College Chemistry Quiz: Mass Spectra Of Elements

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Mass Spectra Of ElementsQuestion 1 of 19

A sample of naturally occurring magnesium is analyzed by mass spectrometry. The spectrum shows three peaks at m/z values of 24, 25, and 26 with relative abundances of 78.99%, 10.00%, and 11.01%, respectively. What is the average atomic mass of this magnesium sample?

24.32 amu

24.99 amu

25.00 amu

25.33 amu

26.00 amu

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College Chemistry Quiz

College Chemistry Quiz: Mass Spectra Of Elements

Practice Mass Spectra Of Elements in College Chemistry 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 Mass Spectra Of Elements, giving you a quick way to practice the rules, question types, and explanations that matter most for College Chemistry.

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

24.32 amu (correct answer)
24.99 amu
25.00 amu
25.33 amu
26.00 amu

Explanation: When you encounter mass spectrometry data for isotopes, you're being asked to calculate the weighted average atomic mass. This means each isotope's mass contributes to the final average based on its relative abundance in nature. To find the average atomic mass, multiply each isotope's mass by its fractional abundance, then sum these products:

\text{Average atomic mass} = (24 \times 0.7899) + (25 \times 0.1000) + (26 \times 0.1101)

= 18.96 + 2.50 + 2.86 = 24.32 \text{ amu}

This confirms that A) 24.32 amu is correct. B) 24.99 amu would result from incorrect calculations or rounding errors. This value is too close to 25, which doesn't reflect the heavy weighting toward the lightest isotope (Mg-24 at 78.99%). C) 25.00 amu represents the simple arithmetic mean of the three masses (24 + 25 + 26) ÷ 3 = 25. This ignores the crucial fact that isotopes don't occur in equal abundances—a common trap in isotope problems. D) 25.33 amu is too high and would occur if you mistakenly weighted the heavier isotopes more heavily than the data shows, or made calculation errors favoring the higher masses. Study tip: Always convert percentages to decimals before calculating, and remember that naturally occurring elements are weighted averages—not simple arithmetic means. The most abundant isotope will pull the average closest to its mass value.

Question 2

An unknown element shows two peaks in its mass spectrum at m/z = 107 and m/z = 109 with approximately equal intensities. Based on this mass spectral data, which element is most likely being analyzed?

Silver (Ag) (correct answer)
Palladium (Pd)
Cadmium (Cd)
Rhodium (Rh)
Indium (In)

Explanation: When you encounter mass spectral data showing two peaks with nearly equal intensities, you're looking at evidence of naturally occurring isotopes. The mass-to-charge ratios (m/z values) tell you the atomic masses of these isotopes, and the relative intensities reveal their natural abundances. The data shows peaks at m/z = 107 and 109 with approximately equal intensities, indicating an element with two major isotopes differing by 2 mass units, each comprising roughly 50% of the natural abundance. Silver (Ag) fits this pattern perfectly. Silver has two naturally occurring isotopes:

^{107}Ag

(51.8% abundance) and

^{109}Ag

(48.2% abundance). These nearly equal abundances would produce mass spectral peaks of similar intensities at exactly m/z = 107 and 109. Let's examine why the other options don't work: Palladium (B) has multiple isotopes with masses around 102-110, but none show the specific 107/109 pattern with equal abundances. Cadmium (C) has several isotopes, but its most abundant ones are at masses 110-116, not 107/109. Rhodium (D) has only one naturally occurring isotope (

^{103}Rh

), so it would show a single major peak, not two. Study tip: Memorize the isotope patterns for common elements that appear frequently in mass spectrometry problems. Silver's nearly 50:50 ratio at 107:109, bromine's pattern at 79:81, and chlorine's 3:1 ratio at 35:37 are classics that appear regularly on chemistry exams.

Question 3

In the mass spectrum of bromine gas ( $Br_2$ ), three peaks are observed. If the natural abundance of $^{79}$ Br is 50.7% and $^{81}$ Br is 49.3%, which peak will have the lowest relative intensity?

