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Chapter 1 – The Early Experiments

Introduction

Chemistry is the science of matter, its properties, and the changes it undergoes. Chemists seek to understand our material universe at a molecular level and to use this understanding to improve our interaction with it, often creating new products that enhance our lives. Chemists often design these products by considering the properties of the desired substance and then proposing reactions of atoms or molecules that might yield the substances of choice. This design process involves particles and processes chemists can imagine but cannot see. They can observe the results of a reaction, such as a color change or the formation of a gas or a solid, but they cannot view directly the collisions of the atoms or molecules in a reaction or the changes these collisions produce. However, chemists are confident that these collisions and changes do take place, and we begin our study of chemistry by examining how we came to the point where we could envision these invisible processes.

1.1 The Scientific Method

Introduction

Chemistry is that branch of science that deals with matter and the changes it undergoes, and science is that branch of knowledge that is gained by the application of the scientific method. In this section, we explain the scientific method and show an example of its use.

Objectives

1.1-1. Scientific Method

The entire body of knowledge called science was achieved through the application of the scientific method.
Our understanding of atoms and molecules was gained by repeated application of the scientific method. There are four steps in the scientific method: The above process is repeated over and over again with new predictions and new tests (experiments). If a hypothesis stands up to many such tests and becomes accepted as the explanation for the observation, then the hypothesis becomes a theory. A theory is an accepted explanation of an observation. Theories can be supported by experiment, but they cannot be proven.

1.1-2. Phlogiston

Phlogiston theory, one of the first theories that attempted to explain a property of matter, was proposed in the late 17th century to explain fire. The following is an application of the scientific method to phlogiston theory. In it, we examine some chemical and physical properties of magnesium.
Observations:

strip of magnesium
Magnesium is a shiny metal.
burning magnesium
Magnesium releases something to produce a brilliant white light when it is burned.
magnesium residue
Burning the magnesium converts it from a shiny metal to a gray powder.
Hypothesis:

Magnesium metal contains a substance that is released when the magnesium is burned. The substance was named "phlogiston." The gray powder was the magnesium without its phlogiston, so it was the "dephlogisticated" form of the metal, which was called "calx." Thus, burning magnesium would have been represented as
magnesium calx of magnesium + phlogiston
Prediction:

If the metal contains both the calx and phlogiston, then the mass of the metal should be greater than the mass of the calx.
Test:

The mass of the calx was greater than the mass of the metal. The phlogiston hypothesis withstood experiment for over 100 years, but this experiment, which was performed by a scientist named Antoine Lavoisier, destroyed the theory. Although some argued that mass was irrelevant to the study, others agreed with Lavoisier's conclusion that phlogiston theory was incorrect.
Thus, application of the scientific method led to the downfall of phlogiston theory, and its continued application to other hypotheses has led us to our current understanding of the universe.

1.2 Lavoisier and the Birth of Modern Chemistry

Introduction

Measurements of the ratios in which the masses of substances combined with one another led to the statements of three laws: conservation of mass, constant composition, and multiple proportions. In this section, we briefly examine these three laws because they give us our next clues about the composition of matter.

Objectives

1.2-1. Laws

Laws summarize observations.
Lavoisier made many careful measurements of the masses of the reactants and products of reactions, and he observed that the total mass (products plus reactants) never changed during a reaction. He summarized his observation in the law of conservation of mass. Note that the above law simply summarizes the observations; it does not explain them. The explanation (theory) would not come for another decade. Lavoisier summarized his observations in the law of conservation of mass, and his work convinced many scientists that phlogiston did not exist. The emerging chemists of the early 19th century began testing the concept that matter consisted of elements and compounds and that mass was indeed relevant to chemistry. After a great number of measurements of relative masses had been performed, the following two laws were also accepted:
Example:
For example, one oxide of iron contains
3.490 g Fe
1 g O
,
while another contains
2.327 g Fe
1 g O
.
Both amounts of iron are combined with the same mass of oxygen (1 g), and the ratio of the masses is determined as follows:
3.490 g Fe
2.327 g Fe
= 1.50 =
3
2
which is a ratio of small whole numbers.

1.2-2. Mass Law Exercise

Exercise 1.1:

Na and O form two compounds. 100 g of each compound contains the following masses of the elements.
  • Compound I: 59 g Na    41 g O
  • Compound II: 74 g Na    26 g O
Show that the data is consistent with the Law of Multiple Proportions.
Step 1. Determine the mass of Na that combines with a fixed mass (1 g) of O in each compound.
Compound I mass Na/mass O = 1.439_0.039__
Compound I =
mass sodium
mass oxygen
=
59 g
41 g
= 1.4 g Na/g O
g Na/g O
Compound II mass Na/mass O = 2.846___
Compound II =
mass sodium
mass oxygen
=
74 g
26 g
= 2.8 g Na/g O
g Na/g O
Step 2. Determine the ratio of the mass of Na combined with 1 g of O in Compound II to Compound I.
Compound II/Compound I = 1.9778___
mass sodium
mass oxygen
=
2.8 g Na/g O
1.4 g Na/g O
= 2.0
Step 3. Express the result of Step 2 as a reduced fraction of small whole numbers.
numerator     2_0__ 2.0 is a ratio of 2/1.
denominator 1_0__ 2.0 is a ratio of 2/1.

1.2-3. Theories

Theories explain laws and observations. They can be accepted but not proven.
Theories are valid only as long as they are not disproven by experiment. In general, theories are continually modified or even discarded as more sophisticated experiments are carried out. The work of Antoine Lavoisier discredited phlogiston, so Lavoisier proposed a new hypothesis, which, after acceptance by others, became known as the theory of combustion. After further experiments, he presented the hypothesis that the mass of a metal increased when burned because burning was the combination of two substances, not the release of one. His combustion hypothesis was that burning is the reaction of a substance with oxygen. He named the product "the oxide" of the substance to indicate that the substance had oxygen added to it. Thus, he would have represented the burning of magnesium as
magnesium + oxygen magnesium oxide
Lavoisier began to classify matter as elements and compounds. Thus, magnesium and oxygen are elements, and magnesium oxide is a compound that is composed of the two elements magnesium and oxygen. Lavoisier's hypothesis was to be accepted by most other scientists and became the chemical theory that was used to explain matter and chemical reactions.

1.3 John Dalton and Atomic Theory

Introduction

We now explore the emergence of atoms and molecules, and examine the meaning of chemical formulas and equations. The atoms of an element had the same mass, and they combined in fixed ratios in compounds. These two facts allowed scientists to determine the relative numbers of atoms of the elements present in a sample; i.e., they were able to determine chemical formulas.

Objectives

1.3-1. Dalton's Atomic Theory

Atomic theory explains the mass relationships in substances and in reactions.
The chemists of the early 19th century had three laws (observations) to explain: conservation of mass, definite proportions, and multiple proportions. It was time for a hypothesis to explain the laws. In 1804, John Dalton suggested the explanation. His hypothesis, now known as Dalton's atomic theory, was based on the assumption that elements consisted of tiny spheres, called atoms. Dalton's atomic theory: The second postulate explains the law of mass conservation and is the basis of balancing chemical equations. The third postulate explains the laws of definite and multiple proportions. Dalton's theory explained all of the observations of his time and was accepted without change for nearly 100 years. Dalton used the term "atom" to refer to both a single atom and a combination of atoms. However, a combination of atoms is now called a molecule. We will use "molecule" to simplify our discussion. Dalton assumed that an atom was the smallest unit of an element, while a molecule was the smallest unit of a compound. This was not quite correct because, as we shall soon see, some elements exist as molecules.

1.3-2. Atom or Molecule Exercise

Exercise 1.2:

Indicate whether each of the following is an atom or a molecule.
    Ar
  • atom Ar is an atom.
  • molecule No. Ar is an atom.
    O3
  • atom No. O3 contains three atoms, so it is a molecule.
  • molecule O3 contains three atoms, so it is a molecule.
    HCl
  • atom No. HCl contains two atoms, so it is a molecule.
  • molecule HCl contains two atoms, so it is a molecule.

1.3-3. Naming Atoms

Dalton suggested a series of symbols to represent atoms and molecules. The symbols suggested by Dalton for some atoms and molecules are shown below.
element and compound symbols
Dalton assumed a 1:1 ratio for the atoms in compounds whose atom ratios were unknown. The figure shows three compounds where this assumption was incorrect.

