2 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 2
The mass spectrum of a sample of neon, Ne, shows three peaks, at masses of 20 amu, 21 amu and 22 amu. Which statement about the atoms in the sample is correct?
Answer and reasoning
AAtoms with 20, 21 and 22 protons per nucleus are present. A student who reads the mass of an isotope as its number of protons picks this. The mass in amu is close to the mass number, protons plus neutrons; an atom with 20 protons would be calcium, not neon.
BNe atoms with 10, 11 and 12 electrons are present. A student who thinks isotopes differ in their numbers of electrons picks this. Neutral atoms of every Ne isotope have 10 electrons; the isotopes differ in neutrons, which changes the mass.
CAtoms with 10, 11 and 12 protons in their nuclei are present. A student who thinks isotopes differ in their numbers of protons picks this. Atoms with 11 or 12 protons would be sodium or magnesium; all isotopes of neon have 10 protons and differ in their numbers of neutrons.
DNe atoms with 10, 11 and 12 neutrons are present.Correct Every Ne atom has 10 protons (atomic number 10). The three peaks are isotopes with mass numbers 20, 21 and 22, so the atoms have 20 − 10 = 10, 11 and 12 neutrons.
The mass spectrum of a sample of a hypothetical element X is shown. Based on the spectrum, which statement about the average atomic mass of X in the sample is correct?
Answer and reasoning
AIt is between 50 amu and 51 amu.Correct 70% of the atoms have a mass of 50 amu and 30% have a mass of 52 amu, so the weighted average is 0.70 × 50 + 0.30 × 52 = 50.6 amu. The average lies closer to the mass of the more abundant isotope, between 50 amu and 51 amu.
BIt is between 51 and 52 amu. A student who thinks the heavier isotope counts for more in a weighted average picks this. Each isotope counts in proportion to its abundance; the lighter isotope is the more abundant (70%), so the average, 50.6 amu, lies closer to 50 amu.
CIt is exactly 51 amu, the midpoint. A student who takes the simple mean of the two isotope masses picks this. 51 amu would be the average only if the isotopes were equally abundant; here 70% of the atoms are the lighter isotope, so the average is 50.6 amu.
DIt is 50 amu, the top peak's mass. A student who takes the mass of the most abundant isotope as the average atomic mass picks this. The 30% of atoms with a mass of 52 amu raise the average above 50 amu, to 50.6 amu.
Working Weighted average = 0.70 × 50 amu + 0.30 × 52 amu = 35.0 + 15.6 = 50.6 amu, which lies between 50 amu and 51 amu.
In preparation: 0 of 2 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
1.2.A.1 Isotopes Fix
Isotopes
Atoms of the same element (the same number of protons) that have different numbers of neutrons and therefore different masses. Isotopes of an element have the same number of electrons in the neutral atom.
Mass number
The total number of protons and neutrons in the nucleus of an atom, written as a superscript before the symbol (²⁰Ne). The mass of an isotope in amu is close to its mass number but, except for ¹²C (exactly 12 amu by definition), is not exactly equal to it.
Mass spectrum of an element
A graph produced by a mass spectrometer showing relative abundance on the vertical axis against mass on the horizontal axis. For a sample of a single element whose atoms are detected as singly charged ions, each peak corresponds to one isotope: its position gives the isotope's mass and its height gives the isotope's relative abundance.
Relative abundance
The proportion of the atoms of an element that are a particular isotope. It may be given as a percentage of all the atoms (the percentages add to 100) or relative to the most abundant isotope, which is then set to 100.
Students often think The relative abundances shown on any mass spectrum are percentages of all the atoms, so they can be divided by 100 to give fractions even when the tallest peak is set to 100. In fact No. Heights given relative to the tallest peak add to more than 100. The fraction of atoms that are each isotope is its height divided by the sum of all the heights.
