Dr. Casagrande’s Chemistry 1 Midterm Exam Review Sheet
Problem Set 1: Chapters 2, 3, 11
Ch. 2: pp. 50-52 #73, 74, 75(a,b), 76(a,b), 77(a-e), 78 (a-c), 80, 82, 83(a-c), 84(a-c), 85(a-d), 87
Ch. 3: pp. 83-84 #61, 62, 63, 65, 66
Ch. 11: pp. 346-347 #72, 74, 75, 77, 78, 82, 83, 89(a,c), 90(a,c), 92(a,c), 93(a,c), 98(a,c), 99(a,c), 101(a,c)103(a,c),
108(a,c), 111(a,b), 112(a,d), 115, 116, 136, 139, 144, 146, 147
Chapter 2
73. An object with a mass of 7.5 g raises the level of water in a graduated cylinder from 25.1 mL to 30.1 mL. What
is the density of the object?
7.5 g
7.5 g
d=
=
= 1.5 g/mL
30.1 mL – 25.1 mL
5.0 g
74. The density of aluminum is 2.7 g/mL. What is the volume of 8.1 g?
8.1 g
8.1 g
1 mL
2.7 g/mL =
; rearrange: V =
= 3.0 mL or V = 8.1 g ×
= 3.0 mL
V
2.7 g/mL
2.7 g
75. Write the following numbers in scientific notation:
a. 0.004 583 4 mm 4.583×10–3 mm
b.
0.03054 g 3.054×10–2 g
76. Write the following numbers in ordinary notation:
a. 8.348×106 km 8 348 000 km
b.
3.402×103 g 3 402 g
77. Complete the following addition and subtraction problems in scientific notation:
a. 6.23×106 kL + 5.34×106 kL
= 11.57×106 kL = 1.157×107 kL
b. 3.1×104 mm + 4.87×105 mm
= 5.18×105 mm
3
2
c. 7.21×10 mg + 43.8×10 mg
= 1.159×104 mg
d. 9.15×10–4 cm + 3.48×10–4 cm
= 12.63×10–4 cm = 1.263×10–3 cm
e. 4.68×10–5 cg + 3.5×10–6 cg
= 5.03×10–5 cg
78. Complete the following multiplication and division problems in scientific notation:
a. (4.8×105 km) × (2.0×103 km)
= 9.6×108 km2
–4
–5
b. (3.33×10 m) × (3.00×10 m)
= 9.99×10–9 m2
c. (1.2×106 m) × (1.5×10–7 m)
= 1.8×10–1 m2
80. Convert the following measurements:
a.
5.70 g to milligrams
b.
4.37 cm to meters
c.
783 kg to grams
d.
45.3 mm to meters
e.
10 m to centimeters
f.
37.5 g/mL to kg/L
? mg = 5.70 g ×
1 mg
= 5.70 × 103 mg
−3
1 × 10 g
1 × 10 −2 m
= 0.0437 m
1 cm
1 × 103 g
5
? g = 783 kg ×
= 783,000 g (7.83 × 10 g)
1 kg
? m = 4.37 cm ×
1 × 10 −3 m
= 0.0453 m
1 mm
1 cm
? cm =10 m ×
= 1000 cm
1 × 10 −2 m
kg 37.5 g
1 kg
1 mL
? =
×
×
= 37.5 kg/L
3
L 1 mL 1 × 10 g 1 × 10 −3 L
? m = 45.3 mm ×
82. The accepted density for copper is 8.96 g/mL. Calculate the percent error for each of these measurements:
a. 8.86 g/mL
c. 9.00 g/mL
0.45%
8.86 g/mL − 8.96 g/mL
1.1%
× 100 =
8.96 g/mL
b. 8.92 g/mL
0.45%
d. 8.98 g/mL
0.22%
83. Round each number to four significant figures:
a. 431 801 kg
431 800 kg
c. 1.0348 m
b. 10 235.0 mg
10 240 mg
1.035 m
84. Round each figure to three significant figures:
a. 0.003 210 g
0.00321 g
c. 219 034 m
b. 3.8754 kg
3.88 kg
219 000 m
85. Round the answers to each of the following problems to the correct number of significant figures:
a. 7.31×104 + 3.23×103
7.63×104
b. 8.54×10–3 – 3.41×10–4
8.20×10–3
74.8 dm3
c. 4.35 dm × 2.34 dm × 7.35 dm
d. 4.78 cm + 3.218 cm + 5.82 cm
13.82 cm
Mass (g)
87. Graph the following data with the volume on the x-axis and the mass on the y-axis. Then calculate the slope of
the line.
