Chapter 2. Components of Matter PDF

2 The Components of Matter 2.1 Elements, Compounds, 2.3 The Observations That Led to the 2.6 Compounds: Introduction and Mixtures: An Atomic Nuc...

2 The Components of Matter 2.1 Elements, Compounds, 2.3 The Observations That Led to the 2.6 Compounds: Introduction and Mixtures: An Atomic Nuclear Atom Model to Bonding Overview Discovery of the Electron Formation of Ionic Compounds Substances Discovery of the Nucleus Formation of Covalent Substances Mixtures 2.4 The Atomic Theory Today 2.7 Compounds: Formulas, Names, Classification and Separation Structure of the Atom and Masses of Mixtures Atomic Number, Mass Number, Ionic Compounds 2.2 The Observations That Led to an and Atomic Symbol Acid Names from Anion Names Atomic View of Matter Isotopes Binary Covalent Compounds Mass Conservation Atomic Masses Straight-Chain Alkanes Definite Composition 2.5 Elements: A First Look at the Molecular Masses Multiple Proportions Periodic Table Formulas and Models Dalton’s Atomic Theory An Overview of the Components of Matter Source: (left): Perry Mastrovito/Image Source/Getty Images; (right): McGraw Hill 40 ISTUDY  Concepts and Skills to Review Before You Study This Chapter › physical and chemical change (Section 1.1) › meaning of a scientific model (Section 1.2) › states of matter (Section 1.1) › SI units and conversion factors (Sections 1.3 and 1.5) › attraction and repulsion between charged particles (Section 1.1) › significant figures in calculations (Section 1.4) T he simple “lead” pencil does not actually contain lead! Instead, a form of carbon known as graphite is responsible for the pencil marks we make on paper. The graphite pencil tip is composed of stacked flat sheets of carbon atoms that are arranged in rings, an arrangement that looks like chicken wire (photo). The carbon sheets can slide past each other, depositing atoms of carbon from the pencil tip to the paper. Atoms not only make up our pencil doo- dles—atoms are the ultimate particles that make up all things. The first record of an atomic theory of matter dates back to ancient Greece when the Greek philosopher Democritus (c. 460–370 bc) reasoned that if matter, such as a piece of aluminum foil, were cut into smaller and smaller pieces, there would eventu- ally be a particle of aluminum so small that it could no longer be cut. Therefore, matter is ultimately composed of indivisible particles with nothing but empty space between them. He called the particles atoms (Greek atomos, “indivisible”). This rev- olutionary idea about matter was suppressed for 2000 years because the influential philosopher Aristotle rejected it. Finally, in the 17th century, the English scientist Robert Boyle argued that, by definition, an element is composed of “simple Bodies, not made of any other Bodies, of which all mixed Bodies are compounded, and into which they are ultimately resolved,” a description remarkably close to our idea of an element, with atoms being the “simple Bodies.” The next two centuries saw rapid progress in chemistry and the development of a “billiard-ball” image of the atom. Then, an early 20th-century burst of creativity led to our current model of an atom with a complex internal structure. IN THIS CHAPTER... We examine the properties and composition of matter on the macro- scopic and atomic scales. › We relate the three types of observable matter—elements, compounds, and mixtures—to the atoms, ions, and molecules they consist of. › We classify mixtures and see how to separate them in the lab. › We see how the defining properties of the types of matter relate to 18th-century laws ­concerning the masses of substances that react with each other. › We examine the 19th-century atomic model proposed to explain these laws. › We describe some 20th- and 21st-century experiments that have led to our current under- standing of atomic structure and atomic mass. › We focus on the general organization of the elements in the periodic table and introduce the two major ways that elements combine. › We derive names and formulas of compounds and calculate their masses on the atomic scale. › We examine some of the ways chemists depict molecules. › We present a final overview of the components of matter. 2.1 ELEMENTS, COMPOUNDS, AND MIXTURES: AN ATOMIC OVERVIEW Based on its composition, matter can be broadly classified into two types—substances and mixtures. A substance is matter with a fixed composition while a mixture consists of two or more substances that are physically combined with a variable composition. 