The peak at m/z = 158 ( $^{79}$ Br- $^{79}$ Br)
The peak at m/z = 160 ( $^{79}$ Br- $^{81}$ Br)
The peak at m/z = 162 ( $^{81}$ Br- $^{81}$ Br) (correct answer)
All peaks have equal intensity due to similar abundances
Cannot be determined from the given information

Explanation: When you encounter mass spectrometry problems involving diatomic molecules with isotopes, you need to consider all possible combinations and calculate their probabilities based on natural abundances. For

Br_2

, three molecular ions are possible:

^{79}

Br-

^{79}

Br (m/z = 158),

^{79}

Br-

^{81}

Br (m/z = 160), and

^{81}

Br-

^{81}

Br (m/z = 162). The relative intensity of each peak corresponds to the probability of forming that particular combination. To find these probabilities, multiply the individual isotope abundances:

$^{79}$ Br- $^{79}$ Br: 0.507 × 0.507 = 0.257 (25.7%)
$^{79}$ Br- $^{81}$ Br: 2 × (0.507 × 0.493) = 0.500 (50.0%)
$^{81}$ Br- $^{81}$ Br: 0.493 × 0.493 = 0.243 (24.3%)

Note that the middle peak gets multiplied by 2 because

^{79}

Br-

^{81}

Br and

^{81}

Br-

^{79}

Br are equivalent but represent two different ways to form the same mass. The

^{81}

Br-

^{81}

Br peak at m/z = 162 has the lowest intensity at 24.3%, making C correct. Option A is incorrect because the

^{79}

Br-

^{79}

Br peak has moderate intensity (25.7%). Option B is wrong since the mixed isotope peak actually has the highest intensity (50.0%). Option D is incorrect because the abundances, while similar, are not identical, leading to different peak intensities. Remember: For diatomic molecules with two isotopes, the mixed isotope peak is always the most intense due to the factor of 2, while the heavier isotope combination is typically the least intense.

Question 4

An element has two isotopes with masses 10 amu and 11 amu. If the average atomic mass is 10.81 amu, what is the percent abundance of the heavier isotope?

19%
39%
61%
81% (correct answer)
91%

Explanation: When you encounter isotope abundance problems, you're dealing with weighted averages. The average atomic mass reflects how much of each isotope exists in nature, not just a simple mathematical average of the masses. To solve this, set up an equation using the fact that abundances must sum to 100%. Let x = percent abundance of the 11 amu isotope, so (100-x) = percent abundance of the 10 amu isotope. The weighted average formula gives us:

10.81 = \frac{x(11) + (100-x)(10)}{100}

Multiplying both sides by 100:

1081 = 11x + 1000 - 10x

1081 = x + 1000

x = 81

So the heavier isotope (11 amu) has 81% abundance. Answer A (19%) represents the abundance of the lighter isotope (100% - 81% = 19%), which is a common mistake when students calculate correctly but report the wrong isotope's abundance. Answer B (39%) might result from incorrectly setting up the equation or making arithmetic errors in the algebra. Answer C (61%) could come from misunderstanding which isotope is "heavier" or from calculation errors in the weighted average setup. Remember this key strategy: always double-check which isotope the question asks about. Many students correctly calculate the abundance but accidentally report the percentage for the wrong isotope. Also, verify your answer makes sense—since 10.81 is much closer to 11 than to 10, the heavier isotope should be more abundant.

Question 5

A researcher analyzes a sample by mass spectrometry and observes peaks at m/z = 20 and m/z = 22 with relative intensities of 90.5% and 9.5%, respectively. What is this sample most likely to be?

Neon gas with natural isotopic abundance (correct answer)
A mixture of neon-20 and neon-22 isotopes
Fluorine gas (F₂) molecules
Doubly charged magnesium ions
Water molecules with different isotopic compositions

Explanation: When you encounter mass spectrometry data showing multiple peaks, you're looking at different masses being detected. The key insight is understanding what creates these mass differences and how the relative intensities relate to natural abundance. The peaks at m/z = 20 and m/z = 22 with intensities of 90.5% and 9.5% perfectly match the natural isotopic composition of neon. Neon has two stable isotopes:

^{20}

Ne (90.48% abundance) and

^{22}

Ne (9.25% abundance). The observed data aligns with these natural percentages, confirming this is simply neon gas as it exists in nature. Let's examine why the other options don't fit. Option B suggests "a mixture" of neon isotopes, but this is misleading because natural neon gas is inherently a mixture of its isotopes—there's no distinction between "natural neon" and "a mixture of neon isotopes." Option C (fluorine gas F₂) would show peaks around m/z = 38 for F₂ molecules, not 20 and 22. Option D (doubly charged magnesium ions) would require Mg²⁺ ions with masses of 40 and 44 to appear at m/z = 20 and 22, but magnesium's isotopic distribution doesn't match the 90.5%/9.5% ratio observed. Study tip: When interpreting mass spec data, always check if the peak intensities match known natural isotopic abundances. This is often the quickest way to identify an element, especially for lighter elements where isotopic patterns are distinctive and well-documented.

Question 6

The mass spectrum of molecular hydrogen ( $H_2$ ) shows peaks at m/z = 2, 3, and 4. Which combination of hydrogen isotopes produces the peak at m/z = 3?

$^1H-^1H$ only
$^1H-^2H$ only (correct answer)
$^2H-^2H$ only
Both $^1H-^1H$ and $^1H-^2H$ combinations
Both $^1H-^2H$ and $^2H-^2H$ combinations

Explanation: When analyzing mass spectra, you need to consider how isotopes combine to form molecular ions. The m/z ratio directly corresponds to the molecular mass of the ion, so a peak at m/z = 3 means you have a molecular hydrogen ion with a total mass of 3 atomic mass units. Let's examine what mass each possible hydrogen isotope combination would produce. Protium (

^1H

) has a mass of 1 amu, while deuterium (

^2H

) has a mass of 2 amu. For

^1H-^2H

, the total molecular mass is 1 + 2 = 3 amu, which perfectly matches the m/z = 3 peak. This confirms that answer B is correct. Now let's see why the other options fail. Option A (

^1H-^1H

only) gives a molecular mass of 1 + 1 = 2 amu, which would produce the peak at m/z = 2, not 3. Option C (

^2H-^2H

only) yields 2 + 2 = 4 amu, corresponding to the m/z = 4 peak. Option D incorrectly suggests that both

^1H-^1H

and

^1H-^2H

contribute to the m/z = 3 peak, but we've shown that

^1H-^1H

produces m/z = 2, not 3. Study tip: For isotope mass spectrum problems, always add up the atomic masses of the individual isotopes to find the molecular mass. Each distinct isotope combination produces its own unique peak, and the m/z value tells you exactly what the total mass must be.

Question 7

A mass spectrum of carbon dioxide shows peaks at m/z = 44, 45, and 46. If the sample contains only $^{12}C$ , $^{13}C$ , and $^{16}O$ , which peak corresponds to $^{13}CO_2$ ?

m/z = 44 only
m/z = 45 only (correct answer)
m/z = 46 only
Both m/z = 44 and 45
Both m/z = 45 and 46

Explanation: Mass spectrometry questions test your ability to calculate molecular masses using isotopes. When you see peaks at different m/z values for the same compound, you're looking at isotopic variants with different total masses. To find which peak corresponds to

^{13}CO_2

, calculate its molecular mass. The molecule contains one

^{13}C

atom (mass = 13) and two

^{16}O

atoms (mass = 16 each). Therefore: 13 + 16 + 16 = 45. This means

^{13}CO_2

appears at m/z = 45. Let's verify by identifying the other peaks. The m/z = 44 peak corresponds to

^{12}CO_2

(12 + 16 + 16 = 44), which would be the most abundant peak since

^{12}C

is the most common carbon isotope. The m/z = 46 peak would correspond to a molecule with an additional mass unit, but since we only have

^{12}C

^{13}C

, and

^{16}O

available, this peak likely represents trace contamination or fragmentation. Answer A is incorrect because m/z = 44 corresponds to

^{12}CO_2

, not

^{13}CO_2

. Answer C is wrong because m/z = 46 would require a molecular mass of 46, which can't be achieved with the given isotopes. Answer D incorrectly suggests

^{13}CO_2

appears at two different masses, but each specific isotopic composition has only one molecular mass. Study tip: Always calculate the exact molecular mass by adding individual isotopic masses. Each isotopic variant of a molecule produces a single, specific m/z peak.