1.3-4. Atom symbols

It is important to know the names and symbols of the fifty most common elements.
Today, the symbols of most of the elements are the first one or two letters of the element's name. You should learn the following table of names and symbols of the fifty most-used elements in this course.

hydrogen H silicon Si cobalt Co cadmium Cd
helium He phosphorus P nickel Ni tin Sn
lithium Li sulfur S copper Cu iodine I
beryllium Be chlorine Cl zinc Zn xenon Xe
boron B argon Ar gallium Ga cesium Cs
carbon C potassium K germanium Ge barium Ba
nitrogen N calcium Ca arsenic As platinum Pt
oxygen O scandium Sc selenium Se gold Au
fluorine F titanium Ti bromine Br mercury Hg
neon Ne vanadium V krypton Kr thallium Tl
sodium Na chromium Cr rubidium Rb lead Pb
magnesium Mg manganese Mn strontium Sr
aluminum Al iron Fe silver Ag

1.3-5. Names from Symbols Exercise

Exercise 1.3:

Write the name of the element from its symbol.
Br o_bromine_ins The answer is: bromine.
Sr o_strontium_ins The answer is: strontium.
Ar o_argon_ins The answer is: argon.
Li o_lithium_ins The answer is: lithium.
Sc o_scandium_ins The answer is: scandium.
H o_hydrogen_ins The answer is: hydrogen.
Ca o_calcium_ins The answer is: calcium.
Cl o_chlorine_ins The answer is: chlorine.
Ag o_silver_ins The answer is: silver.
Be o_beryllium_ins The answer is: beryllium.
Ti o_titanium_ins The answer is: titanium.
Cd o_cadmium_ins The answer is: cadmium.
B o_boron_ins The answer is: boron.
V o_vanadium_ins The answer is: vanadium.
Sn o_tin_ins The answer is: tin.
C o_carbon_ins The answer is: carbon.
Cr o_chromium_ins The answer is: chromium.
I o_iodine_ins The answer is: iodine.
N o_nitrogen_ins The answer is: nitrogen.
Mn o_manganese_ins The answer is: manganese.
Xe o_xenon_ins The answer is: xenon.
O o_oxygen_ins The answer is: oxygen.
Fe o_iron_ins The answer is: iron.
Cs o_cesium_ins The answer is: cesium.
F o_fluorine_ins The answer is: fluorine.
Co o_cobalt_ins The answer is: cobalt.
Ba o_barium_ins The answer is: barium.
Ne o_neon_ins The answer is: neon.
Ni o_nickel_ins The answer is: nickel.
Mg o_magnesium_ins The answer is: magnesium.
Zn o_zinc_ins The answer is: zinc.
Hg o_mercury_ins The answer is: mercury.
Al o_aluminum_ins The answer is: aluminum.
Pt o_platinum_ins The answer is: platinum.
K o_potassium_ins The answer is: potassium.
Si o_silicon_ins The answer is: silicon.
Ge o_germanium_ins The answer is: germanium.
Pb o_lead_ins The answer is: lead.
Rb o_rubidium_ins The answer is: rubidium.
He o_helium_ins The answer is: helium.
Na o_sodium_ins The answer is: sodium.
Cu o_copper_ins The answer is: copper.
Au o_gold_ins The answer is: gold.
Ga o_gallium_ins The answer is: gallium.
Tl o_thallium_ins The answer is: thallium.
P o_phosphorus_ins The answer is: phosphorus.
As o_arsenic_ins The answer is: arsenic.
S o_sulfur_ins The answer is: sulfur.
Se o_selenium_ins The answer is: selenium.
Kr o_krypton_ins The answer is: krypton.

1.3-6. Symbols from Names Exercise

Exercise 1.4:

Write the symbols of the elements given their names.
  • The first letter of each symbol must be upper case, and the second letter in any symbol must be lower case.
bromine o_Br_s The answer is: Br.
argon o_Ar_s The answer is: Ar.
krypton o_Kr_s The answer is: Kr.
vanadium o_V_s The answer is: V.
tin o_Sn_s The answer is: Sn.
thallium o_Tl_s The answer is: Tl.
strontium o_Sr_s The answer is: Sr.
lithium o_Li_s The answer is: Li.
cadmium o_Cd_s The answer is: Cd.
boron o_B_s The answer is: B.
cobalt o_Co_s The answer is: Co.
calcium o_Ca_s The answer is: Ca.
aluminum o_Al_s The answer is: Al.
gallium o_Ga_s The answer is: Ga.
barium o_Ba_s The answer is: Ba.
neon o_Ne_s The answer is: Ne.
carbon o_C_s The answer is: C.
xenon o_Xe_s The answer is: Xe.
oxygen o_O_s The answer is: O.
iron o_Fe_s The answer is: Fe.
cesium o_Cs_s The answer is: Cs.
fluorine o_F_s The answer is: F.
nickel o_Ni_s The answer is: Ni.
platinum o_Pt_s The answer is: Pt.
sodium o_Na_s The answer is: Na.
copper o_Cu_s The answer is: Cu.
gold o_Au_s The answer is: Au.
scandium o_Sc_s The answer is: Sc.
silver o_Ag_s The answer is: Ag.
helium o_He_s The answer is: He.
beryllium o_Be_s The answer is: Be.
titanium o_Ti_s The answer is: Ti.
zinc o_Zn_s The answer is: Zn.
mercury o_Hg_s The answer is: Hg.
silicon o_Si_s The answer is: Si.
germanium o_Ge_s The answer is: Ge.
lead o_Pb_s The answer is: Pb.
phosphorus o_P_s The answer is: P.
arsenic o_As_s The answer is: As.
sulfur o_S_s The answer is: S.
chromium o_Cr_s The answer is: Cr.
iodine o_I_s The answer is: I.
hydrogen o_H_s The answer is: H.
potassium o_K_s The answer is: K.
rubidium o_Rb_s The answer is: Rb.
magnesium o_Mg_s The answer is: Mg.
nitrogen o_N_s The answer is: N.
manganese o_Mn_s The answer is: Mn.
selenium o_Se_s The answer is: Se.
chlorine o_Cl_s The answer is: Cl.

1.3-7. Molecule Symbols

Molecules are represented by their constituent atoms. If there is more than one atom of an element present in a molecule, then the number is given as a subscript in the molecular formula.

water H2O two hydrogen atoms + one oxygen atom
ammonia NH3 one nitrogen atom + three hydrogen atoms
nitrous oxide N2O two nitrogen atoms + one oxygen atom
sugar C12H22O11 12 C atoms + 22 H atoms + 11 O atoms

1.3-8. Writing Formulas Exercise

Exercise 1.5:

Use the ball-and-stick models and the atom color codes below to write chemical formulas for the molecules.
  • light blue: H
  • red: O
  • gray: C
  • purple: N
Write the symbols in the order C, H, N, O. Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+.
two light blue spheres bound together
hydrogen molecule o_H_2_s A hydrogen molecule contains two H atoms, so its chemical formula is H2. Since hydrogen contains two atoms, it is a diatomic molecule.
two light blue spheres bound together
water o_H_2O_s A water molecule contains two H atoms and one oxygen atom.
two light blue spheres bound together
allene o_C_3H_4_s Allene contains three carbon atoms and four hydrogen atoms.
two light blue spheres bound together
TNT o_C_7H_5N_3O_6_s There are 7 C atoms, 5 H atoms, 3 N atoms, and 6 O atoms.

1.3-9. A Common Misconception

As was evident in the previous exercise, hydrogen atoms are present in many compounds, but H2 molecules are present only in the hydrogen molecule. Thus, both H2 and H2O contain two H atoms, but there are two H atoms, not a single H2 molecule in water. As we shall see in the next section, several elements exist as diatomic molecules (H2, N2, O2, F2, Cl2, Br2, and I2). But they are diatomic only as free molecules and never in compounds. Thus, allene (C3H4) contains four H atoms, not two H2 molecules.

1.4 Atoms and Molecules

Introduction

Atoms and molecules are too small to be seen, so scientists work with large numbers of them. The unit we use to describe this large number is called the mole.

Objectives

1.4-1. Laws Concerning Gases

In 1808, Joseph Gay-Lussac published the law of combining volumes: Gay-Lussac's law of combining volumes was soon explained by Amedeo Avogadro. His explanation is now known as Avogadro's law:

1.4-2. Discovering the Formula of Water

We now use the law of combining volumes and Avogadro's law to show how the early scientists discovered the formulas of water, hydrogen, and oxygen.
Initially:

In the absence of any information to the contrary, early scientists assumed that atoms combined in a one-to-one ratio. Thus, water would have been represented as HO. They also believed that the smallest unit of an element was an atom, so the reaction of hydrogen and oxygen to produce water was thought to be the following:
H + O HO
Application of Avogadro's Law:

Experiments showed that the reacting volume of hydrogen was twice that of oxygen. This fact, combined with Avogadro's law, meant that two hydrogen atoms must react with one oxygen atom, so water must be H2O, and the reaction should be written as
2 H + O H2O
Current View:

The volume of water that is produced was later shown to be the same as the volume of hydrogen and one-half the volume of oxygen that react. Therefore, the coefficients of hydrogen and water had to be the same, and each had to be twice that of oxygen. This presented a dilemma until Avogadro concluded that elements did not have to exist as atoms; they could exist as molecules! The equation could be balanced only if hydrogen and oxygen each existed as diatomic molecules; i.e., as molecules with two atoms. The following is the way we now view the reaction of hydrogen and oxygen to produce water:
2 H2 + O2 2 H2O

1.4-3. Elements can be Molecules

Some common elements occur as molecules rather than atoms.
Other elements also exist as diatomic molecules. The diatomic elements are: In addition, some elements exist as molecules with more than two atoms. For example, P4 and S8.

