Chapter 02 Review Problems

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Fundamental Laws

An unused photographic flashbulb contains magnesium and oxygen. After use, the contents are changed to magnesium oxide but the total mass does not change. Which of the following best explains this observation?

1. Law of Constant Composition
2. Law of Multiple Proportions
3. Avogadro’s Law
4. Law of Conservation of Mass

Solution

Answer: D

Concept: fundamental chemical laws

The Law of Conservation of Mass states that in a closed system, matter is neither created nor destroyed during a chemical reaction. The total mass of the reactants must equal the total mass of the products. The flashbulb is a closed system, and the observation that its mass does not change after the reaction is a direct demonstration of this law.

The other laws describe different principles:

• The Law of Constant Composition deals with the fixed mass ratio of elements within a single compound (e.g., water is always 11.2 % hydrogen and 88.8 % oxygen by mass).
• The Law of Multiple Proportions deals with the ratio of masses of elements when they combine to form different compounds (e.g., the ratio of oxygen in H₂O vs. H₂O₂).
• Avogadro’s Law relates the volume of a gas to the number of moles.

Famous Experiments

What quantity did J.J. Thomson determine using a cathode ray tube?

1. the mass of an electron
2. the charge-to-mass ratio of an electron
3. the density of a proton
4. the charge of a neutron
5. the charge-to-mass ratio of a proton

Solution

Answer: B

Concept: famous experiments

In his experiments with cathode ray tubes, J.J. Thomson was able to deflect a beam of electrons using both electric and magnetic fields. By carefully measuring the degree of deflection caused by fields of a known strength, he was able to calculate the charge-to-mass ratio (e/m) of the particles in the beam.

He could not determine the charge (e) or the mass (m) individually, but only the ratio between them. It was Robert Millikan’s later oil drop experiment that determined the charge of a single electron. Once the charge was known, Thomson’s ratio could be used to calculate the electron’s mass.

Periodic Table

Write the group number for each of the following elements.

1. Ba
2. Sb
3. Zr
4. Tl
5. Rn

Solution

Answer:

1. group 2 (formerly 2A)
2. group 15 (formerly 5A)
3. group 4 (formerly 4B)
4. group 13 (formerly 3A)
5. group 18 (formerly 8A)

Concept: periodic table

The group number is determined by locating the element’s column on the periodic table. The IUPAC standard uses a 1–18 numbering system for the groups.

Write the period number for each of the following elements.

1. Hf
2. Am
3. Ga
4. Si
5. He

Solution

Answer:

1. 6
2. 7
3. 4
4. 3
5. 1

Concept: periodic table

The period number is determined by locating the element’s row on the periodic table. It corresponds to the highest principal energy level (n) occupied by electrons in an atom of that element.

Which of the following elements is a metal?

1. Ge
2. Sb
3. H
4. Sn
5. Cl

Solution

Answer: D

Concept: periodic table

Elements on the periodic table are broadly classified as metals, nonmetals, or metalloids based on their properties and position.

• Metals are found on the left side and in the center of the periodic table.
• Nonmetals are found on the upper right side.
• Metalloids are found along the “staircase line” that separates the metals and nonmetals.

Analyzing the options:

• (d) Sn (Tin) is located in Group 14, to the left of the staircase line. It is a metal.

The other options are incorrect:

• (a) Ge (Germanium) and (b) Sb (Antimony) are located directly on the staircase line and are classified as metalloids.
• (c) H (Hydrogen), though located in the upper left corner in Group 1, is a unique element that behaves as a nonmetal under standard conditions.
• (e) Cl (Chlorine) is located in the upper right portion of the periodic table and is a nonmetal.

Which of the following elements is a nonmetal?

1. Ca
2. Si
3. Mo
4. Cs
5. Se

Solution

Answer: E

Concept: periodic table

Elements on the periodic table are broadly classified as metals, nonmetals, or metalloids based on their properties and position. Nonmetals are generally found on the upper right side of the periodic table.

Analyzing the options:

• (e) Se (Selenium) is located in Group 16, to the right of the staircase line. It is a nonmetal.

The other options are incorrect:

• (a) Ca (Calcium), (c) Mo (Molybdenum), and (d) Cs (Cesium) are all located on the left side or in the center of the periodic table and are classified as metals.
• (b) Si (Silicon) is located directly on the staircase line that separates metals and nonmetals and is classified as a metalloid.

