MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4e Atomic Structure Isotopes

20 questions · exam conditions

0:00

4e Atomic Structure IsotopesQuestion 1 of 20

In a metabolic tracing study, a researcher combusts purified glucose isolated from cells fed a mixture of $^{12}\text{CO}_2$ and $^{13}\text{CO}_2$ . The CO $_2$ produced from combustion is analyzed by mass spectrometry, yielding two dominant peaks at 44.0 amu (assigned to $^{12}\text{C}^{16}\text{O}_2$ ) and 45.0 amu (assigned to $^{13}\text{C}^{16}\text{O}_2$ ). If the 45.0 amu peak has 25% of the intensity of the 44.0 amu peak, which statement about the carbon isotopes in the original glucose is most consistent with these results? (Use: $m(^{12}\text{C})=12.000$ amu, $m(^{13}\text{C})=13.003$ amu, $m(^{16}\text{O})=15.995$ amu.)

Approximately 20% of the carbon atoms in the glucose are

^{13}\text{C}

, assuming similar ionization/detection for the two isotopologues.

Approximately 25% of the carbon atoms in the glucose are

^{13}\text{C}

because peak intensity equals isotope fraction directly.

The glucose must contain 25% more total carbon atoms when labeled with

^{13}\text{C}

due to the higher atomic mass.

The glucose contains the same fraction of

^{13}\text{C}

^{12}\text{C}

because isotopes are chemically indistinguishable.

← Back to quizzes

MCAT Chemical and Physical Foundations of Biological Systems Quiz

MCAT Chemical and Physical Foundations of Biological Systems Quiz: 4e Atomic Structure Isotopes

Practice 4e Atomic Structure Isotopes in MCAT Chemical and Physical Foundations of Biological Systems with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on 4e Atomic Structure Isotopes, giving you a quick way to practice the rules, question types, and explanations that matter most for MCAT Chemical and Physical Foundations of Biological Systems.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Approximately 20% of the carbon atoms in the glucose are $^{13}\text{C}$ , assuming similar ionization/detection for the two isotopologues. (correct answer)
Approximately 25% of the carbon atoms in the glucose are $^{13}\text{C}$ because peak intensity equals isotope fraction directly.
The glucose must contain 25% more total carbon atoms when labeled with $^{13}\text{C}$ due to the higher atomic mass.
The glucose contains the same fraction of $^{13}\text{C}$ as $^{12}\text{C}$ because isotopes are chemically indistinguishable.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on isotope abundance calculations from mass spectrometry data. Isotopes are variants of a chemical element that differ in neutron number, and their relative abundances can be determined from mass spectrometry peak intensities. In this scenario, glucose combustion produces CO₂ molecules containing either ¹²C or ¹³C, with the 45.0 amu peak (¹³CO₂) showing 25% of the intensity of the 44.0 amu peak (¹²CO₂). Choice A is correct because when each CO₂ molecule contains one carbon atom, a 25% relative intensity means that for every 100 carbon atoms, approximately 20 are ¹³C (20/(20+80) = 0.20 or 20%). Choice B is incorrect because it confuses relative intensity (25% of the ¹²C peak) with absolute fraction (which would be 20%, not 25%). To avoid similar errors, remember that relative peak intensity must be converted to fractional abundance using the formula: fraction = relative intensity/(1 + relative intensity).

Question 2

In a tracer study of glucose metabolism, a researcher compares two isotopes: $^{14}\mathrm{C}$ (6 protons, 8 neutrons) and $^{14}\mathrm{N}$ (7 protons, 7 neutrons). Which statement about these two nuclides is most consistent with atomic-structure principles?

They are isotopes of the same element because they have the same mass number.
They are isomers because they have the same number of neutrons.
They are isobars because they have the same mass number but different atomic numbers. (correct answer)
They must have identical chemical reactivity because their mass numbers are equal.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on the classification of nuclides with identical mass numbers. Isobars are nuclides that have the same mass number (A = protons + neutrons) but different atomic numbers (different numbers of protons), making them different elements entirely. In this scenario, both ¹⁴C (6 protons + 8 neutrons = 14) and ¹⁴N (7 protons + 7 neutrons = 14) have mass number 14 but different atomic numbers. Choice C is correct because it accurately identifies these nuclides as isobars with the same mass number but different atomic numbers. Choice A is incorrect because isotopes must have the same number of protons, not just the same mass number. To avoid confusion, remember that isotopes share the same element (same protons), while isobars share the same mass number but are different elements.