41 ISTUDY  42   Chapter 2 The Components of Matter Substances Substances can be further subdivided into elements and compounds: 1. Elements. An element is the simplest type of matter with unique physical and chemical properties. The following are some features of elements: ∙ An element consists of only one kind of atom, the smallest unit of an element that retains its chemical nature, and therefore cannot be broken down into a simpler type of matter by any physical or chemical methods. ∙ Each element is unique because the properties of its atoms are unique. An element has a name, such as silicon, oxygen, or copper. A sample of silicon contains only silicon atoms, and a sample of copper contains only copper atoms. The macro- scopic properties of a piece of silicon, such as color, density, and combustibility, are different from those of a piece of copper because the submicroscopic properties of silicon atoms are different from those of copper atoms. A Atoms of an element ∙ In nature, most elements exist as populations of atoms, either separated or in con- tact with each other, depending on the physical state. Figure 2.1A shows atoms of an element in its gaseous state. ∙ Several elements occur in molecular form; a molecule is an independent structure of two or more atoms bound together (Figure 2.1B). Oxygen, for example, occurs in air as diatomic (two-atom) molecules, O2. 2. Compounds. A compound consists of two or more different elements that are bonded chemically (Figure 2.1C). That is, the elements in a compound are not just mixed together: their atoms have joined in a chemical reaction. Many compounds, such as ammonia, water, and carbon dioxide, consist of molecules. But many others, like sodium chloride (which we’ll discuss shortly) and silicon dioxide, do not. B Molecules of an element All compounds have three defining features: ∙ The elements are present in fixed parts by mass (have a fixed mass ratio). This is so because each unit of the compound consists of a fixed number of atoms of each element. For example, consider a sample of ammonia. It is 14 parts nitrogen by mass and 3 parts hydrogen by mass because 1 nitrogen atom has 14 times the mass of 1 hydrogen atom, and each ammonia molecule consists of 1 nitrogen atom and 3 hydrogen atoms: Ammonia gas is 14 parts N by mass and 3 parts H by mass. N 1 N atom has 14 times the mass of 1 H atom. Each ammonia molecule consists of 1 N atom and 3 H atoms. H C Molecules of a compound ∙ A compound’s properties are different from the properties of its elements. Table 2.1 shows a striking example: soft, silvery sodium metal and yellow-green, poisonous chlorine gas are very different from the compound they form—white, crystalline sodium chloride, or common table salt! ∙ A compound, unlike an element, can be broken down into simpler substances—its component elements—by a chemical change. For example, an electric current breaks down molten sodium chloride into metallic sodium and chlorine gas. Mixtures In a mixture, two or more substances (elements and/or compounds) are physically D Mixture of two elements intermingled, not chemically combined. and a compound Mixtures differ from compounds as follows: Figure 2.1 Elements, compounds, and mixtures on the atomic scale. The samples ∙ The components of a mixture can vary in their parts by mass. A mixture of the depicted here are gases, but the types of compounds sodium chloride and water, for example, can have many different parts matter also occur as liquids and solids. by mass of salt to water. ISTUDY  2.1 Elements, Compounds, and Mixtures: An Atomic Overview    43 Table 2.1 Some of the Very Different Properties of Sodium, Chlorine, and Sodium Chloride Property Sodium + Chlorine → Sodium Chloride Melting point 97.8°C −101°C 801°C Boiling point 881.4°C −34°C 1413°C Color Silvery Yellow-green Colorless (white) Density 0.97 g/cm3 0.0032 g/cm3 2.16 g/cm3 Behavior in water Reacts Dissolves slightly Dissolves freely Source: (Sodium, Chlorine, Sodium chloride) Stephen Frisch/McGraw Hill ∙ A mixture retains many of the properties of its components. Saltwater, for instance, is colorless like water and tastes salty like sodium chloride. The component prop- erties are maintained because, on the atomic scale, a mixture consists of the indi- vidual units of its component elements and/or compounds (Figure 2.1D). ∙ Mixtures, unlike compounds, can be separated into their components by physical changes; chemical changes are not needed. For example, the water in saltwater can be boiled off, a physical process that leaves behind solid sodium chloride. Some other separation methods are discussed below. Classification and Separation of Mixtures In the natural world, matter usually occurs as mixtures. A sample of clean air, for exam- ple, consists of many elements and compounds physically mixed together, including oxygen, nitrogen, carbon dioxide, water vapor, and a few