1.4-4. Elements or Compounds

Combining Dalton's atomic theory and Avogadro's suggestion that elements did not have to occur as atoms, we can define elements and compounds in terms of their constituent atoms. Note that it is the number of types of atoms, not the number of atoms, that distinguishes an element from a compound. S8 has more than one atom, so it is a molecule, but it contains only one type of atom (S), so it is an element.

1.4-5. Element or Compound Exercise

Exercise 1.6:

Indicate whether each of the following is an element or a compound.
    Ar
  • element Ar contains only one type of atom, so it is an element.
  • compound No. Ar contains only one type of atom, so it is an element.
    O3
  • element O3 contains three atoms but only one type of atom, so it is an element.
  • compound No. O3 contains three atoms but only one type of atom, so it is an element.
    HCl
  • element No. HCl contains two types of atoms, so it is a compound.
  • compound HCl contains two types of atoms, so it is a compound.

1.4-6. Balancing Chemical Equations Video

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1.4-7. Balancing Chemical Equations

The number of each type of atom must be the same on both sides of a balanced chemical equation.
Dalton's atomic theory indicates that atoms change partners in a chemical reaction, but they do not change their identity. Thus, the number of atoms of each type must be the same on both sides of a chemical equation. In other words, chemical equations must be balanced. However, subscripts cannot be changed as that would change the identity of the molecules, so chemical equations must be balanced by changing only the coefficients of the molecules. The following steps should lead to a balanced equation:
1
Pick the molecule with the greatest number of atoms and make its coefficient one unless another choice is obviously better. For example, sometimes a 2 must be used to assure an even number of one of the atoms.
2
Determine which atoms are fixed by the coefficient used in Step 1, then balance those atoms on the other side of the equation.
3
Determine which atoms are fixed by the coefficient(s) created in Step 2, then balance those atoms on the other side of the equation.
4
Repeat Step 3 until the equation is balanced.
Note that ones are not usually included in the balanced equation.
Example:
As an example, we will follow the steps above to balance the following chemical equation.
     HCl +      MnO2      MnCl2 +      H2O +      Cl2
1
Make the coefficient of either MnO2 or MnCl2 one. We choose MnO2.
     HCl + 1 MnO2      MnCl2 +      H2O +      Cl2
2
The coefficient used in Step 1 fixes the number of Mn atoms at 1 and O atoms at 2, so we balance the Mn atoms with a coefficient of 1 for MnCl2 and the oxygen atoms with a coefficient of 2 for water.
     HCl + 1 MnO2 1 MnCl2 + 2 H2O +      Cl2
3
The coefficient of water fixes the number of hydrogen atoms at four, so we balance the H atoms with a coefficient of 4 for HCl. Note that the coefficient of MnCl2 does not fix the number of Cl atoms because it is not the only source of Cl.
4 HCl + 1 MnO2 1 MnCl2 + 2 H2O +      Cl2
4
The coefficient of HCl fixes the number of chlorine atoms at four, but there are already 2 Cl atoms in 1 MnCl2, so we balance the Cl atoms on the other side of the equation with a coefficient of 1 for Cl2.
4 HCl + 1 MnO2 1 MnCl2 + 2 H2O + 1 Cl2
5
Each side of the equation contains 4 H atoms, 1 Mn atom, 2 O atoms, and 4 Cl atoms. The equation is now balanced, but ones are not usually written. Thus, the balanced equation is usually written as shown in the last step.
4 HCl + MnO2 MnCl2 + 2 H2O + Cl2

1.4-8. Balanced Equations Exercise

Exercise 1.7:

Balance the equations in the activity area with smallest integer coefficients. Click "Submit" after all coefficients have been entered. Use the following steps to help:
  • 1
    Pick the molecule with the greatest number of atoms and make its coefficient one if there is not another obvious choice.
  • 2
    Determine which atoms have been fixed by the coefficient in Step 1, then balance those atoms on the other side of the equation.
  • 3
    Determine which atoms are fixed by the coefficient(s) created in Step 2, then balance those atoms on the other side of the equation.
  • 4
    Continue Step 3 until the equation is balanced.
2_0__ Begin with a coefficient of 2 for N2O5 to make the number of oxygens even on both sides of the equation. This is a useful strategy when balancing equations containing diatomic molecules such as O2. Be sure to use the smallest integer numbers.
 N2    +    
5_0__ Begin with a coefficient of 2 for N2O5 to make the number of oxygens even on both sides of the equation. This is a useful strategy when balancing equations containing diatomic molecules such as O2. Be sure to use the smallest integer numbers.
 O2        
2_0__ Begin with a coefficient of 2 for N2O5 to make the number of oxygens even on both sides of the equation. This is a useful strategy when balancing equations containing diatomic molecules such as O2. Be sure to use the smallest integer numbers.
 N2O5
2_0__ Begin with a coefficient of 2 for N2O5 to make the number of oxygens even on both sides of the equation. This is a useful strategy when balancing equations containing diatomic molecules such as O2. Be sure to use the smallest integer numbers.
 Na  +  
2_0__ 2 Na + 2 H2O 2 NaOH + H2
 H2O    
2_0__ 2 Na + 2 H2O 2 NaOH + H2
 NaOH  +  
1_0__ 2 Na + 2 H2O 2 NaOH + H2
 H2
1_0__ P4 + 6 Cl2 4 PCl3
 P4    +    
6_0__ P4 + 6 Cl2 4 PCl3
 Cl2        
4_0__ P4 + 6 Cl2 4 PCl3
 PCl3
1_0__ C6H12O6 + 6 O2 6 CO2 + 6 H2O
 C6H12O6  +  
6_0__ C6H12O6 + 6 O2 6 CO2 + 6 H2O
 O2    
6_0__ C6H12O6 + 6 O2 6 CO2 + 6 H2O
 CO2  +  
6_0__ C6H12O6 + 6 O2 6 CO2 + 6 H2O
 H2O

1.5 The Mole and Molar Mass

Introduction

Dalton recognized that the mass of an atom is an important characteristic, but individual atoms are much too small to weigh. However, the relative masses of the atoms in a compound can be determined, so Dalton devised a scale of relative masses. In this section, we examine these atomic masses or weights and extend them into molecular masses or weights.

Objectives

1.5-1. Atomic mass

Atomic masses (or weights) describe the relative masses of atoms and molecules.
An atomic mass is simply a number that indicates the relative mass of an atom. Dalton reasoned that H was the lightest element, so he assigned it a relative mass of 1, which he called its atomic weight. Note that atomic mass and atomic weight mean the same thing because the masses are relative, so the two terms are used interchangeably. The atomic weights of the other elements could then be assigned from the experimentally determined masses of the reacting elements and the formulas of the compounds they form. For example, consider the following reaction of H2 and O2 to produce H2O.
2 H2 + O22 H2O
1.0 g + 8.0 g9.0 g
The above data imply that two oxygen atoms (one O2 molecule) has eight times more mass than four hydrogen atoms (two H2 molecules), or 1 O atom is sixteen times heavier than 1 H atom. The relative mass of an H atom was established as 1, so the atomic weight of O must be 16. Similarly, two water molecules have nine times more mass than 4 H atoms, or one water molecule is 18 times heavier than a hydrogen atom. Thus, the relative mass of water is 18. The relative mass of a molecule is called its molecular mass or weight.

1.5-2. Relative Mass Exercise

Exercise 1.8:

Given the atomic weights: H = 1 and O = 16, determine the atomic weight of the element X in each of the following mass relationships.
3 g X + 1 g H2 4 g XH4 12___ The X:H atom ratio in XH4 is 1:4, but the mass ratio of X:H in XH4 is 3:1. Therefore, one atom of X is three times heavier than four atoms of H. The atomic weight of H is 1, so the atomic weight of X is
3(4 × 1) = 12.
Based on its atomic weight, atom X is carbon, and the compound is CH4, which is called methane. The balanced equation for the reaction of carbon and hydrogen (a diatomic gas) to produce methane is
C + 2 H2 CH4.
3 g X + 2 g O2 5 g XO 24___ The X:O atom ratio in XO is 1:1, but the mass ratio of X:O in XO is 3:2. Therefore, one atom of X is 1.5 times heavier than one atom of O. The atomic weight of O is 16, so the atomic weight of X is
1.5(16 × 1) = 24.
Based on its atomic weight, atom X is magnesium, and the compound is MgO, which is magnesium oxide. The balanced equation for the reaction of magnesium and oxygen (a diatomic gas) to produce magnesium oxide is
2 Mg + O2 2 MgO.
9 g X + 8 g O2 17 g X2O3 27___ The X:O atom ratio in X2O3 is 2:3, but the mass ratio of X:O in X2O3 is 9:8. Therefore, two atoms of X are 9/8 = 1.125 times heavier than 3 atoms of O. The atomic weight of O is 16, so the relative mass of two atoms of X is
1.125(3 × 16) = 54.
The atomic weight of X is one-half the relative mass of two atoms, so
atomic weight = (0.50)(54) = 27.
Based on its atomic weight, atom X is aluminum, and the compound is Al2O3, which is aluminum oxide. The balanced equation for the reaction of aluminum and oxygen (a diatomic gas) to produce aluminum oxide is
4 Al + 3 O2 2 Al2O3.