Question 3

A sample contains only two isotopes of lithium: $^{6}\text{Li}$ (6.02 amu) and $^{7}\text{Li}$ (7.02 amu). The measured average atomic mass is 6.94 amu. Which statement about isotopic abundance is most consistent with this result?

$^{6}\text{Li}$ is more abundant than $^{7}\text{Li}$
$^{7}\text{Li}$ is more abundant than $^{6}\text{Li}$ (correct answer)
Both isotopes must be present in equal abundance
Average atomic mass depends only on proton number, so abundance cannot be inferred

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on abundance inference from average mass. Isotopes are variants of a chemical element that differ in neutron number, and average mass reflects their proportional contributions. In this scenario, lithium's average is 6.94 amu, with isotopes at 6.02 and 7.02 amu. Choice B is correct because the average is closer to 7.02 amu, suggesting higher abundance of ^{7}Li. Choice C is incorrect because equal abundance would average to 6.52 amu, not 6.94 amu. To avoid similar errors, perform weighted average calculations for precision. Always base inferences on data rather than assumptions of equality.

Question 4

In an isotopic labeling experiment, glucose is synthesized using water enriched in deuterium ( $^2\text{H}$ ). Compared with protium ( $^1\text{H}$ ), deuterium differs primarily in which subatomic particle count?

One additional neutron (correct answer)
One additional proton
One additional electron in a neutral atom
One additional positron in the nucleus

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on subatomic differences between isotopes. Isotopes are variants of a chemical element that differ in neutron number, affecting mass but not charge or electron count in neutral atoms. In this scenario, deuterium (^2H) is compared to protium (^1H) in a labeling experiment. Choice A is correct because deuterium has one more neutron, increasing its mass by approximately 1 amu. Choice C is incorrect because neutral atoms of hydrogen isotopes both have one electron. To avoid similar errors, focus on nuclear composition for isotopic differences. Always recall that electrons are determined by atomic number, not mass.

Question 5

A lab reports that an unknown element X has two isotopes, $^{62}\text{X}$ and $^{64}\text{X}$ , and the average atomic mass is 63.60 amu. Which isotope is most abundant?

$^{62}\text{X}$ , because it is lighter and thus contributes more strongly to the average
$^{63}\text{X}$ , because average atomic mass must match an existing isotope
$^{64}\text{X}$ , because the average is closer to 64 than to 62 (correct answer)
Both are equally abundant because the average is between them

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on abundance from average mass. Isotopes are variants of a chemical element that differ in neutron number, influencing the weighted average mass. In this scenario, element X has average mass 63.60 amu with isotopes at 62 and 64. Choice C is correct because 63.60 is closer to 64, indicating higher abundance of ^{64}X. Choice D is incorrect because equal abundance would average to 63, not 63.60. To avoid similar errors, use the formula for weighted averages to estimate ratios. Always verify by calculating the deviation from the midpoint.

Question 6

A chemist calculates an element's average atomic mass as 10.81 amu from a natural sample. The element has two stable isotopes: $^{10}\text{B}$ (10.01 amu) and $^{11}\text{B}$ (11.01 amu). Based on the average mass, which isotope is most abundant?

$^{10}\text{B}$ , because it has the smaller mass and thus dominates the average
$^{11}\text{B}$ , because the average is closer to 11.01 amu (correct answer)
They must be 50/50 because the average lies between 10 and 11
Neither; average atomic mass equals the mass number of the most stable isotope

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on determining abundance from average atomic mass. Isotopes are variants of a chemical element that differ in neutron number, and average mass is weighted by their natural abundances. In this scenario, boron's average mass is 10.81 amu, with isotopes at 10.01 and 11.01 amu. Choice B is correct because the average is closer to 11.01 amu, indicating higher abundance of ^{11}B. Choice C is incorrect because a 50/50 ratio would yield an average of 10.51 amu, not 10.81 amu. To avoid similar errors, use the weighted average formula to quantify abundances. Always compare the average to isotopic masses to infer dominance.

Question 7

In a protein turnover experiment, cells are switched from a medium containing mostly $^{14}\text{N}$ to one containing mostly $^{15}\text{N}$ . Over time, newly synthesized proteins show increased mass while maintaining the same amino acid sequence and charge states. Which statement about the isotope labeling is most consistent with these observations?