Atoms and Ions and Ionic Compounds

Which of the following is found in the nucleus?

1. cations
2. neutrons
3. electrons
4. anions
5. none of these

Solution

Answer: B

Concept: atomic structure

An atom is composed of two main regions: the nucleus and the electron cloud.

• The nucleus is the small, dense, positively charged center of the atom. It contains the protons and the neutrons.
• The electron cloud is the region of space surrounding the nucleus where the electrons are found.

Therefore, (b) neutrons are found in the nucleus.

The other options are incorrect: * (c) Electrons are found outside the nucleus. * (a) Cations and (d) Anions are not subatomic particles; they are entire atoms or molecules that have gained or lost electrons to acquire a net charge.

Which of these is NOT a constituent particle found in an atom?

1. electron
2. neutron
3. photon
4. proton
5. all of them are

Solution

Answer: C

Concept: atomic structure

An atom is defined by its three primary constituent subatomic particles:

• Protons and neutrons, which are found in the nucleus.
• Electrons, which are found in the space surrounding the nucleus.

A (C) photon is a fundamental particle, but it is a quantum (or particle) of light. Atoms interact with photons by absorbing or emitting them when electrons change energy levels, but photons are not a permanent, structural component of an atom’s mass or structure. Therefore, a photon is not “found in” an atom in the same way that the other particles are.

True or False: an atom consists mostly of empty space.

1. true
2. false, an atom consists primarily of electrons
3. false, an atom consists primarily of the nucleus
4. false, an atom consists primarily of neutrons
5. false, an atom consists primarily of protons

Solution

Answer: A

Concept: atomic structure

The statement that an atom consists mostly of empty space is true. This was the revolutionary discovery of Ernest Rutherford’s gold foil experiment.

• The Model: An atom consists of a tiny, incredibly dense, positively charged nucleus (containing protons and neutrons) that accounts for nearly all of the atom’s mass. The vast majority of the atom’s volume is the electron cloud, which is the region through which the very small electrons move.
• The Evidence: When Rutherford shot alpha particles at a thin sheet of gold foil, most of the particles passed straight through undeflected, as if they had hit nothing at all. This demonstrated that the atoms in the foil were mostly empty space. Only the very few particles that happened to strike a nucleus directly were deflected at large angles.

Because the nucleus is so small compared to the overall volume of the electron cloud, the atom is, in fact, mostly empty space.

The mass of an atom is primarily determined by the mass of its…

1. protons and electrons.
2. neutrons and electrons.
3. nucleus.
4. electrons.
5. protons.

Solution

Answer: C

Concept: atomic structure

An atom is composed of protons, neutrons, and electrons. While all three have mass, their contributions are vastly different:

• Proton Mass: ~1.007 amu
• Neutron Mass: ~1.009 amu
• Electron Mass: ~0.00055 amu

The nucleus contains all the protons and neutrons, which are the most massive particles. Therefore, the nucleus accounts for more than 99.9% of the total mass of an atom. While the electrons do contribute a very small amount of mass, the atom’s mass is primarily determined by its nucleus.

What is the correct charge of a sulfur atom with 16 protons, 16 neutrons, and 18 electrons?

Solution

Answer: 2−

Concept: ions

The overall charge of an atom or ion is determined by the difference between the number of positively charged protons and negatively charged electrons. Neutrons are neutral and do not affect the charge.

• Charge from Protons: 16 protons = +16
• Charge from Electrons: 18 electrons = −18

The net charge is calculated as: \[ \text{Net Charge} = (\#~\text{of protons}) - (\#~\text{of electrons}) = 16 - 18 = -2 \] By convention, an ionic charge of −2 is written with the numeral first, followed by the sign: 2–. This species is the sulfide anion (S^2–).

What charge will most likely be formed by lithium?

Solution

Answer: 1+

Concept: ions

The likely charge an element will form is determined by its position on the periodic table and its tendency to gain or lose electrons to achieve a stable electron configuration, typically like that of the nearest noble gas (the octet rule).

• Lithium (Li) is in Group 1 of the periodic table.
• Atoms in Group 1 have one valence electron.
• To achieve a stable, noble-gas configuration, the most energetically favorable process for lithium is to lose its one valence electron.

Losing one negatively charged electron leaves the atom with one more proton than electrons, resulting in a net charge of 1+.

Is iodine more likely to gain or lose an electron?