Question 8

An element shows peaks in its mass spectrum at m/z = 54, 56, 57, and 58 with relative intensities of 5.8%, 91.7%, 2.1%, and 0.3%, respectively. Based on this isotopic pattern, what is the identity of this element?

Chromium (Cr)
Iron (Fe) (correct answer)
Nickel (Ni)
Cobalt (Co)
Manganese (Mn)

Explanation: Mass spectrometry questions about isotopic patterns test your ability to match experimental data with known elemental compositions. When you see m/z values with relative intensities, you're looking at the isotopic fingerprint of an element. The key is recognizing that the most abundant peak usually corresponds to the most common isotope of the element. Here, m/z = 56 shows 91.7% relative intensity, making it the dominant isotope. Iron has

^{56}Fe

as its most abundant isotope (91.7% natural abundance), which perfectly matches this data. The other peaks at m/z = 54, 57, and 58 correspond to

^{54}Fe

(5.8%),

^{57}Fe

(2.1%), and

^{58}Fe

(0.3%), respectively, all matching iron's known isotopic distribution. Option A (Chromium) is incorrect because Cr's most abundant isotopes are

^{52}Cr

and

^{53}Cr

, which don't appear in this spectrum. Option C (Nickel) is wrong since Ni's dominant isotope is

^{58}Ni

, but here m/z = 58 shows only 0.3% intensity. Option D (Cobalt) is incorrect because Co has only one stable isotope,

^{59}Co

, so it wouldn't show this multi-peak pattern in the 54-58 range. For isotopic pattern questions, memorize the most abundant isotopes of common transition metals. Focus on both the mass number and the relative abundance percentages—both must match for correct identification. This type of question frequently appears when testing mass spectrometry principles.

Question 9

A mass spectrum of sulfur hexafluoride ( $SF_6$ ) containing natural sulfur and fluorine shows a cluster of peaks around m/z = 146-150. Which isotopic substitution would cause the largest mass shift in the molecular ion?

$^{32}S$ to $^{34}S$ substitution (correct answer)
$^{19}F$ to $^{18}F$ substitution in one position
Multiple $^{19}F$ to $^{18}F$ substitutions simultaneously
All single-atom substitutions cause equal mass shifts
Mass shifts depend on the position of substitution in the molecule

Explanation: When analyzing mass spectra, you need to consider how isotopic substitutions affect the molecular mass of the compound. The key is comparing the actual mass differences between isotopes, not just counting substitutions. For sulfur hexafluoride (

SF_6

), let's examine the mass changes from each possible isotopic substitution. The most common isotopes are

^{32}S

(mass 32),

^{34}S

(mass 34),

^{19}F

(mass 19), and

^{18}F

(mass 18). Answer A is correct because substituting

^{32}S

with

^{34}S

creates a mass shift of +2 atomic mass units (34 - 32 = 2). This is the largest single mass change possible in this molecule. Answer B is incorrect because replacing one

^{19}F

with

^{18}F

only shifts the mass by -1 unit (18 - 19 = -1). While the direction is different, the magnitude of change (1 unit) is smaller than the sulfur substitution. Answer C might seem appealing since multiple substitutions could create larger total shifts, but the question asks about "isotopic substitution" (singular) and which substitution causes the largest shift per substitution event. Answer D is wrong because we've shown that different single-atom substitutions cause different mass shifts: sulfur substitution changes mass by 2 units while fluorine substitution changes it by 1 unit. Remember: when comparing isotopic effects in mass spectrometry, focus on the absolute mass difference between isotopes rather than the number of atoms being substituted.

Question 10

In electron ionization mass spectrometry, an element X forms singly charged ions (X⁺). If the mass spectrum shows the most intense peak at m/z = 120 and this corresponds to the most abundant isotope with 32.4% natural abundance, which statement about element X is most likely correct?