The proteins are isomers formed by rearranging bonds, which changes mass without changing composition.
The mass increases because $^{15}\text{N}$ has one additional neutron per nitrogen atom incorporated, while chemical identity is largely preserved. (correct answer)
The mass increases because $^{15}\text{N}$ has one additional electron per nitrogen atom, increasing molecular weight.
The mass must remain unchanged because isotopes cannot be incorporated into biomolecules during synthesis.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on isotope incorporation in biological molecules. Isotopes are variants of a chemical element that differ in neutron number, and when heavier isotopes are incorporated into biomolecules, they increase molecular mass while preserving chemical identity. In this scenario, cells switch from ¹⁴N to ¹⁵N media, resulting in newly synthesized proteins with increased mass but unchanged sequence and charge. Choice B is correct because ¹⁵N has one additional neutron compared to ¹⁴N (both have 7 protons, but ¹⁵N has 8 neutrons while ¹⁴N has 7), increasing mass by ~1 amu per nitrogen while maintaining identical chemical properties. Choice C is incorrect because isotopes have the same number of electrons, not different numbers. To avoid similar errors, remember that isotopes differ only in neutron number, which affects mass but not chemical behavior.

Question 8

A sample of elemental chlorine from a physiological saline preparation shows two mass spectrometry peaks corresponding to $^{35}\text{Cl}$ and $^{37}\text{Cl}$ . The average atomic mass of chlorine in the sample is measured as 35.45 amu. Using $m(^{35}\text{Cl})=34.97$ amu and $m(^{37}\text{Cl})=36.97$ amu, which isotopic abundance of $^{37}\text{Cl}$ is most consistent with the measurement?

Approximately 22.5% $^{37}\text{Cl}$ (correct answer)
Approximately 45% $^{37}\text{Cl}$
Approximately 77.5% $^{37}\text{Cl}$
Approximately 97% $^{37}\text{Cl}$

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on calculating isotope abundance from average atomic mass. Isotopes are variants of a chemical element that differ in neutron number, and the average atomic mass reflects the weighted average based on natural abundances. In this scenario, chlorine's average atomic mass of 35.45 amu results from the mixture of ³⁵Cl and ³⁷Cl isotopes. Choice A is correct because using the formula: average mass = (fraction of ³⁵Cl × 34.97) + (fraction of ³⁷Cl × 36.97), and letting x = fraction of ³⁷Cl, we get 35.45 = (1-x)(34.97) + x(36.97), which solves to x ≈ 0.24 or 24%, closest to 22.5%. Choice B is incorrect because it represents roughly double the actual abundance, likely from a calculation error. To avoid similar errors, set up the weighted average equation carefully and solve for the unknown fraction systematically.

Question 9

A researcher compares two nuclides used in biochemical experiments: $^{2}\text{H}$ (deuterium) and $^{3}\text{H}$ (tritium). Hydrogen has $Z=1$ . Which statement about their mass difference is most consistent with atomic structure? (Use: $m(^{2}\text{H})=2.014$ amu, $m(^{3}\text{H})=3.016$ amu.)

Their mass difference is primarily due to different numbers of neutrons, not different numbers of electrons. (correct answer)
Their mass difference is primarily due to different numbers of protons, since isotopes differ in proton count.
Their mass difference is primarily due to different electron binding energies, which dominate atomic mass.
Their mass difference is 0.002 amu because isotopes of the same element must have nearly identical masses.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on the source of mass differences between isotopes. Isotopes are variants of a chemical element that differ in neutron number, and this neutron difference accounts for nearly all mass variation between isotopes. In this scenario, deuterium (²H) and tritium (³H) differ by exactly one neutron: ²H has 1 proton and 1 neutron, while ³H has 1 proton and 2 neutrons. Choice A is correct because the mass difference of 1.002 amu (3.016 - 2.014) is primarily due to the additional neutron in tritium, as neutron mass is approximately 1 amu. Choice B is incorrect because isotopes of the same element must have the same number of protons by definition. To avoid similar errors, remember that isotopes differ only in neutron number, and this accounts for their mass differences.

Question 10

In a metabolic tracing study, a researcher compares two carbon isotopes: $^{12}\text{C}$ (12.00 amu) and $^{13}\text{C}$ (13.00 amu). Which statement about these isotopes is most consistent with atomic structure?

