoms. Thus, the mole of H = 4 x {mass of CH 4/molar mass of CH4}.
When converting moles into atoms, it needs to multiply the Avogadro’s number, that is, 6.02x10 23.
Thus the number of H atoms = moles of H atoms x Avogadro’s number = (6.02x10 23)x{4x(48.0/16.0)}
= 7.22x1024
18. What is the mass percent (%) for O in SO2?
(a) 38.09
(b) 45.41
(c) 50.00
(d) 53.86
(e) 56.43
Hint: p. 87. Equation (3.1). Example 3.8. % S = (mass of 2 O/ mass of SO2) x 100% =
{2x16/(1x32+2x16)}x100% = 50.0%
19. Chemical analysis shows the composition of a compound containing carbon, hydrogen, chlorine,
and oxygen, to be 37.84% C, 2.12% H, 55.84% Cl, and 4.20% O. What is its empirical formula?
(a) CHClO
(b) C2HClO4
(c) C12H8Cl6O
(d) C12H8Cl6O4
Hint: p. 88. Example 3.9. Figure 3.5. or p. 109 (3.50). The empirical formula is the simplest integral
ratio of moles among each atom. Here, there are four different kinds of atoms, C, H, Cl and O. Thus
mole of C = 37.84/12 = 3.15; mole of H = 2.12/1 = 2.12; mole of Cl = 55.84/35.45 = 1.58; mole of O =
4.20/16 = 0.26.
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 0.26. So C : H : Cl : O = 3.15/0.26 : 2.12/0.26 : 1.58/0.26 : 0.26/0.26 =
12.1: 8.1 : 6.1: 1 = 12: 8: 6: 1, which indicates that the empirical formula contains 12 C, 8 H, 6 Cl and
1 O. Thus, the empirical formula is written as C12H8Cl6O as 1 is usually not written in the formula.
20. Chemical analysis shows the composition of a compound containing carbon and hydrogen, to be
80.00% carbon and 20% hydrogen and the molar mass is 30 g. What is its molecular formula?
(a) CH
(b) C2H4
(c) C2H6
(d) C6H12
(e) C10H22
Hint: p.p. 91-92. Example 3.11 or p. 109 (3.54). The molecular formula is an integral multiple of
empirical formula. That is, the molar mass = empirical molar mass x integer. From C: H = 80.00/12 :
20.00/1 = 6.66: 20 = 1: 3. So the empirical formula is CH 3 and the empirical molar mass of CH3 =
12x1+1x3 =15. So the integer = 30/15 = 2. Thus there are two empirical formulas in a molecular
formula. Therefore, the molecular formula is C2H6.
21. What are the coefficients respectively when the equation
__PH3 + __O2  __P2O5 + __H2O is balanced?
(a) 2, 2, 1, 3
(b) 2, 1, 3, 4
(c) 2, 3, 1, 2
(d) 2, 4, 1, 3
(e) 1, 1, 3, 3
Hint: p.p. 94-97. Example 3.12 or p. 109 (3.60). Balancing an equation is a process of trial an error.
Sometimes it requires more than one trial. Usually start visual examination and select the most bulky
species, that is, the one with the most different kinds of atoms and number of atoms as the
reference and set its coefficient as one.
In this question, P2O5 is the most bulky one, we put 1 in front of it to remind us we have done
examining P2O5. Now the equation is updated to __PH3 + __O2  1 P2O5 + __H2O
Since P2O5 contains 2 phosphorus atoms, so we need two phosphorus atoms at the left side, which
leads us to put 2 (called coefficient) in front of the PH3.
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Now the equation is updated to 2 PH3 + __O2  1 P2O5 + __H2O
As there are 6 hydrogen atoms in 2 PH3, thus we need to balance the hydrogen atoms at the right side,
which lead us to put 3 in front of the H2O.
Now the equation is updated to 2 PH3 + __O2  1 P2O5 + 3 H2O
Now we need to balance the oxygen atoms. Since there are 1x5+3x1 = 8 oxygen atoms at the right
side, and thus the left side must have the same number of oxygen atoms. That is to say, __x2 = 8. So
__ = 4. Now the equation is updated to 2 PH3 + 4 O2  1 P2O5 + 3 H2O
Since all the atoms of each type has the same amount, this equation is balanced.