Element X is monoisotopic with atomic mass 120 amu
Element X has multiple isotopes with masses near 120 amu (correct answer)
The 32.4% abundance indicates this is a minor isotope
Other isotopes of X must have masses significantly different from 120
The mass spectrum contains mostly molecular ions, not atomic ions

Explanation: When you encounter mass spectrometry questions involving natural abundance, think about what those percentages reveal about an element's isotopic composition. The key insight is that 32.4% abundance for the "most abundant" isotope is actually quite low. For element X showing its most intense peak at m/z = 120 with only 32.4% natural abundance, this tells you that no single isotope dominates the element's composition. If this is the most abundant isotope and it represents less than one-third of all atoms, the remaining ~68% must be distributed among other isotopes. For these other isotopes to appear in the same mass spectrum and contribute significantly to the overall composition, they must have masses close to 120 amu - otherwise, you'd see widely separated peaks with very different abundances. Choice A is incorrect because a monoisotopic element would show 100% abundance for its single isotope, not 32.4%. Choice C misinterprets the data - 32.4% is explicitly stated to be the most abundant isotope, not a minor one. Choice D suggests other isotopes have significantly different masses, but this would create a mass spectrum with widely separated peaks and wouldn't explain why the "most abundant" isotope has such relatively low abundance. The correct answer is B: Element X has multiple isotopes with masses near 120 amu, which explains both the moderate abundance of the most intense peak and the distribution of the remaining abundance among nearby isotopic masses. Study tip: When abundance percentages seem "low" for the most abundant isotope, expect multiple isotopes clustered close together in mass.

Question 11

The mass spectrum of methane ( $CH_4$ ) shows peaks at m/z = 16 and 17. If the natural abundance of $^{13}C$ is 1.1% and $^2H$ is 0.015%, which peak will be more intense: m/z = 16 or m/z = 17?

m/z = 16 will be much more intense due to $^{12}CH_4$ abundance (correct answer)
m/z = 17 will be more intense due to multiple isotopic combinations
Both peaks will have approximately equal intensities
m/z = 17 will be more intense due to $^{13}C$ abundance
The relative intensities cannot be predicted from the given data

Explanation: When analyzing mass spectra with isotopes, you need to calculate the relative probability of each isotopic combination contributing to different m/z peaks. The m/z = 16 peak corresponds to

^{12}CH_4

, where all atoms are the most abundant isotopes. Since

^{12}C

has 98.9% abundance and

^1H

has 99.985% abundance, this combination occurs with probability: 0.989 × (0.99985)⁴ ≈ 0.983 or 98.3%. The m/z = 17 peak results from replacing one light atom with a heavier isotope. This can happen through

^{13}CH_4

(probability: 0.011 × (0.99985)⁴ ≈ 0.011) or

^{12}CH_3D

where one hydrogen is deuterium. For the deuterium case, any of the four hydrogens could be replaced, giving probability: 4 × 0.989 × (0.99985)³ × 0.00015 ≈ 0.0006. The total probability for m/z = 17 is roughly 0.011 + 0.0006 ≈ 0.012 or 1.2%. Answer A is correct because m/z = 16 represents about 98% of molecules while m/z = 17 represents only about 1%. Answer B incorrectly suggests multiple combinations make m/z = 17 more intense—while there are multiple pathways, their combined probability is still much lower. Answer C is wrong because the intensities differ by nearly two orders of magnitude. Answer D overestimates the impact of

^{13}C

abundance alone. Study tip: In isotope pattern problems, always calculate the actual probabilities rather than just counting the number of possible combinations. Abundance percentages matter more than the number of pathways.

Question 12

A mass spectrum shows peaks for an element at m/z values of 63 and 65 with relative intensities of 69.2% and 30.8%, respectively. What is the average atomic mass of this element?

63.31 amu
63.62 amu (correct answer)
64.00 amu
64.31 amu
64.62 amu

Explanation: When you encounter mass spectrometry data showing multiple peaks for an element, you're looking at isotopes—atoms of the same element with different numbers of neutrons. To find the average atomic mass, you need to calculate the weighted average based on each isotope's mass and relative abundance. Here's how to solve this systematically: multiply each isotope's mass by its decimal abundance, then sum the results. For this element, convert the percentages to decimals: 69.2% = 0.692 and 30.8% = 0.308. Average atomic mass =

(63 \times 0.692) + (65 \times 0.308) = 43.596 + 20.02 = 63.616 \text{ amu}

Rounding to two decimal places gives 63.62 amu, confirming answer B is correct. Let's examine why the other options are wrong: Answer A (63.31 amu) would result if you accidentally swapped the abundances, calculating

(63 \times 0.308) + (65 \times 0.692)

. Answer C (64.00 amu) represents a simple arithmetic average of the two masses

\frac{63 + 65}{2}

, which ignores the different abundances entirely—a common mistake. Answer D (64.31 amu) might arise from calculation errors or mishandling the percentage conversions. Remember this key strategy: isotope problems always require weighted averages, not simple averages. The average atomic mass should be closer to the more abundant isotope's mass. Since the 63 amu isotope is more abundant (69.2%), expect an answer closer to 63 than to 65, which helps you quickly eliminate unreasonable choices.