1.5-3. Atomic Masses vs. Molecular Masses

A molecular mass equals the sum of the atomic masses of the atoms that comprise the molecule.
The modern atomic weight scale is no longer based on hydrogen. Instead, it is based on the mass of the most common form of carbon, which is assigned a relative mass (atomic weight) of exactly 12. The masses of individual atoms and molecules are often given in atomic mass units (amu) or Daltons (D). 1 amu = 1 D = 1/12 of the mass of a carbon atom. The atomic masses of the ten lightest atoms are given in the accompanying table.

H 1.01 C 12.01
He 4.00 N 14.01
Li 6.94 O 16.00
Be 9.01 F 19.00
B 10.81 Ne 20.18

Molecular masses (or weights) are simply the sum of the atomic masses of the atoms that make up the molecule. For example, the molecular mass of CO2, M(CO2), would be determined as follows:
M(CO2) = n(C) × M(C) + n(O) × M(O) = 1(12) + 2(16) = 44
where n(C) is the number of C atoms in the molecule and M(C) is the atomic mass of a carbon atom.
Exercise 1.9:

Use the atomic masses above to determine the molecular mass of each of the following.
BF3 67.81___
M(BF3)= n(B) × M(B) + n(F) × M(F) = 1(10.81) + 3(19.00) = 67.81
N2O5 108.02___
M(N2O5) = n(N) × M(N) + n(O) × M(O) = 2(14.01) + 5(16.00) = 108.02
H2C2O4 90.04___
M(H2C2O4) = n(H) × M(H) + n(C) × M(C) + n(O) × M(O) = 2(1.01) + 2(12.01) + 4(16.00) = 90.04

1.5-4. The Mole

Atoms and molecules react in the ratio specified in a balanced chemical equation, and chemists usually mix the reactants in ratios close to those predicted by the equation. However, chemists convert the ratios of atoms and molecules to ratios of mass in order to quickly deliver the correct amounts of material. The conversion between numbers of atoms or molecules and mass is made with the use of the mole. A mole is the number of molecules or atoms in a sample of a compound or element that has a mass equal to its molecular or atomic mass expressed in grams. It is abbreviated mol. It is a number just as a pair is 2 and a dozen is 12. The number of items in a mole is called Avogadro's number, NA.
NA = 6.02 × 1023 mol−1
Avogadro's number is used to convert between a number of moles and a number of molecules or atoms just as 12 is used to convert between a number of dozens and a number of eggs.
Example:
For example, determining the number of C atoms in 2 moles of carbon atoms or in 2 dozen carbon atoms is done exactly the same.
2 mol C ×
6 × 1023 C atoms
1 mol C
= 1.2 × 1024 C atoms
2 doz C ×
12 C atoms
1 doz C
= 24 C atoms
Example:
Similarly, the number of ammonia molecules in 8 moles of ammonia is determined as
8 mol NH3 ×
6 × 1023 NH3 molecules
1 mol NH3
= 4.8 × 1024 NH3 molecules
Avogadro's number is very large. Consider that one mole of dice that are 1/2" on a side would cover the 48 contiguous states of the USA to a height of 100 miles! Molecules are very small. A mole of water molecules has a volume of only 18 mL.

1.5-5. Individual Atoms and Molecules

The molar mass has units of g/mol and is used to convert between mass and moles of a substance.
The mass of a mole is called the molar mass, Mm. Molar mass has units of g/mol. The molar mass of H2 is Mm = 2 g/mol and that of carbon is Mm = 12 g/mol. Thus, 2 g of H2 contain 6.02 × 1023 H2 molecules and 12 g of C contain 6.02 × 1023 C atoms. The factor label method uses conversion factors to convert given quantities into desired quantities. Molar mass has units of g/mol, so it is the conversion factor that is used to convert grams into moles or moles into grams. For example, if compound A has a molar mass Mm, then the number of moles of A present in Z grams of A is
Z g A ×
1 mol A
Mm g A
=
Z
Mm
mol A
Note that the units of the given quantity (g A) cancel with the denominator of the conversion factor to yield the correct units for the answer. We can determine the mass of n moles of A in a similar manner.
n mol A ×
Mm g A
1 mol A
= nMm g A
Again, the units of the given quantity are converted into the units of the desired quantity by the conversion factor (molar mass of A).

1.5-6. Mass and Moles Exercise

Exercise 1.10:

The atomic masses of N, O, and F are 14, 16, and 19, respectively.
The mass of 0.25 mole of N2O3 = 19___ First determine the molar mass of N2O3. The molar masses of nitrogen and oxygen are 14 g/mol and 16 g/mol, respectively, so the molar mass is (2 mol N)(14 g/mol) + (3 mol O)(16 g/mol) = 76 g/mol. Next, use the molar mass to convert moles of sample into grams of sample.
    
0.25 mol N2O3 ×
76 g N2O3
1 mol N2O3
= 19 g N2O3
The mass of a substance can be obtained from the number of moles by multiplying the number of moles by the molar mass. However, the units can be used to help you setup the calculation because the given units (mol N2O3) cancel leaving the desired units. You must only remember that the units of molar mass are g/mol.
g N2O3
The number of moles of NF3 in 11.8 g = 0.166197___ First determine the molar mass of NF3. The molar masses of nitrogen and fluorine are 14 g/mol and 19 g/mol, respectively, so the molar mass is (1 mol N)(14 g/mol) + (3 mol F)(19 g/mol) = 71 g/mol. Next, use the molar mass to convert the mass of sample into moles of sample.
    
11.8 g NF3 ×
1 mol NF3
71 g NF3
= 0.17 mol NF3
The number of moles of a substance can be obtained from the mass by dividing the number of moles by the molar mass. However, the units can be used to help you do the calculation because the given units (g NF3) cancel leaving the desired units. You must only remember that the units of molar mass are g/mol.
mol NF3

1.6 Stoichiometry

Objectives

1.6-1. Stoichiometry Video

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1.6-2. Subscripts as Stoichiometric Factors

The stoichiometric factor used to determine the amount of one atom that is combined with a given mass of another is the ratio of the subscripts of the given to sought elements.
Chemists often need to convert the amount of one substance into the chemically equivalent amount of another substance. The study of such conversions is referred to as stoichiometry. The conversion factor that converts one substance into another is called the stoichiometric factor. In this and the next section, we use the chemical formula of a compound to derive stoichiometric factors for substances. The following conversion factors can be written from the formula Na3PO4.

3 mol Na
1 mol Na3PO4
3 mol Na
4 mol O
4 mol O
1 mol Na3PO4

The above ratios and their reciprocals can be used to convert a given number of moles of one substance into the chemically equivalent number of moles of another substance in the formula. For example, the number of moles of oxygen in a sample of Na3PO4 that contains 6 mol of sodium can be determined as follows.
6 mol Na ×
4 mol O
3 mol Na
= 8 mol O
Thus, a sample of Na3PO4 that contains 6 mol Na also contains 8 mol O. Note how the stoichiometric factor is derived from the subscripts in the chemical formula. Once the number of moles of the desired substance is known, its mass can be determined by multiplying the number of moles by the molar mass.

1.6-3. Coefficients as Stoichiometric Factors

The stoichiometric factor used to determine the amount of one substance that reacts with or is formed from another equals the ratio of the coefficients of the sought to given substances in the balanced equation.
The ratios of the coefficients in balanced chemical equations can be used as stoichiometric factors. Consider the following reaction and the factors derived from the coefficients in the balanced equation.

2 N2 + 3 O2 2 N2O3
2 mol N2 react
2 mol N2O3 forms
3 mol O2 react
2 mol N2O3 forms
2 mol N2 react
3 mol O2 react

Thus, the number of moles of nitrogen that react with 6 mol O2 in the above reaction can be determined using the factor label method and the balanced equation as follows.
6 mol O2 ×
2 mol N2
3 mol O2
= 4 mol N2
Thus, if 6 mol O2 are to react then at least 4 mol N2 must be present. Note that the stoichiometric factor is derived from the coefficients in the balanced equation.

1.6-4. Mole-Mole Substance Stoichiometry Exercise

Exercise 1.11:

How many moles of NH3 are in a sample that contains 27 mol of H atoms?
9.0___ We use the fact that there are three moles of H in each mole of NH3 to create the stoichiometric factor then use the factor label method to convert from the given moles of H into the number of moles of the desired substance, NH3. The factor must be used such that the given number of moles of H cancel with mol H in the factor to leave mol NH3.
    
27 mol H ×
1 mol NH3
3 mol H
= 9.0 mol NH3
mol NH3
A sample of Al2O3 contains 9 mol of O. How many mol of Al does it contain?
6___ We are asked to convert 9 mol O into the equivalent number of mol Al. We use the chemical formula, which indicates that there are 2 mol Al for each 3 mol O, to create the stoichiometric factor. Then we multiply the given number of moles of O by the stoichiometric factor so as to cancel moles of O and leave moles of Al.
    