Density Data
30
Volume (mL)
Mass (g)
Density of Unknown
25
2.0 mL
5.4
4.0 mL
10.8
20
6.0 mL
16.2
15
8.0 mL
21.6
y = 2.70x
10
10.0 mL
27.0
5
The slope of the line is 2.70 g/mL, which is
also the density of the sample.
0
0
2
4
6
8
10
Volume (mL)
Chapter 3
61. A 28.0-g sample of nitrogen gas combines completely with 6.0 g of hydrogen gas to form ammonia. What is
the mass of ammonia formed?
28.0 g N + 6.0 g H = 34.0 g ammonia
62. A substance breaks down into its component elements when it is heated. If 68.0 grams of the substance is
present before it is heated, what is the combined mass of the component elements after heating?
The Law of Conservation of Mass tells us it must be the same: 68.0 g
63. A 13.0-g sample of X combines with a 34.0-g sample of Y to form the compound XY2. What is the mass of the
reactants?
13.0 g X + 34.0 g Y = 47.0 g XY2 (the formula is unimportant)
65. Copper sulfide is formed when copper and sulfur are heated together. In this reaction, 127 g of copper reacts
with 41 g of sulfur. After the reaction is complete, 9g of sulfur remains unreacted. What is the mass of copper
sulfide formed?
127 g + (41 g – 9 g) = 159 g
Chem 1 Midterm Review Set 1 p. 2
66. A 25.3-g sample of an unknown compound contains 0.8 g of oxygen. What is the percent by mass of oxygen in
the compound?
(0.8 g O / 25.3 g) × 100 = 3.2% O
Chapter 11
72. How is a mole similar to a dozen?
A mole, like a dozen, contains a specific number of items.
74. What is molar mass?
Molar mass is the mass in grams of one mole of any element or compound.
75. Which contains more atoms, a mole of silver atoms or a mole of gold atoms? Explain your answer.
It’s a trick question: a mole of anything contains the same amount of representative particles.
77. Which has greater mass, a mole of silver atoms or a mole of gold atoms? Explain your answer.
A mole of Au has more mass since the molar mass of Au, 197.9 g/mol, is larger than that of Ag, 107.9 g/mol.
78. Explain the difference between atomic mass (amu) and molar mass (gram).
Atomic mass (amu) is the mass of an individual particle, while molar mass is the mass of one mole of particles.
82. Which of the following molecules contains the most moles of carbon atoms per mole of the compound:
ascorbic acid (C6H8O6), glycerin (C3H8O3), or vanillin (C8H8O3)? Explain.
The formula for vanillin (C8H8O3) shows that is contains 8 moles of carbon in one mole of the compound,
more than the other two.
83. Explain what is meant by percent composition.
Percent composition is the percent by mass of each element in a compound.
