Skip to content

Quiz: Atomic Structure and Mass Spectrometry

Test your understanding of subatomic particles, isotopes, atomic mass, mass spectrometry, the mole concept, and chemical formulas with these questions.

1. Which subatomic particle defines the identity of an element?

  1. Neutron
  2. Electron
  3. Proton
  4. Positron
Show Answer

The correct answer is C. The number of protons in the nucleus — the atomic number (Z) — uniquely identifies an element. No two elements share the same number of protons. Neutrons (A) determine the isotope, not the element. Electrons (B) can be gained or lost to form ions while the element's identity remains unchanged. Positrons (D) are antimatter particles not found in stable atoms.

Concept Tested: Atomic Number and Elemental Identity

2. Two atoms of the same element have different mass numbers. Which statement best explains this?

  1. They have different numbers of protons.
  2. They have different numbers of neutrons.
  3. They have different numbers of electrons.
  4. One atom has lost electrons to form an ion.
Show Answer

The correct answer is B. Isotopes are atoms of the same element with the same number of protons but different numbers of neutrons, resulting in different mass numbers (A = protons + neutrons). Different proton counts (A) would make them different elements. Electron count (C) determines charge state but not mass number. Ion formation (D) involves electron gain or loss, not a mass number change.

Concept Tested: Isotopes

3. In a mass spectrometer, what property is used to separate ions after they are accelerated by an electric field?

  1. Their mass-to-charge ratio in a magnetic field
  2. Their color as they pass through a prism
  3. Their boiling points as they are heated in a chamber
  4. Their density as they pass through a differentially permeable membrane
Show Answer

The correct answer is A. After acceleration to approximately equal kinetic energies, ions enter a magnetic field. Lighter ions (lower m/z) curve more; heavier ions (higher m/z) curve less. This separation by mass-to-charge ratio produces the mass spectrum. Color (B) is a property of photons, not ions. Mass spectrometers operate on gaseous samples — boiling points (C) are irrelevant. Ion separation by a membrane (D) is not a step in mass spectrometry.

Concept Tested: Mass Spectrometry

4. The average atomic mass of chlorine is 35.45 amu, yet no single chlorine atom has a mass of exactly 35.45 amu. Why?

  1. Chlorine atoms decay radioactively, constantly changing their mass.
  2. The periodic table rounds atomic masses to the nearest integer.
  3. Average atomic mass is a weighted average of isotope masses, calculated using their natural abundances.
  4. The mass spectrometer cannot measure chlorine accurately due to its high reactivity.
Show Answer

The correct answer is C. The value on the periodic table is a weighted average of all naturally occurring isotopes (\(^{35}\)Cl at ~75.77% and \(^{37}\)Cl at ~24.23%), not the mass of any individual atom. Common chlorine isotopes are stable, not radioactive (A). Periodic tables report the calculated weighted average, not a rounded integer (B). Mass spectrometry measures chlorine isotope masses with high precision (D).

Concept Tested: Average Atomic Mass

5. A mass spectrum of neon shows three peaks at m/z = 20, 21, and 22. The tallest peak is at m/z = 20. What does the height of a peak represent?

  1. The atomic number of that isotope
  2. The exact mass in grams of one mole of that isotope
  3. The number of neutrons in that isotope
  4. The relative abundance of that isotope compared to the most abundant species
Show Answer

The correct answer is D. In a mass spectrum, peak height (or area) represents the relative abundance of each isotope — how common it is compared to the most abundant species, which is assigned 100%. The tallest peak at m/z = 20 indicates that \(^{20}\)Ne comprises about 90.5% of natural neon. Peak height does not represent atomic number (A), molar mass in grams (B), or neutron count (C).

Concept Tested: Mass Spectrum Analysis and Relative Abundance

6. How many moles are in 54.0 g of water (\(\ce{H2O}\), molar mass = 18.015 g/mol)?

  1. 3.00 mol
  2. 0.333 mol
  3. 54.0 mol
  4. 18.0 mol
Show Answer

The correct answer is A. Using \(n = m/M\): \(n = 54.0 \text{ g} \div 18.015 \text{ g/mol} \approx 3.00 \text{ mol}\). Dividing 18.015 by 54.0 (the inverse ratio) gives 0.333 mol (B). Treating grams as moles without using molar mass gives 54.0 mol (C). 18.0 mol (D) results from multiplying mass by molar mass instead of dividing.

Concept Tested: Mole Calculations

7. Avogadro's number (\(N_A = 6.022 \times 10^{23}\) mol\(^{-1}\)) is central to the mole concept. Which statement correctly describes its significance?