22. In the reaction of Al(OH)3 with H2SO4, how many moles of water can be produced If the reaction
is begun with 5.500 mole of Al(OH)3?
2Al(OH)3 + 3H2SO4  Al2(SO4)3 + 6H2O
(a) 2.50
(b) 4.75
(c) 6.32
(d) 7.58
(e) 16.50
Hint: p.p. 97-101. Examples 3.13. and 3.14. Figure 3.8. Stoichiometrey. Method One (ratio approach):
From the equation, the involving species, 2 Al(OH)3 and 6 H2O with coefficients 2 and 6 respectively,
tell us that for 2 moles of Al(OH)3 it produces 6 moles of H2O. According to this proportion or ratio,
5.500 mole of Al(OH)3 requires 5.500 x (6/2) = 16.50 moles of H 2O.
Method Two (road map approach): 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.
Since this question starts at moles and thus we only apply (2):
5.500 moles Al(OH)3 x {6 H2O/2 Al(OH)3} = 16.50 moles H2O
23. How many grams of H2O could be formed by the reaction of 16.0 g of CH 4 with 48.0 g of O2?
CH4 + 2O2  CO2 + 2H2O
(a) 27.0
(b) 37.3
(c) 46.8
(d) 54.1
(e) 58.7
Hint: p.p. 97-101. Examples 3.13. and 3.14. Figure 3.8. Stoichiometrey. This question provides two
known quantities of substance A and thus this is the limiting reagent question. So we need to apply
two times of all-four-step-road map to figure out what is the true mass of water. Note that the limiting
reagent limits the (maximum) of product(s) that can be produced. In the road map application, the one
produces the smallest amount of the product is the true limiting reagent. The one with the left over is
the excess reagent.
Assume CH4 (i. e. substance A # 1) is the limiting reagent:
÷ 16 g/mol
x 2H2O/1CH4
x 18 g/mol
16 g CH4 ------------- 1 mole CH4 ---------------- 2 mole H2O ------------- 36 g H2O
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Assume O2 is the limiting reagent:
÷ 32 g/mol
x 2H2O/2O2
x 18 g/mol
48 g O2 ------------- 1.5 mole O2 ---------------- 1.5 mole H2O------------- 27 g H2O
Because O2 produces the least amount of H2O (can compare by using mole, 1.5 < 2, or grams, 27 <
36), it is the true limiting reagent and the maximum amount of water produced is 27 grams. The CH 4
is the excess reagent.
24. What is the excess reagent in the above reaction (Q.23)of CH4 and O2? How many grams of the
excess reagent were consumed?
(a) CH4, 12 grams
(b) O2, 12 grams
(c) CH4, 4 grams
(d) O2, 4 grams
(e) CH4, 8 grams
Hint: p.p. 101-103. Example 3.15. Now the substance A is O2 and CH4 is the substance B because we
need to calculate its amount consumed stoichiometrically. Apply the road map:
÷ 32 g/mol
x 1CH4/2O2
x 16 g/mol
48 g O2 ------------- 1.5 mole O2 ---------------- 0.75 mole CH4------------- 12 g CH4
Thus, there are (16 – 12) = 4 grams of CH4 left.
This is because # grams excess = # grams original – # grams used.
Or say # moles excess = # moles original – # moles used.
25. A 15.6 grams of benzene (C6H6) is mixed with excess HNO3 to prepare nitrobenzene (C6H5NO2).
After the reaction there are 15.6 grams of nitrobenzene produced. What is the percent yield of
nitrobenzene?
C6H6 + HNO3  C6H5NO2 + H2O
(a) 34.3%
(b) 47.6%
(c) 58.9%
(d) 63.4%
(e) 71.2%
Hint: p.p. 103-106. Example 3.16. The amount of product calculated according to the road map is the
theoretical yield and the one obtained by weighing is the actual yield.
% yield = (actual yield/ theoretical yield) x 100%
Now from the road map, we calculate the theoretical yield (note that since the actual yield given is in
unit of grams, so we need to complete the entire road map).
÷ 78 g/mol
x 1C6H5NO2 /1C6H6
x 123g/mol
15.6 gC6H6------------ 0.2mole C6H6 ----------------------- 0.2mole C6H5NO2 --------- 24.6g C6H5NO2
Thus the percent yield = {15.6 g / 24.6 g} x 100% = 63.41%
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