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INSTRUMENTAL ANALYSIS ACS EXAM LATEST
VERSION 2026-2027QUESTIONS WITH CORRECT
SOLUTIONS ALL WITH DETAILED RATIONALES JUST
RELEASED THIS YEAR
Summarized Exam Coverage
Instrumental Analysis ACS exam covers principles, instrumentation, and applications of
spectroscopic methods (UV-Vis, IR, fluorescence, Raman, NMR, atomic absorption/emission),
separation techniques (GC, HPLC, electrophoresis), electrochemical methods (potentiometry,
voltammetry, coulometry), mass spectrometry, thermal analysis, surface characterization, and
data handling including calibration curves, signal-to-noise, and statistical validation of analytical
results.
250 Randomized Multiple-Choice Questions
1. A pharmaceutical lab measures the absorbance of a paracetamol solution at 243 nm and
obtains 0.520 in a 1 cm cuvette. If the molar absorptivity is 8200 M⁻¹cm⁻¹, what is the
concentration in M?
A) 6.34 × 10⁻⁵
B) 1.58 × 10⁻⁴
C) 4.26 × 10⁻⁵
D) 8.90 × 10⁻⁴
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Answer: A
*Using Beer's law A = εbc, c = A/(εb) = 0.520/(8200 × 1) = 6.34 × 10⁻⁵ M.*
2. A researcher using HPLC observes that a non-polar analyte elutes very quickly with a
90% water mobile phase. To increase retention time, what change should be made?
A) Increase flow rate
B) Decrease column temperature
C) Increase organic modifier percentage
D) Decrease organic modifier percentage
Answer: D
In reversed-phase HPLC, decreasing organic solvent (water-rich) increases retention of
non-polar analytes by strengthening hydrophobic interaction with stationary phase.
3. In cyclic voltammetry, a redox couple shows ΔEp of 120 mV at scan rate 50 mV/s and 200
mV at 500 mV/s. This indicates:
A) Nernstian reversible system
B) Quasi-reversible system
C) Irreversible system
D) Adsorption-controlled process
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Answer: B
*Reversible systems have ΔEp ~59 mV independent of scan rate; larger ΔEp increasing
with scan rate indicates quasi-reversible electron transfer kinetics.*
4. A GC-MS analysis of a pesticide shows a molecular ion peak at m/z 304 and a base peak
at m/z 125. The base peak likely results from:
A) Loss of a small neutral fragment
B) Complete fragmentation of the molecule
C) Electron ionization producing intact radical cation
D) Rearrangement to a stable ion
Answer: A
*Base peak is most abundant fragment; loss of a neutral fragment (e.g., C₆H₁₁O₃) from
molecular ion gives stable ion at m/z 125.*
5. An analyst uses external calibration for Pb determination by flame AAS. A blank gives
0.004 absorbance, and three standards at 1.0, 2.0, 5.0 ppm give 0.102, 0.201, 0.498
respectively. The sample absorbance is 0.315. What is the concentration (ppm)?
A) 2.50
B) 3.12
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C) 3.15
D) 3.20
Answer: C
*Calibration slope = (0.498-0.102)/(5-1)=0.099, intercept ≈0.003, c = (0.315-0.003)/0.099
= 3.15 ppm.*
6. In FTIR spectroscopy, what is the primary advantage of using a Michelson interferometer
over a dispersive grating?
A) Higher wavelength resolution
B) Fellgett’s advantage (multiplex)
C) Elimination of water vapor interference
D) Lower cost of components
Answer: B
Fellgett's advantage allows simultaneous detection of all frequencies, improving signal-
to-noise ratio for a given measurement time.
7. A student analyzes caffeine in coffee by HPLC-UV and obtains peak area of 12500 for
sample and 15200 for a 50 mg/L standard. If sample was diluted 1:10 before injection,
what is original concentration (mg/L)?
4
SUCCESS!