9 mol O ×
2 mol Al
3 mol O
= 6 mol Al
mol Al

1.6-5. Mass-Mass Stoichiometry Exercise

Exercise 1.12:

What is the mass of sulfur is in a 5.00 g sample of Ca2S3? The atomic mass of Ca is 40 and that of S is 32.
First, convert the amount of given mass into moles.
molar mass of Ca2S3 = 176___ The atomic masses are: Ca = 40.0 and S = 32.0.
    Mm = 2 mol Ca × 40.0 g/mol + 3 mol S × 32.0 g/mol = 176 g/mol
g/mol
number of moles in a 5.00 g sample = 0.028409___
5.00 g Ca2S3 ×
1 mol Ca2S3
176 g Ca2S3
= 0.0284 mol Ca2S3
mol Ca2S3
Next, convert the moles of Ca2S3 into the chemically equivalent number of moles of sulfur.
0.085227___
0.0284 mol Ca2S3 ×
3 mol S
1 mol Ca2S3
= 0.0852 mol S
mol S
Finally, convert the number of moles of sulfur into its mass in grams to obtain the answer.
2.72727___
0.0852 mol S ×
32 g S
1 mol S
= 2.73 g S
g S

1.6-6. Mole-Mole Reaction Stoichiometry Exercise

Exercise 1.13:

How many moles of each reactant are required to produce 68 g of NH3 by the following reaction? The atomic weights of N and H are 14 and 1, respectively.
N2 + 3 H2 2 NH3
1. First convert the given amount into moles! The number of moles of NH3 to be prepared is
4.0___ The molar mass of
NH3 = n(N) × M(N) + n(H) × M(H) = 1(14) + 3(1) = 17 g/mol,
which can be used to convert a mass in grams to the number of moles as follows.
    
68 g NH3 ×
1 mol NH3
17 g NH3
= 4.0 mol NH3
Note that g NH3 cancel in the multiplication. The above is an example of the factor label method for doing problems because the units of the given quantity (g NH3) are converted to the units of the desired quantity (mol NH3) with a stoichiometric factor (mol NH3/g NH3).
mol NH3
2. Once the number of moles of the ammonia are known, the number of moles of each reactant can be determined. We first determine the number of moles of H2 that would be required.
6.0___ Convert the moles of the given substance (NH3) into moles of the desired substance (H2) by using the factor label method and the stoichiometric factor derived from the balanced equation.
    
4.0 mol NH3 ×
3 mol H2
2 mol NH3
= 6.0 mol H2
Note that mol NH3 cancel in the multiplication.
mol H2
3. Finally, determine the number of moles of N2 that are needed.
2.0___ We use the 4.0 mol NH3 determined in Step 1 and the balanced equation, which shows that 1 mol N2 is required for every 2 mol NH3.
    
4.0 mol NH3 ×
1 mol N2
2 mol NH3
= 2.0 mol N2
mol N2

1.6-7. Mole-Mass Reaction Stoichiometry Exercise

Exercise 1.14:

Once the numbers of moles of reactants and products have been determined, the masses can be obtained by multiplying by the molar masses. In this example, we determine the number of moles of O2 that are required to react with 0.30 mol K and the mass of K2O that is produced in the following reaction.
4 K + O2 2 K2O
The number of moles of O2 required = 0.075___
0.30 mol K ×
1 mol O2
4 mol K
= 0.075 mol O2
mol
The number of moles of K2O produced = 0.15___
0.30 mol K ×
2 mol K2O
4 mol K
= 0.15 mol K2O
mol
The mass of potassium oxide that forms = 14.1___
0.15 mol K2O ×
94 g K2O
1 mol K2O
= 14 g K2O
g

1.6-8. Limiting Reactants Video

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1.6-9. Limiting Reactants

When reactants are not added in the required stoichiometric ratio, the amount of reaction is dictated by the limiting reactant.
Chemists seldom add chemicals in the exact stoichiometric ratio, so the amount of product that forms depends upon the amount of the reactant, called the limiting reactant, that is consumed first. In the following, we examine the reaction of 5 mol S and 6 mol O2 to produce SO3.
5 yellow spheres and 6 pairs of blue spheres
Initially there are 5 mol S, the five yellow spheres in the figure, and 6 mol O2, the six pairs of blue spheres.
5 yellow spheres and 12 blue spheres
There are 12 mol O atoms in 6 mol O2, so there are 5 mol S and 12 mol O.
4 yellow spheres with three blue spheres around them plus on lone yellow sphere.
Each mole of SO3 requires one mole of S, so there is enough S present to produce 5 mol SO3. Each mole of SO3 also requires 3 mol of O, so there are enough oxygen atoms to make only 12/3 = 4 mol SO3. Since less SO3 can be made with the oxygen, it is the limiting reactant. Consumption of 6 mol O2 produces 4 mol SO3, so that is all that can be made. Initially, there were 5 mol S, but making 4 mol SO3 requires only 4 moles. Therefore, there is 1 mol S left over.

1.6-10. Limiting Reactants 2

Typically, the limiting reactant is determined by applying the stoichiometric factors for the reaction to the given numbers of moles of reactant to determine which reactant produces the least amount of product; i.e., to determine the limiting reactant. Consider the reaction of 5 mol S with 6 mol O2 as discussed in the previous section. The chemical equation for the reaction is 2 S + 3 O2 2 SO3. The amount of SO3 that can be produced from 5 mol S and 6 mol O2 is then determined to be
5 mol S ×
2 mol SO3
2 mol S
= 5 mol SO3
and
6 mol O2 ×
2 mol SO3
3 mol O2
= 4 mol SO3
Oxygen produces less product, so it is the limiting reactant and the amounts of all products that form and the amounts of reactants that react are based on the number of moles of O2 that react.

1.6-11. Limiting Reactant Exercise

Exercise 1.15:

What mass of Al2O3 can be produced from 10.0 mol Al and 9.0 mol O2?
4 Al + 3 O2 2 Al2O3
The number of moles of Al2O3 produced from 10.0 mol Al = 5.0___
10.0 mol Al ×
2 mol Al2O3
4 mol Al
= 5.0 mol Al2O3
mol
The number of moles of Al2O3 produced from 9.0 mol O2 = 6.0___
9.0 mol O2 ×
2 mol Al2O3
3 mol O2
= 6.0 mol Al2O3
mol
The limiting reactant is o_Al_s Which substance (Al or O2) produces less product? .
The maximum number of moles of Al2O3 that can be produced = 5.0___ How many moles can the limiting reactant produce? mol
The molar mass of Al2O3 = 102___ 2 mol Al + 3 mol O = 2(27 g/mol) + 3(16 g/mol) g/mol
The mass of Al2O3 that can be produced = 510___ The limiting reactant is Al, and it can produce only 5.0 mol Al2O3.
5.0 mol Al2O3 ×
102 g Al2O3
1 mol Al2O3
= 510 g Al2O3
g

1.7 Energy

Introduction

Chemical processes are driven by energy differences, which result from changes in the interactions between charges. We now examine two types of energy and the energy of interaction between charged particles.

Objectives

1.7-1. Energy

There are two forms of energy: kinetic energy and potential energy.
In simple terms, energy is the capacity to move something. The energy of an object is the sum of two terms. An object that is moving can make another object move by colliding with it. The kinetic energy of a particle of mass m moving with a velocity v is
KE =
1
2
mv2.
Molecules and atoms are in constant motion, and their speed (kinetic energy) is dictated by their temperature.
An object that has the capacity to move due to its position has potential energy. Two examples of potential energy: Molecules and atoms have potential energy because they interact with one another. The change in potential energy caused by these interactions is responsible for the formation of chemical bonds and the condensation and freezing of molecules.

1.7-2. Energy Change

Systems in nature seek to minimize their energy, so energy changes for most processes are negative.
The energy of an object is a relative quantity and cannot be measured absolutely. For example, a ball at rest on the floor has neither kinetic nor potential energy relative to the floor, but it does have kinetic energy relative to the sun and potential energy relative to the center of the Earth. Consequently, it is energy change, not absolute energy, that is typically measured. The symbol Δ (Delta) is used to represent change and the symbol E is used for energy, so the energy change is expressed as ΔE. By convention, energy change is the final energy minus the initial energy.
ΔE = EfinalEinitial
Thus, ΔE < 0 means that the object loses energy because its final energy is less than its initial energy. A very important concept in chemistry is that systems naturally seek the position of lowest energy; i.e., nature favors processes for which ΔE < 0. Although energy change in chemical processes is not discussed in detail until Chapter 9, we will use the concept that molecules undergo chemical processes in order to reduce their potential energy in every chapter of this text.
Example:
As an example of the amount of energy that can be released in a common reaction, consider the combustion of octane, a component of gasoline. The chemical equation for the combustion is
C8H18 + 12.5 O2 8 CO2 + 9 H2O
ΔE < 0 for this reaction, and the energy that is released is used to power many automobiles. The amount of energy released in the reaction can be better appreciated by considering the following:
  • In order to deliver the same amount of energy as all of the kinetic energy of a 2200 pound (1000 kg) car moving at 40 mph
    (18 m/s),
    you would have to burn only 3 g of octane (~0.0008 gal).