  1. It equals the number of neutrons in exactly one gram of hydrogen.
  2. It represents the number of molecules that fit in one liter of any gas at STP.
  3. It is the number of atoms in one mole of hydrogen gas (\(\ce{H2}\)).
  4. It links the atomic mass scale (amu) to the macroscopic mass scale (grams per mole).
Show Answer

The correct answer is D. Avogadro's number bridges the atomic and laboratory scales: the molar mass in grams per mole is numerically equal to the atomic or molecular mass in amu, with \(6.022 \times 10^{23}\) particles per mole as the conversion factor. One gram of hydrogen contains approximately \(5.97 \times 10^{23}\) hydrogen atoms, not Avogadro's number of neutrons (A). One mole of gas at STP occupies 22.4 L, not 1 L (B). One mole of \(\ce{H2}\) gas contains \(6.022 \times 10^{23}\) molecules but \(1.204 \times 10^{24}\) atoms (C).

Concept Tested: Avogadro's Number and the Mole Concept

8. A compound is found to contain 40.00% C, 6.71% H, and 53.29% O by mass. What is the empirical formula?

  1. \(\ce{C6H12O6}\)
  2. \(\ce{CHO}\)
  3. \(\ce{CH2O}\)
  4. \(\ce{C2H4O2}\)
Show Answer

The correct answer is C. Assuming 100 g gives 40.00 g C, 6.71 g H, and 53.29 g O. Converting to moles: C = 3.330 mol, H = 6.657 mol, O = 3.331 mol. Dividing by the smallest (3.330) yields C:H:O = 1:2:1, giving \(\ce{CH2O}\). \(\ce{C6H12O6}\) (A) is a molecular formula corresponding to n = 6 (glucose). \(\ce{CHO}\) (B) would require equal moles of all three elements, which the data do not support. \(\ce{C2H4O2}\) (D) corresponds to n = 2, a valid molecular formula but not the simplest ratio.

Concept Tested: Empirical Formula from Percent Composition

9. A compound with empirical formula \(\ce{CH2O}\) has a measured molar mass of 60.05 g/mol. What is its molecular formula?

  1. \(\ce{CH2O}\)
  2. \(\ce{C2H4O2}\)
  3. \(\ce{C3H6O3}\)
  4. \(\ce{C6H12O6}\)
Show Answer

The correct answer is B. The molar mass of the empirical formula \(\ce{CH2O}\) is \(12.011 + 2(1.008) + 15.999 = 30.026\) g/mol. The multiplier is \(n = 60.05 \div 30.026 \approx 2\), so the molecular formula is \(2 \times \ce{CH2O} = \ce{C2H4O2}\). \(\ce{CH2O}\) (A) corresponds to n = 1 (~30 g/mol). \(\ce{C3H6O3}\) (C) corresponds to n = 3 (~90 g/mol). \(\ce{C6H12O6}\) (D) is glucose, with n = 6 (~180 g/mol).

Concept Tested: Molecular Formula from Empirical Formula and Molar Mass

10. In combustion analysis, a 0.200 g organic sample (C, H, and O only) produces 0.293 g \(\ce{CO2}\) and 0.120 g \(\ce{H2O}\). What is the correct method for finding the mass of oxygen in the original sample?

  1. Multiply the mass of \(\ce{CO2}\) by the molar mass ratio of oxygen to carbon dioxide.
  2. Add the masses of \(\ce{CO2}\) and \(\ce{H2O}\) and subtract from the sample mass.
  3. Divide the mass of \(\ce{H2O}\) by Avogadro's number to convert to moles of oxygen.
  4. Subtract the masses of C and H (calculated from the combustion products) from the original sample mass.
Show Answer

The correct answer is D. Oxygen mass is found by difference: calculate the mass of C from the \(\ce{CO2}\) produced (\(m_C = m_{\ce{CO2}} \times 12.011/44.009\)), calculate the mass of H from the \(\ce{H2O}\) produced (\(m_H = m_{\ce{H2O}} \times 2.016/18.015\)), then subtract both from the original sample mass: \(m_O = m_{sample} - m_C - m_H\). Multiplying \(\ce{CO2}\) mass by the O/CO₂ ratio (A) would give the oxygen inside the \(\ce{CO2}\), not the oxygen originally in the compound. Adding both combustion product masses and subtracting (B) gives grams of CO₂ and H₂O lost, not elemental oxygen. Dividing by Avogadro's number (C) converts particles to moles, which is an unrelated operation here.

Concept Tested: Combustion Analysis