They have the same number of protons but different numbers of neutrons (correct answer)
They have different numbers of protons but the same number of neutrons
They have different electron configurations in neutral atoms because mass differs
They are structural isomers of carbon that differ in bonding arrangement

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on the definition and differences between isotopes. Isotopes are variants of a chemical element that differ in neutron number, while sharing the same number of protons and thus the same atomic number. In this scenario, ^{12}C and ^{13}C are compared in a metabolic study, highlighting their nuclear differences. Choice A is correct because it accurately states they have the same protons but different neutrons, consistent with isotopic properties. Choice D is incorrect because it confuses isotopes with structural isomers, which differ in atomic connectivity rather than nuclear composition. To avoid similar errors, remember that isotopes are defined by neutron variance within the same element. Always distinguish between nuclear and molecular structural concepts in atomic analyses.

Question 11

A mass spectrometer shows two peaks for bromine at 78.92 amu and 80.92 amu with nearly equal intensities. Assume these correspond to $^{79}\text{Br}$ and $^{81}\text{Br}$ , respectively. Which statement about the sample is most consistent with the experimental results?

The sample contains mostly $^{79}\text{Br}$ because its peak is at lower m/z
The isotopes are present in roughly a 1:1 ratio (correct answer)
The peaks represent two different elements with the same atomic number
Equal peak height implies the isotopes have identical masses

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on interpreting mass spectrometry data for abundance. Isotopes are variants of a chemical element that differ in neutron number, and mass spectrometry peaks reflect their abundances through intensity. In this scenario, bromine shows two peaks of equal intensity at 78.92 and 80.92 amu. Choice B is correct because equal peak heights indicate approximately equal abundances of ^{79}Br and ^{81}Br. Choice D is incorrect because equal heights do not imply identical masses, as peaks are at different m/z values. To avoid similar errors, interpret peak intensity as a direct measure of relative abundance. Always confirm that peaks correspond to isotopes of the same element, not different elements.

Question 12

A sample of boron from two suppliers is tested for its average atomic mass. Supplier X reports $10.81\,\mathrm{amu}$ , while Supplier Y reports $10.20\,\mathrm{amu}$ . Given $m(^{10}\mathrm{B})=10.013\,\mathrm{amu}$ and $m(^{11}\mathrm{B})=11.009\,\mathrm{amu}$ , which statement is most consistent with the difference?

Supplier Y's boron has a higher fraction of $^{11}\mathrm{B}$ than Supplier X.
Supplier Y's boron has a higher fraction of $^{10}\mathrm{B}$ than Supplier X. (correct answer)
Supplier X's boron must be a different element because its average mass is not an integer.
Both suppliers must have identical isotopic composition because isotopes have identical chemistry.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on how isotopic composition affects average atomic mass. A lower average atomic mass indicates a higher proportion of the lighter isotope, while a higher average mass indicates more of the heavier isotope. In this scenario, Supplier Y's boron (10.20 amu) is closer to the mass of ¹⁰B (10.013 amu) than Supplier X's (10.81 amu), indicating Y has more ¹⁰B. Choice B is correct because it accurately states that Supplier Y's boron has a higher fraction of the lighter ¹⁰B isotope. Choice A is incorrect because it reverses the relationship - lower average mass means more light isotope, not more heavy isotope. To interpret average atomic masses, remember that values closer to a specific isotope's mass indicate higher abundance of that isotope.

Question 13

A clinical lab measures the average atomic mass of chlorine in a saline sample using mass spectrometry and reports 35.45 amu. Assume chlorine exists only as $^{35}\text{Cl}$ (34.97 amu) and $^{37}\text{Cl}$ (36.97 amu). Based on the data, which isotope is most abundant in the sample?

$^{37}\text{Cl}$ , because the average mass is closer to 37 amu
$^{35}\text{Cl}$ , because the average mass is closer to 35 amu (correct answer)
$^{35}\text{Cl}$ and $^{37}\text{Cl}$ are equally abundant, because the average is between them
Neither isotope is more abundant; isotopes differ only in electron number

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on isotopic abundance determination from average atomic mass. Isotopes are variants of a chemical element that differ in neutron number, and the average atomic mass is a weighted average reflecting their relative abundances. In this scenario, the average mass of chlorine is given as 35.45 amu, with isotopes at 34.97 amu and 36.97 amu. Choice B is correct because the average is closer to 35 amu, indicating higher abundance of ^{35}Cl to pull the average downward. Choice A is incorrect because it misinterprets the proximity, assuming closeness to 37 amu despite the data showing otherwise. To avoid similar errors, ensure isotopic abundances are inferred by comparing the average mass to the isotopic masses quantitatively. Always verify calculations for weighted averages to confirm abundance interpretations.