1.8 Electromagnetism and Coulomb's Law

Introduction

The attraction of opposite charge and the repulsion of like charge are the result of the electromagnetic force. The electromagnetic force, which is responsible for all of the interactions between atoms and molecules discussed in this course, is the topic of this section.

Objectives

1.8-1. Electromagnetism and Coulomb's Law

Coulomb's law describes the force exerted between two charged particles. The Coulombic force is responsible for the forces in molecules.
In the 1800's, scientists recognized that charged particles interacted, but Charles Coulomb was the first to measure the electromagnetic force.
photo of Charles Coulomb
He observed that the force of interaction between two charged particles was His observations are summarized by Coulomb's law: Two particles with charges q1 and q2 and separated by a distance r experience the following force F.
( 1.1 )
F =
kq1q2
εr2
The Coulombic Force

1.8-2. Energy of Interaction

Opposite charges attract and like charges repel because doing so lowers their potential energy (energy of interaction).
Energy is a force exerted through a distance
(E = Fr).
Consequently, the potential energy of two charged particles separated by a distance r is determined by multiplying Equation 1.1
F =
kq1q2
εr2
by r:
( 1.2 )
ΔE =
kq1q2
εr
Energy of Interaction of 2 Charged Particles
ΔE, which is referred to as the energy of interaction, is the potential energy of the two particles separated by a distance r relative to their potential energy when they are separated by an infinite distance (i.e., not interacting).
ΔE = ErE = Er − 0 = Er = E
Consequently, the energy of interaction is often written without the Δ. Note that the energy change is negative (the energy decreases) as r decreases when q1 and q2 have opposite signs, therefore particles of opposite charge lower their energy as they get closer. Systems strive to lower their energy, so particles of opposite sign are attracted. However, the energy change is positive (the energy increases) when q1 and q2 have the same sign, which means that the energy of two particles of the same sign increases as they get closer. Consequently, particles of the same charge move apart to lower their energy; i.e., particles of like charge are repelled. The energy of interaction of two charged particles as a function of the distance between them is shown in Figure 1.1.
Figure 1.1
Energy of Interaction: The energy of like-charged particles increases as they get closer, while the energy of oppositely-charged particles decreases as they get closer. Thus, particles of like charge avoid (are repelled by) one another, while particles of opposite charge are attracted to one another.

1.8-3. Coulomb's Law Exercise

Exercise 1.16:

Pick the system in each pair that has the lower potential energy.
  • A sphere with a charge of +2 is separated from a sphere with a charge of -3 by a distance of 3.
  • A sphere with a charge of +2 is separated from a sphere with a charge of -2 by a distance of 3.
    Both are attractive due to opposite charges, so the first system is the lower energy because the magnitude of the product of the charges is greater.
  • A sphere with a charge of +2 is separated from a sphere with a charge of +3 by a distance of 3.
    Both are repulsive due to like charges, so the second system is the lower energy because the magnitude of the product of the charges is smaller.
  • A sphere with a charge of +2 is separated from a sphere with a charge of +2 by a distance of 3.
  • A sphere with a charge of +2 is separated from a sphere with a charge of +3 by a distance of 3.
    Both interactions are repulsive due to like charges, and the magnitude of the charges is the same. Thus, the second system is the lower energy system because the charges are farther apart.
  • A sphere with a charge of +2 is separated from a sphere with a charge of +3 by a distance of 4.
  • A sphere with a charge of -1 is separated from a sphere with a charge of -2 by a distance of 2.
  • A sphere with a charge of +2 is separated from a sphere with a charge of +2 by a distance of 3.
    ΔE
    q1q2
    r
    ;
    (−1)(−2)
    2
    = 1.0;
    (+2)(+2)
    3
    = 1.3
    1.0 < 1.3, so the first system has the lower energy.
  • A sphere with a charge of +2 is separated from a sphere with a charge of -3 by a distance of 3.
  • A sphere with a charge of +2 is separated from a sphere with a charge of -3 by a distance of 4.
    Both interactions are attractive due to opposite charges, and the magnitude of the charges is the same. Thus, the first system is the lower energy system because the charges are closer.
  • A sphere with a charge of +3 is separated from a sphere with a charge of -3 by a distance of 4.
  • A sphere with a charge of +1 is separated from a sphere with a charge of -1 by a distance of 2.
    ΔE
    q1q2
    r
    ;
    (+3)(−3)
    4
    = −2.25;
    (+1)(−1)
    2
    = −0.5
    −2.25 < −0.5,
    so the first system has the lower energy.

1.9 Atomic Structure

Introduction

New technology led to new experiments at the end of the 19th century, experiments that Dalton's atomic theory could not explain. We now examine three of the most important of those experiments and take the atom to the next step.

Objectives

1.9-1. Thomson Experiment

photo of J. J. Thomson
In 1897, the British physicist J. J. Thomson discovered the electron while he was exploring the nature of the "cathode rays." In his experiment, which is demonstrated schematically below, he examined the effect of electric fields on cathode rays.
cathode ray tube
Figure 1.2a: Cathode Ray Tube
Two metal plates, which are connected to a power supply, are sealed into an evacuated glass tube. One of the plates has a small hole in its center. When a high voltage is applied across the metal plates, a "ray" originating at the cathode (–) (hence their name "cathode rays") passes through the hole in the other plate, which is the anode (+) and hits the end of the tube, which is coated with ZnS. ZnS glows when struck with cathode rays to produce the dot in the end of the tube. The yellow line represents the trajectory of the ray, but cathode rays are not visible, which is why ZnS was necessary.
deflected cathode ray
Figure 1.2b: Deflected Cathode Ray
The "ray" was deflected when an electric field was applied. The extent of deflection is given by the value of Δ, the amount by which the spot moves from its position in the absence of the electric field. Thomson noted the direction of the deflection and carefully measured the value of Δ at several field strengths.

1.9-2. Thomson Observations

Cathode rays are beams of negatively-charged particles.
Thomson made the following observations and conclusions:
Observation:
  • The "rays" were deflected by electric and magnetic fields.
Conclusion:
  • Cathode rays are not light rays because light rays are not deflected by these fields. Therefore, the "rays" were actually charged particles, which are now called electrons.
Observation:
  • The deflection was away from the negative plate and toward the positive plate.
Conclusion:
  • The particles were negatively charged.
Observation:
  • The deflection was large.
Conclusion:
  • The amount of deflection depended upon the strength of the field and on the charge and mass of the particles. The greater the charge on the particle (q), the greater the deflection, but the greater its mass (m), the smaller the deflection. Thus, Δq/m. Thus, the large deflection meant that either the particle had a very high charge or a very small mass. He was familiar with charges observed by others in different experiments and reasoned that the charge could not be the reason for the large deflection. He concluded that the mass of the particle must be less than one-thousandth that of the hydrogen atom. He was shocked! This meant that the hydrogen atom was not the smallest unit of matter as Dalton had suggested nearly a century earlier.
Observation:
  • The value of Δ
Conclusion:
  • Thomson used the experimental value of Δ as shown in Figure 1.2b and the strength of the electric field that caused the deflection to determine the charge-to-mass ratio for the electron.
qe
me
= −1.76 × 1011 kg · C−1

1.9-3. Millikan Experiment

Robert Millikan, an American physicist at the University of Chicago, was the first to accurately determine the charge on the electron.
drawing of oil drop experiment
Figure 1.3: Oil Drop Experiment

1.9-4. Millikan Observations

The mass of an electron is only 1/1800 the mass of a hydrogen atom.
photo of Robert Millikan
At the point where the particle moved neither up nor down, the electrostatic force (qE) that pulled the droplet up equaled the gravitational force (mg) that pulled it down, so
q =
mg
E
.
E, m, and g were all known, so he was able to determine the charge on the droplet. Various experiments yielded different values of q for different droplets, but all of the measured charges were multiples of the same charge, –1.6 × 10–19 C . Millikan reasoned that the charges on the droplets were different because each droplet had a different number of electrons; i.e.,
q = nqe,
where n is the number of electrons and qe is the charge on each electron. In this way he was able to determine that the charge on an electron must be qe = 1.6 × 10–19 C. With this charge and Thomson's charge-to-mass ratio, Millikan was able to determine the mass of the electron to be 9.1 × 10–31 kg, which is approximately 1/1800 the mass of the hydrogen atom.


charge on the electron mass of electron
qe = 1.6 × 10−19 C
me = 9.1 × 10−31 kg

1.9-5. Kelvin-Thomson Model

Clearly, it was time to refine Dalton's atomic model. One proposed model was the "raisin pudding" model of Lord Kelvin and J. J. Thomson. They reasoned that because atoms are uncharged themselves, they must contain enough positive charge to balance the negative charge of the electrons. In the Kelvin-Thomson model, the atom resembled raisin pudding with the negatively-charged electrons (the raisins) embedded in a mass of diffuse positive charge (the pudding). In the accompanying figure of the raisin pudding model, six electrons or raisins are embedded in a mass of positive charge. The raisins carry very little mass but all of the negative charge, while the pudding carries almost all of the mass and all of the positive charge in a (grey area). Since the positive charge was thought to be spread over a relatively large volume, it was thought to be diffuse. Atoms are neutral, so the positive charge in the bulk of the mass must equal the negative charge of the electrons.
raisin pudding model
Raisin-Pudding model of an atom with six electrons.