Question 13

An element X has three isotopes with masses 28, 29, and 30 amu. If the mass spectrum shows peaks with relative intensities in the ratio 92.2:4.7:3.1, what is the identity of element X?

Silicon (Si) (correct answer)
Phosphorus (P)
Sulfur (S)
Aluminum (Al)
Magnesium (Mg)

Explanation: When you encounter a mass spectrum problem with isotope data, you need to calculate the weighted average atomic mass and match it to known elements from the periodic table. To find the average atomic mass, multiply each isotope's mass by its relative abundance (converted to decimal form), then sum the results:

\text{Average mass} = (28 \times 0.922) + (29 \times 0.047) + (30 \times 0.031)

= 25.816 + 1.363 + 0.930 = 28.109 \text{ amu}

This calculated mass of 28.109 amu matches silicon's atomic mass of 28.09 amu from the periodic table, confirming that (A) Silicon is correct. Let's examine why the other options don't work: (B) Phosphorus has an atomic mass of about 31 amu, which is significantly higher than our calculated value. (C) Sulfur has an atomic mass of about 32 amu, even further from our result. (D) Aluminum has an atomic mass of about 27 amu, which is close but still doesn't match our calculation of 28.109 amu. The key insight is that the most abundant isotope (mass 28 with 92.2% abundance) heavily influences the average, pulling it close to 28 but slightly higher due to the contributions from the heavier isotopes. Study tip: For isotope problems, always convert percentages to decimals, calculate the weighted average carefully, and remember that the average will be closest to the mass of the most abundant isotope. Keep a periodic table handy to compare your calculated mass with known atomic masses.

Question 14

In mass spectrometry, why do molecular ions of the same compound but different isotopic compositions appear at different m/z values?

Different isotopes have different ionization energies
Different isotopes are accelerated to different velocities
Different isotopes have different masses but the same charge (correct answer)
Different isotopes have different charges but the same mass
Different isotopes fragment differently in the mass spectrometer

Explanation: When you encounter mass spectrometry questions, focus on the fundamental principle: the m/z ratio (mass-to-charge ratio) determines where peaks appear in the spectrum. Different isotopes of the same element have different numbers of neutrons, giving them different atomic masses while maintaining the same number of protons and electrons. In mass spectrometry, molecular ions are typically formed by removing one electron, giving most ions a +1 charge. When a molecule contains different isotopes—like ¹²C versus ¹³C, or ¹H versus ²H—the resulting molecular ions will have identical charges but different total masses. Since the detector separates ions based on their m/z values, these isotopic variants appear as distinct peaks separated by the mass difference of the isotopes involved. Choice C correctly identifies that isotopic molecular ions have different masses but the same charge, directly explaining why they appear at different m/z values. Choice A incorrectly suggests ionization energy differences matter—while isotopes do have slightly different ionization energies, this doesn't affect where they appear in the spectrum once ionized. Choice B misunderstands the separation mechanism; in most mass spectrometers, ions are accelerated to the same kinetic energy regardless of mass. Choice D reverses the actual relationship—isotopes have the same nuclear charge and typically form ions with identical charges. Remember this key principle: in mass spectrometry, isotope patterns arise from mass differences at constant charge. Look for isotope-related peaks that are shifted by whole mass units (like +1 for ¹³C or +2 for ²H incorporation).

Question 15

A researcher observes that the mass spectrum of an organic compound shows molecular ion peaks at m/z = 78 and 79 with relative intensities of 92.1% and 7.9%, respectively. If the compound contains only C, H, and one other element X, what is the most likely identity of element X?