Solution

Answer: Gain an electron (to increase its stability)

Concept: ions

The tendency for an element to gain or lose electrons is determined by its goal of achieving a stable electron configuration, typically a full valence shell of eight electrons (an octet), like the nearest noble gas.

• Iodine (I) is in Group 17 (the halogens) of the periodic table.
• Atoms in Group 17 have seven valence electrons.
• To achieve a stable octet, the most energetically favorable process for iodine is to gain one electron to complete its valence shell.

Gaining one negatively charged electron gives the atom one more electron than protons, resulting in the iodide anion (I⁻) with a 1− charge. Losing seven electrons would require a vast amount of energy and is not a favorable process.

What is the approximate atomic mass (in u) of a magnesium ion containing 12 protons, 13 neutrons, and 10 electrons?

Solution

Answer: 25 u

Concept: atomic mass

The atomic mass of an atom or ion is determined by the total mass of its constituent protons, neutrons, and electrons. However, the mass of an electron (~0.00055 u) is negligible compared to the mass of a proton (~1 u) or a neutron (~1 u).

Therefore, for the purpose of calculating a whole-number approximate mass, we use the mass number (A), which is the sum of only the protons and neutrons.

\[ \begin{align*} A &= (\#~\text{of protons}) + (\#~\text{of neutrons}) \\[1.5ex] A &= 12 + 13 = 25 \end{align*} \]

The approximate atomic mass of this ion is 25 u. The number of electrons (10) is not used to calculate the mass number; it is used to determine the ion’s charge (12 protons - 10 electrons = 2+ charge).

Write the chemical formula for a compound formed between aluminum and sulfur.

Solution

Answer: Al₂S₃

Concept: writing formulas for binary ionic compounds

To write the formula for an ionic compound, we must first determine the charge of the ions each element is likely to form and then find the simplest whole-number ratio that results in a neutral compound.

1. Determine the Ions: We find the likely charge of each element based on its position on the periodic table.
   • Aluminum (Al): Located in Group 13. It will lose its three valence electrons to form a Al³⁺ cation.
   • Sulfur (S): Located in Group 16. It will gain two electrons to complete its octet, forming a S²⁻ anion.
2. Determine the Formula by Balancing Charge: The total positive charge must equal the total negative charge. We need to find the least common multiple of the charge magnitudes (3 and 2), which is 6.
   • To get a total positive charge of 6+, we need two Al³⁺ ions (2 × +3 = +6).
   • To get a total negative charge of 6−, we need three S^2– ions (3 × −2 = −6).
   This gives us a ratio of 2 aluminum ions to 3 sulfide ions. Therefore, the correct formula is Al₂S₃.

In which set do all elements tend to form cations in binary ionic compounds?

1. Na, Al, S
2. Ca, Mn, Sn
3. P, As, Sb
4. S, Cl, Br

Solution

Answer: B

Concept: predicting ion formation from the periodic table

The tendency to form a cation (a positive ion) is the defining chemical property of metals. Cations are formed when an atom loses one or more valence electrons. Nonmetals, in contrast, tend to gain electrons to form anions (negative ions).

We must find the set that contains only metallic elements.

• (b) Ca, Mn, Sn: This is the correct answer. Calcium (Ca) is an alkaline earth metal, Manganese (Mn) is a transition metal, and Tin (Sn) is a post-transition metal. All are metals and will form cations.

The other sets are incorrect because they contain nonmetals or metalloids that tend to form anions: * (a) Na, Al, S: Sulfur (S) is a nonmetal and forms the sulfide anion (S²⁻). * (c) P, As, Sb: Phosphorus (P) and Arsenic (As) are a nonmetal and a metalloid, respectively, and both typically form anions (e.g., P³⁻). * (d) S, Cl, Br: All three are nonmetals and form anions (S²⁻, Cl⁻, Br⁻).

Isotopes

Which of the statements about two isotopes is false?

1. They will have the same charge on the nucleus.
2. They will have different numbers of neutrons.
3. They will have essentially the same chemical reactivity.
4. They will have the same atomic numbers.
5. They will have the same atomic weights.

Solution

Answer: E

Concept: isotopes and atomic structure

Isotopes are atoms of the same element that have the same number of protons but different numbers of neutrons.

Statement (E) is false. Because isotopes have different numbers of neutrons, their mass numbers (total protons + neutrons) are different. This means their individual atomic masses (or atomic weights) are also different. For example, carbon-12 has a different atomic mass than carbon-14.