1.9-6. Rutherford Experiment

Almost all of the mass and all of the positive charge in an atom is concentrated in its nucleus.
photo of Ernest Rutherford
In another classic experiment, Ernest Rutherford tested the raisin pudding model in 1911 and discovered the nucleus of the atom. Rutherford bombarded a very thin gold foil with α particles (particles with atomic mass = 4 and charge = +2) moving at 10,000 mi/s. If the raisin pudding model was correct, most of the particles would pass through the foil undeflected because the positive charge of the atom was assumed to encompass the entire atom, which would make it diffuse. Some particles were expected to experience minor deflection. Consistent with the predictions, most of the particles did indeed pass through undeflected or with only minor deflection. However, a few (1 in 20,000) were deflected back at acute angles, and these few showed that the raisin pudding model could not be correct!
drawing of Rutherford's experiment
Figure 1.4: The Rutherford Experiment

1.9-7. Rutherford Observations

His experiment showed that the raisin pudding model was incorrect, so Rutherford had to view his observations in light of a new model. The following observations had to be explained.
Observation:
  • Most particles passed through undeflected.
Explanation:
  • Most of the volume of the atom contained very little mass to deflect the massive, positively-charged α particles.
Observation:
  • Some minor deflections were observed.
Explanation:
  • Small deflections of the α particles were caused by near misses with a massive particle within the atom.
Observation:
  • One particle in 20,000 was deflected at an acute angle.
Explanation:
  • Only a very massive and highly, positively-charged particle could cause the high-energy, positively-charged α particles to reverse direction. The fact that only one particle in 20,000 was deflected meant that the cross-sectional area of this positive charge was only 1/20,000 of the cross section of the atom. Thus, most of the mass and all of the positive charge of the atom was concentrated in a very small particle, which is called the nucleus.
Figure 1.5
Rutherford's Nuclear Model of the Atom: The positive charge and most of the mass resides in the nucleus as represented by the black dots. Only near-collisions with the very small, massive, and highly-positively-charged nuclei resulted in acute deflections.

1.9-8. Nuclear Model of the Atom

As a result of the work of Thomson, Millikan, and Rutherford, we had a very different view of the atom. It was no longer the "billiard ball" put forth by Dalton 100 years earlier. Based on his experiments and those of his contemporaries, Rutherford presented his nuclear model of the atom: All of the positive charge and most of the mass of the atom was contained in the very small nucleus at its center. The negative charge, but almost none of the mass, was carried by electrons, which orbited the nucleus much like the planets orbit the sun. His model was a giant step forward, but Rutherford was not satisfied with it because he could not understand why the electromagnetic force between the negatively-charged electrons and the positively-charged nucleus would not cause them to combine. The answer, which would dramatically alter the way we understand matter and the universe, would not be forthcoming for another decade.

1.10 Subatomic Particles, Isotopes, and Ions

Introduction

A third subatomic particle, the neutron, was discovered about 20 years after Rutherford introduced the nuclear model of the atom. The atom now consisted of three particles: neutrons and protons in the nucleus and electrons surrounding it. We now examine the effect that each of these particles has on the characteristics of the atom.

Objectives

1.10-1. Subatomic Particles

The major subatomic particles are:

particle mass (amu) charge
electron
5 × 10−4
–1
proton 1.0073 +1
neutron 1.0087 0

The proton is the source of the positive charge that balances the negative charge of the electrons to produce neutral atoms; i.e., the number of electrons equals the number of protons in an atom. The neutron is slightly more massive than the proton, but it carries no charge. Neutrons somehow keep the positively charged protons together in the nucleus.

1.10-2. Atomic Number, Z

It is the number of protons in the nucleus (the atomic number) that characterizes an atom.
The atomic number, Z, is the number of protons in the nucleus. It is the number that characterizes the element. Sulfur is the element that has sixteen protons in its nucleus; i.e., Z = 16 for sulfur. The number of neutrons determines the mass, and the number of electrons determines its charge. The atomic numbers of the first ten elements are:

H He Li Be B C N O F Ne
1 2 3 4 5 6 7 8 9 10

1.10-3. Mass Number, A

The mass number, A, is the number of protons plus neutrons in the nucleus.
The mass number, A, is the number of protons plus the number of neutrons in the nucleus. It is sometimes given as a superscript preceding the symbol. For example, an atom of 19F (read "fluorine-19") has 9 protons because it is fluorine and 10 neutrons because its mass is 19 (10 + 9 = 19). The mass of a neutron and a proton are each ~1 amu, so the atomic mass is close to 19 amu, the mass number. Note that the symbol 19F does not include the atomic number because the number of protons is known if the element (F) is known. However, it is sometimes included to aid in balancing nuclear reactions. In these cases, the atomic number is found as a subscript preceding the symbol.
A
Z
X =
number of protons & neutrons
number of protons
Symbol

1.10-4. Isotopes

Isotopes have the same Z but different A.
Isotopes are atoms with the same atomic number but different mass numbers. For example, naturally occurring chlorine is a mixture of two isotopes: 35Cl and 37Cl (chlorine-35 and chlorine-37). They are both chlorine atoms, so they both contain 17 protons, but they differ in the number of neutrons. Elements that have atomic masses that are not nearly integers exist in more than one isotope.
Example:
For example, the atomic mass of chlorine is 35.5 because naturally occurring chlorine is 75.8% 35Cl and 24.2% 37Cl. That is, one mole of chlorine contains 0.758 mol 35Cl and 0.242 mol 37Cl. The mass of one mole of chlorine atoms is therefore         
0.758 mol 35Cl
mol Cl
×
35.0 g Cl
mol 35Cl
+
0.242 mol 37Cl
mol Cl
×
37.0 g Cl
mol 37Cl
=
35.5 g Cl
mol Cl

1.10-5. Ions

An ion is a charged particle. The charge is shown as a superscript after the symbol.
Atoms are neutral because they have an equal number of protons and electrons. However, electrons can be added to or removed from atoms to produce ions.
Ion charge = number of protons − number of electrons
The charge of the ion is given as a superscript with the number and then the sign as in the F1– anion and the Ca2+ cation. (The "1" in 1– and 1+ ions is normally not written; however, it is included in this course because a superscript of "–" is too easily missed.) Remember that it is the number of protons that characterizes the element, and the number of protons does not change when an ion is formed from its elements. Thus, F and F1– each have 9 protons, but F has 9 electrons while F1– has 10. Similarly, Ca and Ca2+ each have 20 protons, but Ca has 20 electrons while Ca2+ has only 18 electrons.

1.10-6. Exercise on Particles in Ions

Exercise 1.17:

Indicate whether each species is an atom, a cation, or an anion, and give the number of electrons, protons, and neutrons present.
The atomic numbers of the elements can be found in the Elements resource.
    27Al3+
    Type:
  • atom The species is positively charged, so it is a cation.
  • cation
  • anion The species is positively charged, so it is a cation.
Protons = 13_0__ The atomic number of aluminum is 13, so there are 13 protons.
Neutrons = 14_0__ The number of neutrons equals the mass number minus the atomic number
= AZ = 27 − 13 = 14 neutrons.
Electrons = 10_0__ The +3 charge indicates that the number of electrons is three less than the number of protons, so Al3+ has 10 electrons.
    32S2–
    Type:
  • atom The species is negatively charged, so it is an anion.
  • cation The species is negatively charged, so it is an anion.
  • anion
Protons = 16_0__ The atomic number of sulfur is 16, so there are 16 protons.
Neutrons = 16_0__ The number of neutrons equals the mass number minus the atomic number
= AZ = 32 − 16 = 16 neutrons.
Electrons = 18_0__ The –2 charge indicates that the number of electrons is two greater than the number of protons, so S2– has 18 electrons.
    56Fe2+
    Type:
  • atom The species is positively charged, so it is a cation.
  • cation
  • anion The species is positively charged, so it is a cation.
Protons = 26_0__ The atomic number of iron is 26, so there are 26 protons.
Neutrons = 30_0__ The number of neutrons equals the mass number minus the atomic number
= AZ = 56 − 26 = 30 neutrons.
Electrons = 24_0__ The +2 charge indicates that the number of electrons is two less than the number of protons, so Fe2+ has 24 electrons.
    80Br
    Type
  • atom
  • cation The species is neutral, so it is an atom.
  • anion The species is neutral, so it is an atom.
Protons = 35_0__ The atomic number of bromine is 35, so there are 35 protons.
Neutrons = 45_0__ The number of neutrons equals the mass number minus the atomic number
= AZ = 80 − 35 = 45 neutrons.
Electrons = 35_0__ The number of electrons in an atom equals the number of protons.