Nitrogen (N), which is monoisotopic (correct answer)
Oxygen (O), which is essentially monoisotopic
Fluorine (F), which is monoisotopic
Chlorine (Cl), which has two isotopes
Bromine (Br), which has two isotopes

Explanation: When you see mass spectra showing molecular ion peaks with different intensities, you're observing isotope effects. The key is analyzing the intensity ratio to determine which element's isotopic pattern matches the data. The molecular ion appears at m/z = 78 and 79 with a 92.1% to 7.9% intensity ratio, which is approximately 92:8 or about 11.6:1. This pattern indicates the presence of an element with a heavy isotope that's about 8% abundant relative to the light isotope. Nitrogen fits this pattern perfectly. ¹⁴N makes up about 99.6% of natural nitrogen, while ¹⁵N accounts for about 0.4%. However, when nitrogen is incorporated into an organic molecule, the M+1 peak (from ¹⁵N) typically shows an intensity around 7-8% relative to the molecular ion peak due to the statistical probability of having one ¹⁵N atom in the molecule. This matches your observed 7.9% intensity at m/z = 79. Choice B (oxygen) is incorrect because oxygen is essentially monoisotopic - ¹⁸O is only 0.2% abundant, which wouldn't produce a significant M+1 peak. Choice C (fluorine) is wrong since fluorine is completely monoisotopic (only ¹⁹F exists naturally), so no M+1 peak would appear. Choice D (chlorine) is incorrect because chlorine's isotope pattern shows ³⁵Cl and ³⁷Cl in roughly 3:1 ratio, which would give M+2 peaks separated by 2 mass units, not the M+1 pattern you observe. Remember: nitrogen-containing compounds characteristically show M+1 peaks around 7-8% intensity due to ¹⁵N contribution.

Question 16

A mass spectrum of chlorine gas ( $Cl_2$ ) shows peaks at m/z = 70, 72, and 74. If the natural abundance of $^{35}$ Cl is 75.8% and $^{37}$ Cl is 24.2%, what is the expected relative intensity of the peak at m/z = 72?

18.4%
36.7% (correct answer)
48.4%
57.4%
75.8%

Explanation: When you encounter mass spectrometry problems involving diatomic molecules with isotopes, you need to consider all possible isotopic combinations and their statistical probabilities. Chlorine gas (

Cl_2

) contains two chlorine atoms, each of which can be either

^{35}

Cl or

^{37}

Cl. This creates three possible molecular combinations:

^{35}

Cl-

^{35}

Cl (m/z = 70),

^{35}

Cl-

^{37}

Cl (m/z = 72), and

^{37}

Cl-

^{37}

Cl (m/z = 74). For the peak at m/z = 72, you have molecules containing one

^{35}

Cl and one

^{37}

Cl atom. Since there are two ways this can occur (

^{35}

Cl-

^{37}

Cl or

^{37}

Cl-

^{35}

Cl), you multiply the individual probabilities by 2: Relative intensity = 2 × (0.758) × (0.242) = 0.367 = 36.7% Choice A (18.4%) represents the calculation without the factor of 2, missing that there are two equivalent ways to form the mixed isotope molecule. Choice C (48.4%) appears to use an incorrect probability calculation, possibly confusing the abundance values. Choice D (57.4%) likely results from incorrectly calculating the probability of having at least one

^{37}

Cl atom. Remember that for diatomic molecules in mass spectrometry, mixed isotope peaks always have a factor of 2 in their intensity calculations because there are two equivalent ways to arrange different isotopes. Always identify all possible isotopic combinations first, then apply the multiplication principle with the correct stoichiometric factors.

Question 17

In a time-of-flight mass spectrometer, ions are accelerated through the same potential difference. Which statement correctly describes the relationship between ion mass and flight time?