other gaseous elements. The oceans are mixtures of numerous substances dissolved in water, and living things contain thousands of substances­—carbohydrates, proteins, and nuclei acids, to name a few. The two broad classes of mixtures—heterogeneous mixtures and homogeneous mixtures (solutions)—are detailed in Table 2.2. Table 2.2 Heterogeneous and Homogeneous Mixtures Heterogeneous Mixtures Homogeneous Mixtures (Solutions) ∙ One or more visible boundaries among ∙ Has no visible boundaries the components ∙ Has a uniform composition ∙ The composition is not uniform through- ∙ The components of a homogenous out the mixture, but rather varies from mixture are uniformly intermingled on one region of the mixture to another a molecular level ∙ In some heterogeneous mixtures like ∙ Cannot visually tell whether a sample milk, the boundaries can only be seen of matter is a homogeneous mixture or with a microscope an element or compound. ∙ Example: sand (colored blue) mixed ∙ Example: blue copper(II) sulfate mixed with water with water Source: Holly Curry/McGraw Hill; Matt Meadows/McGraw Hill ISTUDY  44   Chapter 2 The Components of Matter MIXTURE The difference between a mixture and a compound is well illustrated using iron and sulfur as components (Figure 2.2). Any proportions of iron metal filings and pow- dered sulfur form a mixture. The components can be separated with a magnet because iron metal is magnetic. But if we heat the container strongly, the components will form the compound iron(II) sulfide (FeS). The magnet can then no longer remove the iron because it has undergone a chemical change and lost its magnetic properties. Chemists have devised many procedures for separating a mixture into its compo- nents; the techniques depend on the differences in the physical properties of the ­substances in the mixture; no chemical changes occur: S8 ∙ Filtration is based on differences in particle size and is often used to separate a solid from a liquid, which flows through the tiny holes in filter paper as the solid is retained. In vacuum filtration, reduced pressure below the filter speeds the flow Fe of the liquid through it. ∙ Crystallization is based on differences in solubility. The solubility of a substance is A the amount that dissolves in a fixed volume of solvent at a given temperature. Since solubility often increases with temperature, the impure solid is dissolved in hot sol- COMPOUND vent and, when the mixture cools, the purified compound solidifies (crystallizes). ∙ Distillation separates components through differences in volatility, the tendency of a substance to become a gas. Simple distillation separates components with large differences in volatility, such as water from dissolved ionic compounds. The mixture is boiled, producing a vapor that is richer in the more volatile component—in this case, water. The water boils off and is condensed and collected separately, leaving the solid compound behind. Fractional distillation (discussed in Chapter 13) uses many vaporization-condensation steps to separate components with small volatility differences, such as those in petroleum. ∙ Chromatography is also based on differences in solubility. The mixture is dissolved S2– in a gas or liquid, and the components are separated as this solution, called the mobile phase, flows through a solid (or viscous liquid) called the stationary phase. Fe2+ A component with lower solubility in the stationary phase moves through it faster than a component with a higher solubility. In one technique, called gas-liquid chro- matography (GLC), the mobile phase is an inert gas, such as helium, that carries a B gaseous mixture of components into a long tube packed with the stationary phase Figure 2.2 The distinction between (Figure 2.3A). The components emerge separately and reach a detector. A chromato- mixtures and compounds. A, A mixture gram has numerous peaks, each representing the amount of a specific component of iron and sulfur consists of the two (Figure 2.3B). High-performance (high-pressure) liquid chromatography (HPLC) is ­elements. B, The compound iron(II) sulfide similar to GLC, but the mixture need not be vaporized, since the mobile phase is consists of an array of Fe2+ and S2− ions. a liquid, so more heat-sensitive components can be separated. Source: Stephen Frisch/McGraw Hill Solubility in stationary phase 20 18 16 Detector response 14 12 10 8 6 Time 4 He gas He gas He gas He gas 2 0 Gaseous mixture (blue and red) is carried Component (blue) that is more soluble 22 24 26 28 30 32 34 36 A into column with mobile phase. in stationary phase moves slower. B Time (minutes) Figure 2.3 Principle of gas-liquid chromatography (GLC). The stationary phase is shown as a viscous