1.10-7. Exercise on Writing Symbols

Exercise 1.18:

Determine the symbol for the listed species.
The atomic numbers of the elements can be found in the Elements resource.
Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, ^1H^1+ for 1H1+.
6 protons, 8 electrons, 6 neutrons o_^12C^2-_ins The element that contains 6 protons is carbon. The mass number = 6 protons + 6 neutrons = 12. The charge = 6 protons – 8 electrons = –2. So the species is 12C2–, an anion.
28 protons, 26 electrons, 32 neutrons o_^60Ni^2+_s The element that contains 28 protons is nickel. The mass number = 28 protons + 32 neutrons = 60. The charge = 28 protons – 26 electrons = +2. So the species is 60Ni2+, a cation.
32 protons, 32 electrons, 40 neutrons o_^72Ge_s The element that contains 32 protons is germanium. The mass number = 32 protons + 40 neutrons = 72. The charge = 32 protons – 32 electrons = 0. So the species is 72Ge, an atom..
15 protons, 18 electrons, 16 neutrons o_^31P^3-_s The element that contains 15 protons is phosphorus. The mass number = 15 protons + 16 neutrons = 31. The charge = 15 protons – 18 electrons = –3. So the species is 31P3–, an anion.

1.10-8. Exercise on Protons, Neutrons and Electrons

Exercise 1.19:

Determine the number of protons, neutrons and electrons for the species below.
(a)    16O2–
Protons = 8_0__ The number of protons in the nucleus is given by the atomic number, which is the number above the element's symbol on the periodic chart.
Neutrons = 8_0__ The number of neutrons equals the mass number minus the number of protons.
16 − 8 = 8.
Electrons = 10_0__ The number of electrons equals the number of protons minus the charge on the ion.
8 + 2 = 10.
(b)    27Al3+
Protons = 13_0__ The number of protons in the nucleus is given by the atomic number, which is the number above the element's symbol on the periodic chart.
Neutrons = 14_0__ The number of neutrons equals the mass number minus the number of protons.
27 − 13 = 14.
Electrons = 10_0__ The number of electrons equals the number of protons minus the charge on the ion.
13 − 3 = 10.
(c)    25Mg
Protons = 12_0__ The number of protons in the nucleus is given by the atomic number, which is the number above the element's symbol on the periodic chart.
Neutrons = 13_0__ The number of neutrons equals the mass number minus the number of protons.
25 − 12 = 13.
Electrons = 12_0__ The number of electrons equals the number of protons in an atom.
(d)    19F
Protons = 9_0__ The number of protons in the nucleus is given by the atomic number, which is the number above the element's symbol on the periodic chart.
Neutrons = 10_0__ The number of neutrons equals the mass number minus the number of protons.
19 − 9 = 10.
Electrons = 9_0__ The number of electrons equals the number of protons in an atom.
(e)    48Ti4+
Protons = 22_0__ The number of protons in the nucleus is given by the atomic number, which is the number above the element's symbol on the periodic chart.
Neutrons = 26_0__ The number of neutrons equals the mass number minus the number of protons.
48 − 22 = 26.
Electrons = 18_0__ The number of electrons equals the number of protons minus the charge on the ion.
22 − 4 = 18.

1.11 Dimitri Mendeleev and The Periodic Law

Introduction

As the number of known elements and the properties grew, it became clear that there had to be some way of organizing all of the information if chemistry was to grow as a science. The organization came in what is now called the periodic law. We now turn our study to this very powerful tool.

Objectives

Periodic Trends

Arranged in the order of their atomic numbers, the elements exhibit periodicity in their chemical and physical properties.
photo of Dimitri Mendeleev
Sixty years after Dalton published his atomic theory but 28 years before Thomson's discovery of the electron, Dimitri Mendeleev, a Russian chemist, was writing a textbook. As he tried to determine how best to break the book into chapters, he placed all of the elements in order of increasing atomic mass. He noticed that both the physical and chemical properties of the elements varied in a periodic manner. In order to maintain the periodicity, he had to reverse order of two elements (Te and I) and insert some spaces for yet undiscovered elements. His observations were summarized as the periodic law in 1869: The elements, if arranged in an order that closely approximates that of their atomic weights, exhibit an obvious periodicity in their properties. The reason some elements had to be reversed is that it is the atomic number, not atomic weight, that characterizes an element, but atomic numbers were unknown when he discovered this relationship. Today, the periodic law is stated slightly differently. Mendeleev arranged the elements in rows of a length such that elements of similar properties fell directly beneath one another. Elements that fell in the same column had similar properties and formed chemical families or groups. Elements that fell in the same row formed a period. The properties of the elements in a period changed gradually in going from left to right in the period. This behavior is demonstrated for a physical property (melting points) and a chemical property (formula of oxides) in the following section.

1.11-2. Examples

plot of physical properties
Figure 1.6: Periodicity of Physical Properties
Figure 1.6 shows a plot of the melting points of the elements. Elements in the same family or group share the same color. Note that K and Ar have been reversed from the order of mass.
plot of chemical properties
Figure 1.7: Periodicity of Chemical Properties
Figure 1.7 represents the maximum number of oxygen atoms that combine with one atom of each element. For example, nitrogen is at 2.5 because it forms an oxide with the formula N2O5, so there are 2.5 oxygen atoms for each nitrogen atom. Oxygen itself has been omitted. Also, K and Ar have been reversed from the order of mass.

1.11-3 Periodicity Exercise

Exercise 1.20:

Identify each formula. Indicate any subscripted characters with an underscore (_) and any superscripted characters with a carat (^). For example, NH_4^1+ for NH41+.
What is the formula of potassium sulfide if that of sodium sulfide is Na2S?
o_K_2S_s Na and K are in the same chemical group, so they have similar chemical properties.
What is the formula of calcium chloride if that of sodium chloride is NaCl and that of aluminum chloride is AlCl3?
o_CaCl_2_s Na, Ca, and Al are in the same period, so their chemical properties vary continuously.
What is the formula of fluoride of nitrogen if those of carbon and oxygen are CF4 and OF2?
o_NF_3_s The number of fluorines combined to elements in the same period varies continuously.
What is the formula of aluminum oxide given the formulas Na2O, MgO and SiO2?
o_Al_2O_3_s The number oxygen atoms per metal atom is in the order 0.5, 1, x, 2, so x must be 1.5 (3/2).
What is the formula of phosphorus oxide given the information in the previous question?
o_P_2O_5_s The next number in the oxygen per metal atom series is 2.5 (5/2).

1.11-4. Metals and Nonmetals

Elements on the left side of the periodic table are metallic, and elements on the right side are nonmetallic.
Elements fall into three classes: The elements on the left side of a period are metallic, those on the right are nonmetallic, and those lying between the two broader classes are metalloid. Thus, the elements start out metallic and become less metallic and more nonmetallic as you move from lower order number to higher order number (left to right) within a period.
color bar
In the above image, the gradual change of colors from blue to yellow represents the manner in which the elements change gradually in a period from metallic on the left to nonmetallic on the right. There are one or two elements between the two larger classes in each period that have properties of both. These elements are called metalloids.

1.11-5. Information Available in the Periodic Table

The periodic table not only gives the order of the elements, it also gives important information about each element. Indeed, some periodic tables can show a large amount of data for each element. However, the periodic table found in the resource titled Periodic Table gives the information as shown in the figure.
atomic number, chemical symbol, atomic mass

1.11-6. Other Periodic Tables

Mendeleev's arrangement of the elements has become known as the periodic chart or the periodic table and is the source of a great deal of information about the physical and chemical properties of the elements. The columns define groups, which consist of elements with similar properties. The rows define periods, which contain elements whose properties change gradually. There are two methods of numbering the groups in the periodic table, the American method, 1A–8A and 1B–8B, and the newer method, which numbers the groups as 1–18. Both numbering schemes are shown on the Periodic Table. However, we use the older method in discussions in this text. The elements discussed in this course can also be classified in the following way: A current periodic table that includes chemical symbols, atomic masses, and atomic numbers can be found by going to the "Resources Menu" above and selecting "Periodic Table." For more information about the elements, click "WebElements" below and to see the use of the elements in comic books, click "Comic Books."

1.11-7. Some Commonly Named Groups

Several chemical groups have common names that we will use throughout the course. Note that the last element in each group is radioactive and not included in the figure.
common groups
Figure 1.8: Common Named Chemical Groups

1.11-8. Periodic Table Exercise

Exercise 1.21:

Use the Periodic Table to identify each element by symbol.
The halogen in the third period
o_Cl_ins The halogens are the members of Group 7A.
The alkaline earth in the fourth period
o_Ca_ins The alkaline earths are the members of Group 2A.
The Group 4 metalloid in the fourth period
o_Ge_s There are two metalloids in Group 4A and two in the fourth period. All are used in the semiconductor industry.
The noble gas in the fifth period
o_Xe_s The noble gases are the members of Group 8A.
The third transition metal in the first row transition metals
o_V_s The transition metals occupy the B Groups. The first row transition metals cover Sc through Zn.

1.12 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.