Heavier ions travel faster and have shorter flight times
Lighter ions travel faster and have shorter flight times (correct answer)
All ions travel at the same speed regardless of mass
Flight time is inversely proportional to the square root of mass
Flight time is directly proportional to mass

Explanation: Time-of-flight mass spectrometry relies on a fundamental principle: when ions with different masses are accelerated through the same potential difference, they acquire the same kinetic energy but different velocities. When an ion is accelerated through a potential difference V, it gains kinetic energy equal to

qV = \frac{1}{2}mv^2

, where q is the charge, m is mass, and v is velocity. Since all ions experience the same potential difference and typically have the same charge, they all gain the same kinetic energy. Solving for velocity gives us

v = \sqrt{\frac{2qV}{m}}

. This shows that velocity is inversely proportional to the square root of mass - lighter ions move faster than heavier ones. Since the ions travel the same distance to reach the detector, and time equals distance divided by velocity, lighter ions with higher velocities will arrive first (shorter flight times), while heavier ions with lower velocities will arrive later (longer flight times). Looking at the wrong answers: A) incorrectly states that heavier ions travel faster, which contradicts our velocity equation. C) suggests all ions travel at the same speed, which would only be true if kinetic energy weren't conserved or if mass didn't matter - both false. D) states that flight time is inversely proportional to the square root of mass, but since

t = \frac{d}{v}

and

v \propto \frac{1}{\sqrt{m}}

, flight time is actually directly proportional to the square root of mass. Remember: same energy input + different masses = different speeds, with lighter particles always moving faster.

Question 18

A synthetic sample contains only two isotopes of element Z in a 3:1 molar ratio. If the lighter isotope has mass 50 amu and the heavier has mass 54 amu, what is the average atomic mass of this sample?

50.0 amu
51.0 amu (correct answer)
52.0 amu
53.0 amu
54.0 amu

Explanation: When you encounter isotope problems, you're working with weighted averages. The average atomic mass depends not just on the masses of each isotope, but on how abundant each one is in the sample. To find the weighted average, you multiply each isotope's mass by its relative abundance, then sum these products. Here, you have a 3:1 molar ratio of lighter to heavier isotope, meaning for every 4 atoms total, 3 are the lighter isotope (50 amu) and 1 is the heavier isotope (54 amu). Convert the ratio to fractions: the lighter isotope represents

\frac{3}{4} = 0.75

of the sample, while the heavier represents

\frac{1}{4} = 0.25

. Calculate the weighted average:

(0.75 × 50) + (0.25 × 54) = 37.5 + 13.5 = 51.0

amu. Choice A (50.0 amu) would be correct only if the sample contained exclusively the lighter isotope. Choice C (52.0 amu) represents the simple arithmetic mean of 50 and 54, ignoring the abundance ratio entirely—a common trap. Choice D (53.0 amu) would result from incorrectly assuming a 1:3 ratio (more heavy isotope than light), which is the reverse of what's given. Remember: average atomic mass always falls between the isotope masses, but closer to whichever isotope is more abundant. Since the lighter isotope dominates this 3:1 mixture, expect the average to be closer to 50 than to 54.

Question 19

A sample contains equal molar amounts of $^{20}Ne$ and $^{22}Ne$ . What would be the calculated average atomic mass of this artificial mixture?

20.0 amu
21.0 amu (correct answer)
22.0 amu
21.5 amu
20.18 amu

Explanation: This question tests your understanding of weighted average atomic mass, a fundamental concept in chemistry. When you encounter isotope mixtures, you need to calculate the average based on both the mass of each isotope and its relative abundance. To find the average atomic mass, you multiply each isotope's mass by its fractional abundance, then sum the results. Since this sample contains equal molar amounts of

^{20}Ne

and

^{22}Ne

, each isotope represents 50% (or 0.5) of the mixture. Average atomic mass = (mass₁ × abundance₁) + (mass₂ × abundance₂) Average atomic mass = (20.0 amu × 0.5) + (22.0 amu × 0.5) Average atomic mass = 10.0 amu + 11.0 amu = 21.0 amu This confirms that answer B (21.0 amu) is correct. Looking at the wrong answers: A (20.0 amu) represents only the lighter isotope's mass, ignoring the heavier one entirely. C (22.0 amu) makes the opposite error, considering only the heavier isotope. D (21.5 amu) might tempt you if you mistakenly think the average should fall closer to the heavier isotope, but with equal abundances, the average falls exactly halfway between the two masses. Remember this key principle: when isotopes are present in equal amounts, the average atomic mass will always be the arithmetic mean of their individual masses. For unequal abundances, the average shifts toward whichever isotope is more abundant.