CHEM 120 Ultimate Exam Review Bundle | Practice
Questions, Verified Answers & Step-by-Step
Solutions – Latest Update 2026/2027
QUESTION 1
What is the empirical formula of a compound containing 40.0% carbon,
6.7% hydrogen, and 53.3% oxygen by mass? (Molar masses: C = 12.01
g/mol, H = 1.008 g/mol, O = 16.00 g/mol)
• A. 𝐶𝐻𝑂
• B. 𝐶𝐻2 𝑂
• C. 𝐶2 𝐻4 𝑂2
• D. 𝐶𝐻3 𝑂
Correct Answer: B. 𝐶𝐻2 𝑂
Detailed Rationale: Assuming 100 g of the compound, there are 40.0 g
of C, 6.7 g of H, and 53.3 g of O. Convert to moles: moles C =
40.0/12.01 = 3.33 mol; moles H = 6.7/1.008 = 6.65 mol; moles O =
53.3/16.00 = 3.33 mol. Divide by the smallest number of moles (3.33)
to find the simplest whole-number ratio: C : H : O = 1: 2: 1, yielding the
empirical formula 𝐶𝐻2 𝑂.
QUESTION 2
A gaseous mixture consists of 2.0 moles of helium (𝐻𝑒) and 3.0 moles
of neon (𝑁𝑒) at a total pressure of 5.0 atm. What is the partial pressure
of neon?
, • A. 2.0 atm
• B. 2.5 atm
• C. 3.0 atm
• D. 5.0 atm
Correct Answer: C. 3.0 atm
Detailed Rationale: According to Dalton's law of partial pressures, the
partial pressure of a gas is equal to its mole fraction multiplied by the
total pressure. The mole fraction of neon is 3.0 mol/(2.0 mol +
3.0 mol) = 3.0/5.0 = 0.60. The partial pressure of neon = 0.60 ×
5.0 atm = 3.0 atm.
QUESTION 3
What is the standard enthalpy change (Δ𝐻 ∘ ) for the combustion of
methane: 𝐶𝐻4 (𝑔) + 2𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 2𝐻2 𝑂(𝑙), given standard
enthalpies of formation (Δ𝐻𝑓∘ ): 𝐶𝐻4 (𝑔) = −74.8 kJ/mol, 𝐶𝑂2 (𝑔) =
−393.5 kJ/mol, 𝐻2 𝑂(𝑙) = −285.8 kJ/mol?
• A. −890.3 kJ/mol
• B. −815.5 kJ/mol
• C. +890.3 kJ/mol
• D. −604.5 kJ/mol
Correct Answer: A. −890.3 kJ/mol
Detailed Rationale: Use Δ𝐻 ∘ = ∑ Δ 𝐻𝑓∘ (products) − ∑ Δ 𝐻𝑓∘ (reactants).
Δ𝐻 ∘ = [(−393.5) + 2(−285.8)] − [(−74.8) + 0] = [−393.5 −
571.6] + 74.8 = −965.1 + 74.8 = −890.3 kJ/mol.
,QUESTION 4
Which of the following sets of quantum numbers (𝑛, 𝑙, 𝑚𝑙 , 𝑚𝑠 ) is
invalid?
• A. 3,2, −1, +1/2
• B. 4,0,0, −1/2
• C. 2,2,0, +1/2
• D. 5,1, −1, +1/2
Correct Answer: C. 2,2,0, +1/2
Detailed Rationale: The angular momentum quantum number 𝑙 can
have integer values ranging from 0 to 𝑛 − 1. For 𝑛 = 2, the maximum
value of 𝑙 is 1. Therefore, an orbital with 𝑛 = 2 and 𝑙 = 2 cannot exist,
making set C invalid.
QUESTION 5
Which of the following elements has the highest first ionization energy
(𝐼𝐸1 )?
• A. 𝑁𝑎
• B. 𝑀𝑔
• C. 𝑃
• D. 𝐶𝑙
Correct Answer: D. 𝐶𝑙
Detailed Rationale: First ionization energy generally increases across a
period from left to right due to increasing effective nuclear charge (𝑍eff )
, which holds valence electrons more tightly. Chlorine (𝐶𝑙) is located
furthest to the right in Period 3 among the given choices.
QUESTION 6
What is the formal charge on the central sulfur atom in the sulfate ion
(𝑆𝑂42− ), when drawn with standard single bonds to all four oxygen
atoms?
• A. 0
• B. +1
• C. +2
• D. −2
Correct Answer: C. +2
Detailed Rationale: Formal charge = Valence electrons −
1
Non-bonding electrons − (Bonding electrons). Sulfur has 6 valence
2
electrons, zero non-bonding electrons, and 8 bonding electrons (4
single bonds). Formal charge = 6 − 0 − 4 = +2.
QUESTION 7
According to VSEPR theory, what is the molecular geometry of chlorine
trifluoride (𝐶𝑙𝐹3 )?