89. Determine the number of representative particles in each of the following:
6.02 × 10 23 atoms Ag
a. 0.250 mol silver ×
= 1.51×1023 atoms Ag
1 mol Ag
c. 35.3 mol carbon dioxide ×
6.02 × 10 23 molecules CO2
= 2.13×1025 molecules CO2
1 mol CO2
90. Determine the number of moles in each of the following:
1 mol Pb
= 5.39×10–4 mol Pb
a. 3.25×1020 atoms lead ×
23
6.02 × 10 atoms Pb
1 mol NaOH
= 2.59×10–1 mol NaOH
c. 1.56×1023 formula units sodium hydroxide ×
23
6.02 × 10 f.un. NaOH
92. How many molecules are contained in each of the following?
a. 1.35 mol CS2 × 6.02 × 10 molecules CS2 = 8.13×1023 molecules CS2
1 mol CS2
6.02 × 10 23 molecules H 2O
= 7.53×1023 molecules water
c. 1.25 mol water ×
1 mol H 2O
23
Chem 1 Midterm Review Set 1 p. 3
93. How many moles contain each of the following?
1 mol CO2
= 2.08×10–9 mol CO2
6.02 × 10 23 molecules CO2
1 mol CaCO3
= 4.80×103 mol CaCO3
c. 2.89×1027 formula units calcium carbonate ×
23
6.02 × 10 f.un. CaCO3
a. 1.25×1015 molecules carbon dioxide ×
98. Calculate the mass of the following:
4.00 g He
a. 5.22 mol He ×
= 20.9 g He
1 mol He
47.88 g Ti
c. 2.22 mol Ti ×
=106 g Ti
1 mol Ti
99. Make the following conversions:
9.65 g
a. 3.50 mol Li to g Li ×
= 24.3 g Li
1 mol
1 mol
c. 5.62 g Kr to Mol Kr ×
=0.0671 mol Kr
83.80 g
101. Convert the following to mass in grams.
1 mol U
238.0 g U
×
=1.67 × 10 –6 g U
a. 4.22×1015 atoms U ×
23
6.02 × 10 atoms U 1 mol U
1 mol O
16.00 g O
×
= 0.332 g O
c. 1.25×1022 atoms O ×
23
6.02 × 10 atoms O 1 mol O
103. Calculate the number of atoms in each of the following:
1 mol Hg
6.02 × 10 23 atoms Hg
a. 25.8 g Hg ×
×
= 7.74×1022 atoms Hg
200.59 g Hg
1 mol Hg
1 mol Ar 6.02 × 10 23 atoms Ar
c. 150 g Ar ×
×
= 2.3×1024 atoms Ar
39.95 g Ar
1 mol Ar
108. How many moles of oxygen atoms are contained in the following?
a. 2.50 mol KMnO4 × 4 mol O = 10.0 mol O
1 mol KMnO4
(4+5) mol O
= 0.112 mol O
c. 1.25×10–2 mol CuSO4•5H2O ×
1 mol CuSO4 • 5H 2O
111. Determine the molar mass of each of the following.
a. nitric acid (HNO3) 1.01 g + 14.01 g + 3(16.00 g) = 63.02 g HNO3/mol HNO3
b. ammonium nitrate (NH4NO3) 2(14.01 g) + 4(1.01 g) + 3(16.00 g) = 80.06 g NH4NO3/mol NH4NO3
112. Calculate the molar mass of each of the following:
a. ascorbic acid (C6H8O6) 6×12.01+8×1.01+6×16.00=176.12 g/mol
d. saccharin (C7H5NO3S) 7(12.01 g) + 5(1.01 g) + 14.01 g + 3(16.00 g) + 32.07 g = 183.2 g C7H5NO3S/mol
C7H5NO3S
Chem 1 Midterm Review Set 1 p. 4
115. What is the mass of each of the following?
a. 4.50×10–2 mol CuCl2 MM = 63.55 g + 2(35.45 g) = 135.45 g CuCl2/mol CuCl2
135.45 g CuCl 2
−2
? g CuCl 2 = 4.50 × 10 mol CuCl 2 ×
= 6.05 g CuCl2
1 mol CuCl 2
b. 1.25×102 mol Ca(OH)2 MM = 40.08 g + 2(16.00 g) + 2(1.01 g) = 74.10 g Ca(OH)2/mol Ca(OH)2