Students often think The mass of an isotope in amu (its peak position in a mass spectrum) equals the number of protons in its atoms. In fact No. The mass of an isotope in amu is approximately its mass number, the total number of protons and neutrons. The number of protons is the atomic number, which is the same for every isotope of the element.
1.2.A.2 Average atomic mass Fix
Average atomic mass
The weighted average of the masses of an element's isotopes, each mass multiplied by the fraction of atoms that are that isotope: average mass = Σ(fraction × isotopic mass). It is the atomic mass printed on the periodic table for a typical natural sample.
Weighted average
An average in which each value counts in proportion to how often it occurs. For an element with two isotopes, the average atomic mass lies closer to the mass of the more abundant isotope.
Students often think The average atomic mass of an element is the simple mean of its isotope masses, found by adding the masses and dividing by the number of isotopes. In fact No. It is a weighted average: each isotope's mass counts in proportion to the fraction of atoms that are that isotope. The simple mean equals the average atomic mass only if all the isotopes are equally abundant.
Students often think The average atomic mass of an element is the mass of its most abundant isotope, because the tallest peak represents the element. In fact No. The most abundant isotope pulls the average toward its mass, but every isotope contributes in proportion to its abundance, so the average lies between the masses of the lightest and heaviest isotopes. For an element with two isotopes it is not equal to the mass of either one.
5 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 5
The mass spectrum of a sample of a hypothetical element Z is shown, with the height of each peak given relative to the tallest peak. What is the average atomic mass of Z in the sample, in amu?
Answer and reasoning
A28.0 A student who takes the mass of the most abundant isotope as the average atomic mass picks this. The small peaks at 29 amu and 30 amu raise the average slightly, to 28.1 amu.
B28.1Correct The heights are relative to the tallest peak, so the fraction of atoms that are each isotope is its height divided by the total, 112. Average = (28 × 100 + 29 × 8.0 + 30 × 4.0)/112 = 3152/112 = 28.1 amu.
C29.0 A student who takes the simple mean of the three isotope masses picks this. The isotope of mass 28 amu makes up most of the atoms (100 of every 112), so the average is much closer to 28 amu.
D31.5 A student who treats heights given relative to the tallest peak as percentages picks this, dividing 3152 by 100. The heights add to 112, not 100; dividing by 112 gives 28.1 amu, which lies between the isotope masses, as an average must.
Working Sum of heights = 100 + 8.0 + 4.0 = 112. Fractions: 100/112, 8.0/112, 4.0/112. Average = (28 × 100 + 29 × 8.0 + 30 × 4.0)/112 = 3152/112 = 28.1 amu.
Boron has two stable isotopes, ¹⁰B (10.01 amu) and ¹¹B (11.01 amu). A sample of boron is prepared in which 90.0% of the atoms are ¹¹B. Which value is closest to the average atomic mass of boron in this sample, in amu?
Answer and reasoning
A10.5 A student who takes the simple mean of the two isotope masses picks this. The isotopes are not equally abundant: 90.0% of the atoms are ¹¹B, so the average is much closer to 11.01 amu.
B10.8 A student who thinks the average atomic mass is fixed for an element picks this, the periodic-table value for natural boron. That value applies to natural boron (about 80% ¹¹B); this sample, with 90.0% ¹¹B, has a greater average atomic mass.
C10.9Correct The weighted average for this sample is 0.900 × 11.01 + 0.100 × 10.01 = 10.91 amu. It is greater than the periodic-table value for boron, 10.81, because this sample contains a larger fraction of the heavier isotope than natural boron.
D11.0 A student who takes the mass of the most abundant isotope as the average atomic mass picks this. The 10.0% of atoms that are ¹⁰B lower the average to 10.9 amu.
A geochemist wants to find out whether the lithium extracted from a mineral contains the isotopes ⁶Li and ⁷Li in the same relative abundances as ordinary lithium. Which procedure is best aligned with this aim?