liquid (gray circles) coating the solid beads (yellow) of an inert packing. ISTUDY  2.1 Elements, Compounds, and Mixtures: An Atomic Overview    45 The following sample problem will help you to differentiate elements, compounds, and mixtures. SAMPLE PROBLEM 2.1 Distinguishing Elements, Compounds, and Mixtures at the Atomic Scale Problem The scenes below represent atomic-scale views of three samples of matter: (a) (b) (c) Describe each sample as an element, a compound, or a mixture. Plan We have to determine the type of matter by examining the component particles. If a sample contains only one type of particle, it is either an element or a compound; if it contains more than one type, it is a mixture. Particles of an element have only one kind of atom (one color of sphere in an atomic-scale view), and particles of a compound have two or more kinds of atoms. Solution (a) Mixture: there are three different types of particles. Two types contain only one kind of atom, either green or purple, so they are elements (composed of diatomic molecules), and the third type contains two red atoms for every one yellow, so it is a compound. (b) Element: the sample consists of only blue atoms. (c) Compound: the sample consists of molecules that each have two black and six blue atoms. FOLLOW-UP PROBLEMS Brief Solutions for all Follow-up Problems appear at the end of the chapter. 2.1A Does each of the following scenes best represent an element, a compound, or a mixture? (a) (b) (c) 2.1B Describe the following representation of a reaction in terms of elements, compounds, and mixtures. SOME SIMILAR PROBLEMS 2.3, 2.4, and 2.14 ISTUDY  46   Chapter 2 The Components of Matter Matter Separation by Mixtures Substances physical methods Homogeneous Heterogeneous Separation by Compounds Elements mixtures mixtures chemical methods Figure 2.4 The classification of matter. Figure 2.4 summarizes the classification of matter that we have discussed in this section. › Summary of Section 2.1 › All matter exists as either elements, compounds, or mixtures. › Every element or compound is a substance, matter with a fixed composition. › An element consists of only one type of atom and occurs as a collection of individual atoms or molecules; a molecule consists of two or more atoms chemically bonded together. › A compound contains two or more elements chemically combined and exhibits different properties from its component elements. The elements occur in fixed parts by mass because each unit of the compound has a fixed number of each type of atom. Only a chemical change can break down a compound into its elements. › A mixture consists of two or more substances mixed together, not chemically combined. The components exhibit their individual properties, can be present in any proportion, and can be separated by physical changes. › Heterogeneous mixtures have visible boundaries among the components. › Homogeneous mixtures (solutions) have no visible boundaries because mixing occurs at the molecular level. › Separation methods are based on differences in physical properties and include filtration (particle size), crystallization (solubility), distillation (volatility), and chromatography (solubility). 2.2 THE OBSERVATIONS THAT LED TO AN ATOMIC VIEW OF MATTER Any model of the composition of matter had to explain three so-called mass laws: the law of mass conservation, the law of definite (or constant) composition, and the law of multiple proportions. Mass Conservation The first mass law, stated by Lavoisier on the basis of combustion experiments, was the most fundamental chemical observation of the 18th century: ∙ Law of mass conservation: the total mass of substances does not change during a chemical reaction. ISTUDY  2.2 The Observations That Led to an Atomic View of Matter    47 Solid lead chromate in Lead nitrate Sodium sodium nitrate solution chromate solution solution Mass BEFORE reaction...... equals mass AFTER reaction. Figure 2.5 The law of mass conservation. Source: Stephen Frisch/McGraw Hill The number of substances may change and, by definition, their properties must, but the total amount of matter remains constant. Figure 2.5 illustrates mass conservation because the lead nitrate and sodium chromate solutions (left) have the same mass as the solid lead chromate in sodium nitrate solution (right) that forms after their reaction. Even in a complex biochemical change, such as the metabolism of the sugar glucose, which involves many reactions, mass is conserved. For example, in the reac- tion of, say, 180 g of glucose, with oxygen, we have 180 g glucose + 192 g oxygen gas ⟶ 264 g carbon dioxide + 108 g water 372 g material before ⟶ 372 g material after Mass conservation means that, based on all chemical