Question 14

A hospital uses potassium chloride (KCl) for IV solutions. Consider two potassium isotopes: $^{39}\text{K}$ (38.96 amu) and $^{41}\text{K}$ (40.96 amu). Which isotope would you expect to have the greatest mass difference from $^{39}\text{K}$ ?

$^{40}\text{K}$ , because it differs by 1 neutron from $^{39}\text{K}$
$^{41}\text{K}$ , because it differs by 2 amu from $^{39}\text{K}$ (correct answer)
$^{39}\text{K}^+$ , because losing an electron changes mass by 1 amu
$^{39}\text{K}$ in a different compound, because chemical environment changes atomic mass

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on mass differences between isotopes. Isotopes are variants of a chemical element that differ in neutron number, leading to variations in atomic mass while maintaining the same atomic number. In this scenario, potassium isotopes ^{39}K and ^{41}K are considered, with masses differing by approximately 2 amu. Choice B is correct because ^{41}K shows a 2 amu difference from ^{39}K, which is greater than potential differences from other options. Choice C is incorrect because electron mass is negligible and does not significantly alter atomic mass. To avoid similar errors, calculate mass differences based on given isotopic masses directly. Always consider that chemical environment does not change atomic mass.

Question 15

A lab uses $^{18}\mathrm{O}$ -labeled water to track oxygen incorporation into a peptide during synthesis. Two oxygen isotopes are considered: $^{16}\mathrm{O}$ and $^{18}\mathrm{O}$ . Which statement is most consistent with their behavior in chemical bonding during the experiment?

$^{18}\mathrm{O}$ forms stronger covalent bonds than $^{16}\mathrm{O}$ because it has more neutrons.
$^{16}\mathrm{O}$ and $^{18}\mathrm{O}$ have the same number of protons and electrons, so their bonding preferences are very similar. (correct answer)
$^{18}\mathrm{O}$ has a different atomic number than $^{16}\mathrm{O}$ , so it forms different functional groups.
$^{16}\mathrm{O}$ cannot be incorporated into peptides because it is the lighter isotope.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on how isotopes behave in chemical bonding. Isotopes of the same element have identical numbers of protons and electrons (in neutral atoms), which determines their electronic structure and thus their chemical bonding behavior. In this scenario, ¹⁶O and ¹⁸O both have 8 protons and 8 electrons, differing only in neutron number (8 vs 10). Choice B is correct because it accurately states that these oxygen isotopes have the same number of protons and electrons, resulting in very similar bonding preferences. Choice A is incorrect because neutron number has minimal effect on bond strength, which is determined by electron interactions. To verify isotopic behavior, remember that chemical properties depend on electron configuration, which is identical for all isotopes of an element.

Question 16

A mass spectrometry method distinguishes ions that differ only by isotopic composition. Consider singly charged chloride ions containing either $^{35}\mathrm{Cl}$ or $^{37}\mathrm{Cl}$ (atomic masses: 34.969 amu and 36.966 amu, respectively). Which observation is most consistent with isotopic mass differences in the spectrum?

The $^{37}\mathrm{Cl}^-$ peak appears at a higher $m/z$ than the $^{35}\mathrm{Cl}^-$ peak by about 2 units. (correct answer)
The two peaks have identical $m/z$ because isotopes have identical chemistry.
The $^{35}\mathrm{Cl}^-$ peak appears at higher $m/z$ because lighter isotopes deflect less.
The peaks differ in $m/z$ by about 37 units because the atomic number changes.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on mass spectrometry detection of isotopes. Mass spectrometry separates ions based on their mass-to-charge ratio (m/z), and isotopes of the same element have different masses due to different neutron numbers. In this scenario, ³⁵Cl⁻ and ³⁷Cl⁻ ions differ in mass by approximately 2 amu (36.966 - 34.969 ≈ 2), and since both carry the same -1 charge, their m/z values differ by about 2 units. Choice A is correct because it accurately describes that the heavier ³⁷Cl⁻ isotope appears at a higher m/z value by about 2 units. Choice B is incorrect because isotopes have different masses despite identical chemistry. To interpret mass spectra correctly, remember that isotope peaks are separated by their mass difference when charge states are identical.

Question 17

A simplified stability screen flags nuclides with unusually high neutron-to-proton ratios ( $N/Z$ ) as more likely to undergo beta decay to move toward a more stable ratio. Two nuclides are compared: $^{3}\mathrm{H}$ (1 proton, 2 neutrons) and $^{3}\mathrm{He}$ (2 protons, 1 neutron). Which statement is most consistent with this screening principle?