The other statements are all true:

• (A, D) Isotopes of an element always have the same number of protons. Therefore, they have the same atomic number (Z) and the same positive charge on the nucleus.
• (B) The difference in the number of neutrons is the defining characteristic of isotopes.
• (C) Chemical reactivity is determined by the arrangement of electrons, which in a neutral atom is equal to the number of protons. Since isotopes have the same number of protons and electrons, they exhibit nearly identical chemical reactivity.

What is the mass number for ²²Na?

Solution

Answer: 22

Concept: isotope notation

The standard notation for an isotope is ^A_ZX, where:

• A is the mass number (the total number of protons and neutrons), written as a superscript.
• Z is the atomic number (the number of protons), written as a subscript, and is often omitted.
• X is the symbol for the element.

In the given isotope, ²²Na, the number in the superscript position represents the mass number. Therefore, the mass number for this isotope of sodium is 22.

Which of the following represents isotopes?
 I. \(^{32}_{16}\mathrm{X}\)
 II. \(^{32}_{15}\mathrm{X}\)
 III. \(^{34}_{16}\mathrm{X}\)
 IV. \(^{34}_{17}\mathrm{X}\)

1. I and II
2. I and III
3. I and IV
4. III and IV

Solution

Answer: B

Concept: identifying isotopes

The definition of isotopes is atoms that have the same number of protons but a different number of neutrons. In isotope notation (\(^{A}_{Z}\mathrm{X}\)), this means we are looking for a pair of atoms that have:

1. The same atomic number (Z), the subscript.
2. Different mass numbers (A), the superscript.

Let’s analyze the given options:

• I: Z = 16, A = 32
• II: Z = 15, A = 32
• III: Z = 16, A = 34
• IV: Z = 17, A = 34

Only the pair I and III have the same atomic number (Z = 16) but different mass numbers (A = 32 and A = 34). Therefore, they are isotopes of the element sulfur.

The other pairs are incorrect:

• I and II are isobars (same mass number, different atomic number).
• I and IV have different atomic numbers and different mass numbers.
• III and IV have different atomic numbers.

Complete the table.

Solution

Answer:

Concept: atomic structure; isotopes

An element consists of four naturally occurring isotopes. The percent abundance and isotopic mass for each are given as:

• X₁: 1.40 %; 203.973 u
• X₂: 24.10 %; 205.9745 u
• X₃: 22.10 %; 206.9759 u
• X₄: 52.40 %; 207.9766 u

Calculate the average atomic mass (in u) of the element.

Solution

Answer: 207.2 u

Concept: average atomic mass and isotopes

The average atomic mass of an element is the weighted average of the masses of its naturally occurring isotopes. It is calculated by multiplying the mass of each isotope by its fractional abundance and then summing the results.

\[ \begin{align*} \bar{m}_{\mathrm{a}}(\mathrm{E}) &= \sum_{i=1}^{n} \left[ x(X_i) \cdot m(X_i) \right] \\[1.5ex] &= [0.014\bar{0} \cdot 203.97\bar{3}~\mathrm{u}] + [0.241\bar{0} \cdot 205.974\bar{5}~\mathrm{u}] + \\[1ex] &\phantom{=}~~~[0.221\bar{0} \cdot 206.975\bar{9}~\mathrm{u}] + [0.52\bar{4}0 \cdot 207.976\bar{6}~\mathrm{u}] \\[1.5ex] &= [2.8\bar{5}56~\mathrm{u}] + [49.6\bar{4}00~\mathrm{u}] + [45.7\bar{4}16~\mathrm{u}] + [109.\bar{0}0~\mathrm{u}] \\[1.5ex] &= 207.\bar{2}37~\mathrm{u} \\[1.5ex] &= 207.2~\mathrm{u} \end{align*} \] The precision of the final answer is limited by the addition rule. The least precise term is the last one (109.0 u), which is significant to the tenths place. Therefore, the sum must be rounded to the tenths place.

Note:

You will encounter several terms for atomic mass that are often used interchangeably, but which have distinct formal definitions.

• Average Atomic Mass (m_a): This is a true mass, representing the weighted average mass of an element’s atoms. It has units, typically the unified atomic mass unit (u). This is the quantity you calculate in a weighted-average problem.

• Atomic Weight (or Relative Atomic Mass, A_r): This is a dimensionless ratio. It is the ratio of the average atomic mass of an element to the unified atomic mass unit.