INSTRUMENTAL ANALYSIS ACS EXAM LATEST
VERSION 2026-2027QUESTIONS WITH CORRECT
SOLUTIONS ALL WITH DETAILED RATIONALES JUST
RELEASED THIS YEAR
Summarized Exam Coverage
Instrumental Analysis ACS exam covers principles, instrumentation, and applications of
spectroscopic methods (UV-Vis, IR, fluorescence, Raman, NMR, atomic absorption/emission),
separation techniques (GC, HPLC, electrophoresis), electrochemical methods (potentiometry,
voltammetry, coulometry), mass spectrometry, thermal analysis, surface characterization, and
data handling including calibration curves, signal-to-noise, and statistical validation of analytical
results.
250 Randomized Multiple-Choice Questions
1. A pharmaceutical lab measures the absorbance of a paracetamol solution at 243 nm and
obtains 0.520 in a 1 cm cuvette. If the molar absorptivity is 8200 M⁻¹cm⁻¹, what is the
concentration in M?
A) 6.34 × 10⁻⁵
B) 1.58 × 10⁻⁴
C) 4.26 × 10⁻⁵
D) 8.90 × 10⁻⁴
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,Page 2 of 143
Answer: A
*Using Beer's law A = εbc, c = A/(εb) = 0.520/(8200 × 1) = 6.34 × 10⁻⁵ M.*
2. A researcher using HPLC observes that a non-polar analyte elutes very quickly with a
90% water mobile phase. To increase retention time, what change should be made?
A) Increase flow rate
B) Decrease column temperature
C) Increase organic modifier percentage
D) Decrease organic modifier percentage
Answer: D
In reversed-phase HPLC, decreasing organic solvent (water-rich) increases retention of
non-polar analytes by strengthening hydrophobic interaction with stationary phase.
3. In cyclic voltammetry, a redox couple shows ΔEp of 120 mV at scan rate 50 mV/s and 200
mV at 500 mV/s. This indicates:
A) Nernstian reversible system
B) Quasi-reversible system
C) Irreversible system
D) Adsorption-controlled process
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Answer: B
*Reversible systems have ΔEp ~59 mV independent of scan rate; larger ΔEp increasing
with scan rate indicates quasi-reversible electron transfer kinetics.*
4. A GC-MS analysis of a pesticide shows a molecular ion peak at m/z 304 and a base peak
at m/z 125. The base peak likely results from:
A) Loss of a small neutral fragment
B) Complete fragmentation of the molecule
C) Electron ionization producing intact radical cation
D) Rearrangement to a stable ion
Answer: A
*Base peak is most abundant fragment; loss of a neutral fragment (e.g., C₆H₁₁O₃) from
molecular ion gives stable ion at m/z 125.*
5. An analyst uses external calibration for Pb determination by flame AAS. A blank gives
0.004 absorbance, and three standards at 1.0, 2.0, 5.0 ppm give 0.102, 0.201, 0.498
respectively. The sample absorbance is 0.315. What is the concentration (ppm)?
A) 2.50
B) 3.12
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C) 3.15
D) 3.20
Answer: C
*Calibration slope = (0.498-0.102)/(5-1)=0.099, intercept ≈0.003, c = (0.315-0.003)/0.099
= 3.15 ppm.*
6. In FTIR spectroscopy, what is the primary advantage of using a Michelson interferometer
over a dispersive grating?
A) Higher wavelength resolution
B) Fellgett’s advantage (multiplex)
C) Elimination of water vapor interference
D) Lower cost of components
Answer: B
Fellgett's advantage allows simultaneous detection of all frequencies, improving signal-
to-noise ratio for a given measurement time.
7. A student analyzes caffeine in coffee by HPLC-UV and obtains peak area of 12500 for
sample and 15200 for a 50 mg/L standard. If sample was diluted 1:10 before injection,
what is original concentration (mg/L)?
4
SUCCESS!