Answer and reasoning
ARecord a mass spectrum of each sample and compare the relative heights of the two peaksCorrect A mass spectrum separates the atoms by mass: the ⁶Li and ⁷Li peaks give the relative abundance of each isotope, so comparing the peak heights for the two samples answers the question directly.
BDo a flame test on each lithium sample and compare the colors of the two flames produced A student who thinks isotopes have noticeably different chemical properties picks this. Both isotopes have three electrons arranged in the same way, so both samples give the same red flame whatever their isotopic composition.
CFind the number of protons in the atoms of each lithium sample and compare the two numbers A student who thinks isotopes differ in their numbers of protons picks this. Every lithium atom has three protons, so the comparison cannot show any difference in isotopic composition.
DWeigh a piece of each lithium sample and compare the masses shown on the balance A student who thinks the mass of a sample shows the masses of its atoms picks this. The mass of a piece depends on how much lithium it contains, so the two masses say nothing about which isotopes are present.
The diagram represents ten atoms in a sample of copper, Cu, which has atomic number 29. Which statement best describes the mass spectrum of a sample of copper with this composition?
Answer and reasoning
ATwo peaks, at 34 amu and 36 amu, with heights in the ratio 7 : 3 A student who reads the number of neutrons as the mass of the isotope picks this. The 29 protons in each nucleus also contribute about 1 amu each, so the peaks are at about 63 amu and 65 amu.
BOne peak, at 29 amu, because each atom in the box has 29 protons A student who thinks the position of a peak is the number of protons picks this. The mass of an atom in amu is close to its number of protons plus neutrons, which is 63 for seven of the atoms and 65 for the other three.
CTwo peaks, at 63 amu and 65 amu, with heights in the ratio 7 : 3Correct Each atom has 29 protons, so the atoms with 34 and 36 neutrons have masses of about 63 amu and 65 amu and give two peaks. The height of a peak shows the relative abundance of its isotope: seven atoms to three.
DOne peak, at 63.6 amu, that is the average mass of the ten atoms A student who thinks every atom has the average atomic mass picks this. 63.6 amu is the weighted average for these ten atoms, but none of them has that mass; the spectrum shows a peak for each isotope, at 63 amu and at 65 amu.
Working Mass of each atom ≈ protons + neutrons: 29 + 34 = 63 amu and 29 + 36 = 65 amu, so there are two peaks. Seven of the ten atoms have a mass of 63 amu and three have a mass of 65 amu, so the peak heights are in the ratio 7 : 3.
A laboratory prepares a sample of magnesium in which the proportions of the three isotopes of magnesium differ from those in natural magnesium. Which information is needed to calculate the average atomic mass of magnesium in this sample?
Answer and reasoning
AThe masses of all three isotopes and the number of isotopes A student who takes the average atomic mass to be the simple mean of the isotope masses picks this, planning to add the three masses and divide by three. That gives the average only if the three isotopes are equally abundant; each mass must be weighted by its abundance in the sample.
BThe atomic mass given for magnesium on the periodic table A student who thinks the average atomic mass is fixed for an element picks this. The printed value, 24.30 amu, is the weighted average for the natural mixture of isotopes; a sample with different proportions has a different average.
CThe mass of the isotope giving the tallest mass-spectrum peak A student who takes the mass of the most abundant isotope as the average atomic mass picks this. The other two isotopes also contribute to the average, in proportion to their abundances in the sample.
DThe mass and the relative abundance of each isotope in the sampleCorrect The average atomic mass is the weighted average of the isotopic masses: each isotope's mass is multiplied by its fractional abundance in the sample and the products are added. Both the masses and the abundances in this sample are needed.
Working Average atomic mass = Σ (fractional abundance of isotope in the sample × mass of isotope). The calculation needs the mass of each of the three isotopes and the relative abundance of each in this sample; the periodic-table value, 24.30 amu, applies to natural magnesium.
Compiled from the AP Chemistry Course and Exam Description (effective Fall 2024) and our question bank · Specialist review in progress. How these pages are made · Free, no account