74.10 g Ca(OH)2
? g Ca(OH)2 = 1.25 × 10 2 mol Ca(OH)2 ×
= 9.26×103 g Ca(OH)2
1 mol Ca(OH)2
116. Determine the number of moles in each of the following.
a. 1.25×102 g Na2S MM = 2(22.99 g) + 32.07 g = 78.05 g Na2S/mol Na2S
1 mol Na 2S
? Na 2S = 1.25 × 10 2 g Na 2S ×
= 1.60 mol Na 2S
78.05 g Na 2S
b. 0.145 g H2S MM = 2(1.01 g) + 32.07 g = 34.09 g H2S/mol H2S
1 mol H 2S
? mol H 2S = 0.145 g H 2S ×
= 4.25 × 10 –3 mol H 2S
34.09 g H 2S
136. Express the composition of each of the following as the mass percent of its elements (percent composition).
a. sucrose (C12H22O11) MM = 12×12.01 g C + 22×1.008 g H + 11×16.00 g O = 342.30 g C12H22O11
144.12 g C
22.18 g H
× 100 = 42.10% C;
× 100 = 6.480 % H;
342.30 g C12 H 22O11
342.30 g C12 H 22O11
176.00 g O
× 100 = 51.42% O
342.30 g C12 H 22O11
b. magnetite (Fe3O4) MM = 3×55.85 g Fe + 4×16.00 g O = 231.55 g Fe3O4
167.55 g Fe
64.00 g O
× 100 = 72.36% Fe;
× 100 = 27.64% O
231.55 g Fe3O4
231.55 g Fe3O4
c. aluminum sulfate (Al2(SO4)3) MM = 2×26.98 g Al + 3×32.07 g S + 12×16.00 g O = 342.17 g
53.96 g Al
96.21 g S
× 100 = 15.77% Al;
× 100 = 28.12% S;
342.17 g Al 3 (SO4 )3
342.17 g Al 3 (SO4 )3
192.00 g O
× 100 = 56.11% O
342.17 g Al 3 (SO4 )3
139. Caffeine, a stimulant found in coffee, has the chemical formula C8H10N4O2.
a. Calculate the molar mass of caffeine. 8(12.01 g) + 10(1.01 g) + 4 (14.01 g) + 2(16.00 g) = 194.22 g
C8H10N4O2/mol C8H10N4O2
b. Determine the percent composition of caffeine.
84.08 g
10.1 g
C:
× 100 = 49.47% C H:
× 100 = 5.19% H
194.22 g
194.22 g
N:
56.04 g
32.00 g
× 100 = 28.86% NO:
× 100 = 16.48% O
194.22 g
194.22 g
Chem 1 Midterm Review Set 1 p. 5
144. Monosodium glutamate (MSG) is sometimes added to food to enhance flavor. Analysis determined this
compound to be 35.5% C, 4.77% H, 8.29% N, 13.6% Na and 37.9% O. What is the empirical formula for MSG?
Element
%
mass
actual # moles
relative # moles
1 mol C
=
C
35.5 % 35.5 g ×
2.96 mol C
÷ 0.592 =
5
12.01 g C
1 mol H
=
1.01 g H
4.72 mol H
÷ 0.592 =
7.97 ≈ 8
1 mol N
=
14.01 g N
0.592 mol N
÷ 0.592 =
1
1 mol Na
=
22.99 g Na
0.592 mol Na
÷ 0.592 =
1
1 mol O
=
16.00 g O
2.37 mol O
÷ 0.592 =
4
H
4.77 % 4.77 g
×
N
8.29 % 8.29 g
×
Na
13.6 % 13.6 g
×
O
37.9 % 37.9 g
×
Empirical Formula = C5H8NNaO4
146. Vanadium oxide is used as an industrial catalyst. The percent composition of this oxide is 56.0% vanadium and
44.0% oxygen. Determine the empirical formula for vanadium oxide.
relative
relative
Element
%
mass
actual # moles
# moles
# moles
1 mol V
=
V
56.0 % 56.0 g ×
1.10 mol V
÷ 1.10 =
1
2
×2
50.94 g V
O
44.0 % 44.0 g
×
1 mol O
=
16.00 g O
2.75 mol O
÷ 1.10 =
2.5
×2
5
Empirical Formula = V2O5
Note: the correct name for this compound is vanadium(V) oxide, as we learned in chapter 8, but that would
have given away the empirical formula, which is also the formula unit since this is an ionic compound.
147. What is the empirical formula of a compound that contains 10.52 g Ni, 4.38 g C, and 5.10 g N?
Element mass
actual # moles
relative # moles
1 mol Ni
= 0.179 mol Ni ÷ 0.179 =
Ni
10.52 g ×
1
58.69 g Ni
C
4.38 g
×
1 mol C
=
12.01 g C
0.365 mol C
÷ 0.179 =
2.04 ≈ 2
N
5.10 g
×
1 mol N
=
14.01 g N
0.364 mol N
÷ 0.179 =
2.03 ≈ 2
Note: this is actually Ni(CN)2; what is the name of this compound?
Chem 1 Midterm Review Set 1 p. 6
Empirical Formula = NiC2N2