$^{3}\mathrm{He}$ is more likely to beta decay because it has the larger $N/Z$ .
$^{3}\mathrm{H}$ is more likely to beta decay because it has the larger $N/Z$ . (correct answer)
Both are equally likely to beta decay because they have the same mass number.
Neither can undergo beta decay because isotopes of the same mass number are always stable.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on neutron-to-proton ratios and nuclear stability. Beta decay typically occurs when a nucleus has too many neutrons relative to protons, converting a neutron to a proton to achieve a more stable ratio. In this scenario, ³H has N/Z = 2/1 = 2.0, while ³He has N/Z = 1/2 = 0.5, making tritium's ratio much higher. Choice B is correct because it accurately identifies ³H (tritium) as having the larger N/Z ratio and thus being more likely to undergo beta decay. Choice A is incorrect because it miscalculates the N/Z ratios - helium-3 actually has the smaller ratio. To assess beta decay likelihood, calculate N/Z ratios and remember that unusually high values indicate neutron excess and potential beta decay.

Question 18

In a PET imaging protocol, a radiochemist considers using either $^{18}\mathrm{F}$ (9 protons, 9 neutrons) or stable $^{19}\mathrm{F}$ (9 protons, 10 neutrons) in the same molecular scaffold. Which statement is most consistent with how these two fluorine nuclides differ at the atomic level?

They have different atomic numbers, so they are different elements.
They have the same number of protons but different numbers of neutrons, so they are isotopes of fluorine. (correct answer)
They have the same number of neutrons, so they are isotopes of fluorine.
They differ in electron count in neutral atoms, so they must have different bonding valences.

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on the definition of isotopes. Isotopes are variants of the same element that have identical numbers of protons but different numbers of neutrons, resulting in different mass numbers. In this scenario, both ¹⁸F and ¹⁹F have 9 protons (making them both fluorine), but ¹⁸F has 9 neutrons while ¹⁹F has 10 neutrons. Choice B is correct because it accurately identifies these nuclides as isotopes of fluorine with the same proton number but different neutron numbers. Choice C is incorrect because isotopes differ in neutron number, not have the same neutron number. To identify isotopes correctly, verify that the atomic number (proton count) is identical while the mass number differs.

Question 19

A lab measures an average atomic mass of 107.87 amu for silver in a reagent bottle. Assume silver exists only as $^{107}\text{Ag}$ (106.91 amu) and $^{109}\text{Ag}$ (108.91 amu). Which statement is most consistent with this average?

$^{109}\text{Ag}$ is slightly more abundant than $^{107}\text{Ag}$
$^{107}\text{Ag}$ is slightly more abundant than $^{109}\text{Ag}$ (correct answer)
Only $^{108}\text{Ag}$ is present because the average is near 108 amu
The isotopes must be 1:1 because 107.87 is between 106.91 and 108.91

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on abundance from average atomic mass. Isotopes are variants of a chemical element that differ in neutron number, influencing the calculated average. In this scenario, silver's average is 107.87 amu with isotopes at 106.91 and 108.91 amu. Choice B is correct because 107.87 is slightly closer to 106.91, indicating more ^{107}Ag. Choice D is incorrect because a 1:1 ratio would average to 107.91 amu, not 107.87 amu. To avoid similar errors, find the midpoint and assess deviation. Always perform precise calculations for small differences in averages.

Question 20

A mass spectrometry report for magnesium shows three peaks corresponding to $^{24}\text{Mg}$ , $^{25}\text{Mg}$ , and $^{26}\text{Mg}$ . If the $^{24}\text{Mg}$ peak is largest, which conclusion is most consistent with the results?

$^{26}\text{Mg}$ is most abundant because it has the largest mass number
$^{24}\text{Mg}$ is most abundant because it produces the highest-intensity peak (correct answer)
All isotopes are equally abundant because magnesium has atomic number 12
Peak height reflects electron count, not isotope abundance

Explanation: This question tests understanding of atomic structure and isotopic behavior, focusing on mass spectrometry for isotopic abundance. Isotopes are variants of a chemical element that differ in neutron number, with peak sizes indicating relative amounts. In this scenario, magnesium shows peaks for ^{24}Mg, ^{25}Mg, and ^{26}Mg, with ^{24}Mg largest. Choice B is correct because the highest peak intensity reflects greatest abundance of ^{24}Mg. Choice A is incorrect because abundance is not determined by mass number size. To avoid similar errors, associate peak height directly with abundance. Always remember that atomic number is shared among isotopes, not influencing abundance.