• A. Trigonal planar
• B. T-shaped
• C. Trigonal pyramidal
• D. See-saw
Correct Answer: B. T-shaped
Questions, Verified Answers & Step-by-Step
Solutions – Latest Update 2026/2027
QUESTION 1
What is the empirical formula of a compound containing 40.0% carbon,
6.7% hydrogen, and 53.3% oxygen by mass? (Molar masses: C = 12.01
g/mol, H = 1.008 g/mol, O = 16.00 g/mol)
• A. 𝐶𝐻𝑂
• B. 𝐶𝐻2 𝑂
• C. 𝐶2 𝐻4 𝑂2
• D. 𝐶𝐻3 𝑂
Correct Answer: B. 𝐶𝐻2 𝑂
Detailed Rationale: Assuming 100 g of the compound, there are 40.0 g
of C, 6.7 g of H, and 53.3 g of O. Convert to moles: moles C =
40.0/12.01 = 3.33 mol; moles H = 6.7/1.008 = 6.65 mol; moles O =
53.3/16.00 = 3.33 mol. Divide by the smallest number of moles (3.33)
to find the simplest whole-number ratio: C : H : O = 1: 2: 1, yielding the
empirical formula 𝐶𝐻2 𝑂.
QUESTION 2
A gaseous mixture consists of 2.0 moles of helium (𝐻𝑒) and 3.0 moles
of neon (𝑁𝑒) at a total pressure of 5.0 atm. What is the partial pressure
of neon?
, • A. 2.0 atm
• B. 2.5 atm
• C. 3.0 atm
• D. 5.0 atm
Correct Answer: C. 3.0 atm
Detailed Rationale: According to Dalton's law of partial pressures, the
partial pressure of a gas is equal to its mole fraction multiplied by the
total pressure. The mole fraction of neon is 3.0 mol/(2.0 mol +
3.0 mol) = 3.0/5.0 = 0.60. The partial pressure of neon = 0.60 ×
5.0 atm = 3.0 atm.
QUESTION 3
What is the standard enthalpy change (Δ𝐻 ∘ ) for the combustion of
methane: 𝐶𝐻4 (𝑔) + 2𝑂2 (𝑔) → 𝐶𝑂2 (𝑔) + 2𝐻2 𝑂(𝑙), given standard
enthalpies of formation (Δ𝐻𝑓∘ ): 𝐶𝐻4 (𝑔) = −74.8 kJ/mol, 𝐶𝑂2 (𝑔) =
−393.5 kJ/mol, 𝐻2 𝑂(𝑙) = −285.8 kJ/mol?
• A. −890.3 kJ/mol
• B. −815.5 kJ/mol
• C. +890.3 kJ/mol
• D. −604.5 kJ/mol
Correct Answer: A. −890.3 kJ/mol
Detailed Rationale: Use Δ𝐻 ∘ = ∑ Δ 𝐻𝑓∘ (products) − ∑ Δ 𝐻𝑓∘ (reactants).
Δ𝐻 ∘ = [(−393.5) + 2(−285.8)] − [(−74.8) + 0] = [−393.5 −
571.6] + 74.8 = −965.1 + 74.8 = −890.3 kJ/mol.
,QUESTION 4
Which of the following sets of quantum numbers (𝑛, 𝑙, 𝑚𝑙 , 𝑚𝑠 ) is
invalid?
• A. 3,2, −1, +1/2
• B. 4,0,0, −1/2
• C. 2,2,0, +1/2
• D. 5,1, −1, +1/2
Correct Answer: C. 2,2,0, +1/2
Detailed Rationale: The angular momentum quantum number 𝑙 can
have integer values ranging from 0 to 𝑛 − 1. For 𝑛 = 2, the maximum
value of 𝑙 is 1. Therefore, an orbital with 𝑛 = 2 and 𝑙 = 2 cannot exist,
making set C invalid.
QUESTION 5
Which of the following elements has the highest first ionization energy
(𝐼𝐸1 )?
• A. 𝑁𝑎
• B. 𝑀𝑔
• C. 𝑃
• D. 𝐶𝑙
Correct Answer: D. 𝐶𝑙
Detailed Rationale: First ionization energy generally increases across a
period from left to right due to increasing effective nuclear charge (𝑍eff )
, which holds valence electrons more tightly. Chlorine (𝐶𝑙) is located
furthest to the right in Period 3 among the given choices.
QUESTION 6
What is the formal charge on the central sulfur atom in the sulfate ion
(𝑆𝑂42− ), when drawn with standard single bonds to all four oxygen
atoms?
• A. 0
• B. +1
• C. +2
• D. −2
Correct Answer: C. +2
Detailed Rationale: Formal charge = Valence electrons −
1
Non-bonding electrons − (Bonding electrons). Sulfur has 6 valence
2
electrons, zero non-bonding electrons, and 8 bonding electrons (4
single bonds). Formal charge = 6 − 0 − 4 = +2.
QUESTION 7
According to VSEPR theory, what is the molecular geometry of chlorine
trifluoride (𝐶𝑙𝐹3 )?
• A. Trigonal planar
• B. T-shaped
• C. Trigonal pyramidal
• D. See-saw
Correct Answer: B. T-shaped