Because of the way the unit u is defined, these two quantities are always numerically identical. For this reason, the terms are often used synonymously in introductory texts. In this book, we will use “average atomic mass” when referring to a calculated value with units, and “atomic weight” when referring to the dimensionless values found on the periodic table.

A hypothetical element (E) consists of two isotopes, X₁ and X₂. The relative isotopic mass and percent abundance of each isotope are as follows:

• X₁: 64.23; 26.0 %
• X₂: 65.32

What is the average atomic mass (in u) of the element?

Solution

Answer: 65.0 u

Concept: weight; percent abundance

\[ \begin{align*} \bar{m}_{\mathrm{a}}(\mathrm{E}) &= \sum_{i=1}^n ~\biggl[ x\!\left (^A\mathrm{E} \right ) ~ A_{\mathrm{r}}\!\left (^A\mathrm{E} \right ) \left ( m_{\mathrm{u}} \right ) \biggr]_i \\[1.5ex] &= \left ( \dfrac{x \left ( \mathrm{X}_1 \right )\%}{100~\%} \right ) ~ A_{\mathrm{r}}(\mathrm{X}_1) ~ m_{\mathrm{u}} ~~+ \left ( \dfrac{x \left ( \mathrm{X}_2 \right )\%}{100~\%} \right ) ~ A_{\mathrm{r}}(\mathrm{X}_2) ~ m_{\mathrm{u}}~~+ \\[2ex] &= \left ( \dfrac{26.\bar{0}~\%}{100~\%} \right ) ~ \left ( 64.2\bar{3} \right ) ~(\mathrm{1~u}) ~~+ \left ( \dfrac{(100-26.\bar{0})~\%}{100~\%} \right ) ~ \left ( 65.3\bar{2} \right ) ~(\mathrm{1~u}) \\[2ex] &= (16.\bar{6}99~\mathrm{u}) + (48.\bar{3}36~\mathrm{u}) \\[1.5ex] &= 65.\bar{0}35~\mathrm{u} \\[1.5ex] &= 65.0~\mathrm{u} \end{align*} \]

The element rhenium (Re) has a standard atomic weight of 186.207 and exists as two naturally occurring isotopes. Given are the percent abundance and relative isotopic mass for each isotope.

1. ¹⁸⁵Re
2. ¹⁸⁷Re: 62.60 %; 186.956

Find the atomic weight of rhenium-185.

Solution

Answer: 185

Concept: percent abundance; isotopes

\[ \begin{align*} A_{\mathrm{r}} \! \left ( \mathrm{E} \right ) &= \sum_{i=1}^n ~\biggl[ x\!\left (^A\mathrm{E} \right ) ~ A_{\mathrm{r}}\!\left (^A\mathrm{E} \right ) \biggr]_i \\[1.5ex] A_{\mathrm{r}}^{\circ} \left ( \mathrm{Re} \right ) &= x \left (\mathrm{^{185}Re} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{185}Re} \right ) + x \left (\mathrm{^{187}Re} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{187}Re} \right ) \longrightarrow \\[1.5ex] A_{\mathrm{r}} \left ( \mathrm{^{185}Re} \right ) &= \dfrac{ A_{\mathrm{r}}^{\circ} \left ( \mathrm{Re} \right ) ~-~ \Biggr [ x \left (\mathrm{^{187}Re} \right ) ~ A_{\mathrm{r}} \left ( \mathrm{^{187}Re} \right ) \Biggr] } {x \left (\mathrm{^{185}Re} \right )}\\[2ex] &= \dfrac{ 186.20\bar{7} ~-~ \Biggr[ \left ( \dfrac{62.6\bar{0}~\%}{100~\%} \right ) \left ( 186.95\bar{6} \right ) \Biggr] } { \Biggr[ \dfrac{\left ( 100~\%-62.6\bar{0}~\%\right ) }{100~\%} \Biggr ] }\\[2ex] &= \dfrac{ \left ( 186.20\bar{7} - 117.\bar{0}34 \right ) } { \left ( \dfrac{37.4\bar{0}}{100~\%} \right ) } \\[2ex] &= \dfrac{ \left ( 186.20\bar{7} - 117.\bar{0}34 \right ) } { 0.374\bar{0} } \\[2ex] &= \dfrac{69.\bar{1}73} {0.374\bar{0}} \\[2ex] &= 18\bar{4}.95 \\[0.5ex] &= 185 \end{align*} \]