experience, matter cannot be created or destroyed. To be precise about it, however, we now know, based on the work of Albert Einstein (1879–1955), that the mass before and after a reaction is not exactly the same. Some mass is converted to energy, or vice versa, but the difference is too small to measure, even with the best balance. For example, when 100 g of carbon burns, the carbon dioxide formed weighs 0.000000036 g (3.6×10−8 g) less than the sum of the carbon and oxygen that reacted. Because the energy changes of chemical reactions are so small, for all practical purposes, mass is conserved. Later in the text, you’ll see that energy changes in nuclear reactions are so large that mass changes are easy to measure. Definite Composition We stated earlier that there is a fixed proportion of elements in a particular compound; we can express that fixed proportion in two ways: ∙ The fraction by mass (mass fraction) is the part of the compound’s mass that each element contributes. It is obtained by dividing the mass of each element in the compound by the mass of the compound: mass of element X in compound A ​​ _____________________________ Mass fraction =          ​​ mass of compound A 1.0 g 1.0 g 1.0 g ∙ The percent by mass (mass percent, mass %) is the fraction by mass expressed 2.0 g 2.0 g as a percentage (multiplied by 100): 3.0 g 3.0 g 3.0 g Mass percent = mass fraction × 100 For an everyday example, consider a box that contains three types of marbles: 16.0 g marbles yellow ones weigh 1.0 g each, purple 2.0 g each, and red 3.0 g each. Each type ISTUDY  48   Chapter 2 The Components of Matter No matter what makes up a fraction of the total mass of marbles, 16.0 g. The mass fraction of yellow the source of a marbles is their number times their mass divided by the total mass: compound... no. of yellow marbles × mass of yellow marbles Mass fraction of yellow marbles = ​​ _______________________________________           ​​ total mass of marbles 3 ​marbles​ × 1.0 ​ g​/ ​marble​ _______________________ = ​​         ​​ = 0.19 16.0 ​g​ The mass percent (parts per 100 parts) of yellow marbles is 0.19 × 100 = 19% by mass. Similarly, the purple marbles have a mass fraction of 0.25 and represent 25% CALCIUM CARBONATE 40 mass % calcium by mass of the total, and the red marbles have a mass fraction of 0.56 and represent 12 mass % carbon 56% by mass of the total. 48 mass % oxygen Now let’s apply this idea to the compound calcium carbonate, composed of three elements—calcium, carbon, and oxygen. Calcium carbonate is the major compound in seashells, marble, and coral. Let’s consider a 20.0 g sample of calcium carbonate from a marble statue (Sample I) and a 45.0 g sample from coral (Sample II):... its elements occur in the same proportion by mass. Analysis by Mass Mass Fraction Mass Percent 8.0 g calcium _____________________ Sample I 8.0 g calcium ​​       ​​= 0.40 0.40 × 100 = 40.% 20.0 g 20.0 g calcium carbonate 2.4 g carbon _____________________ 2.4 g carbon ​​       ​​= 0.12 0.12 × 100 = 12% 20.0 g calcium carbonate Figure 2.6 The law of definite compo- sition. Calcium carbonate occurs in many 9.6 g oxygen _____________________ 9.6 g oxygen ​​       ​​= 0.48 0.48 × 100 = 48% forms (such as marble, top, and coral, 20.0 g calcium carbonate bottom). 18.0 g calcium _____________________ Source: (top): Matthia Cortesi (Roman Sample II 18.0 g calcium ​​       ​​= 0.40 0.40 × 100 = 40.% sculptor active 100–150 CE)/atthiam/123RF; 45.0 g 45.0 g calcium carbonate (bottom): Alexander Cherednichenko/ 5.4 g carbon _____________________ Shutterstock 5.4 g carbon ​​       ​​= 0.12 0.12 × 100 = 12% 45.0 g calcium carbonate 21.6 g oxygen 21.6 g oxygen ​​ _____________________       ​​= 0.48 0.48 × 100 = 48% 45.0 g calcium carbonate We can draw two conclusions from this data: ∙ The sum of the mass fractions (or mass percents) equals 1.00 part (or 100%) by mass. ∙ Both samples of this compound contain 40.% calcium, 12% carbon, and 48% oxygen by mass (Figure 2.6). The second conclusion is expressed as the second mass law: ∙ Law of definite (or constant) composition: no matter what its source, a particular compound is composed of the same elements in the same parts (fractions) by mass. Because a given element always constitutes the same mass fraction of a given compound, we can use that mass fraction to find the actual mass of the element in any sample of the compound: Mass of element in sample mass of element in compound = mass of compound in sample × _________________________     ​​      ​​ (2.1) mass of compound For example, knowing that a 20.0-g sample of calcium carbonate contains 8.0 g of calcium, the mass of calcium in a 75.0-g sample of calcium carbonate can be calculated: 8.0 g calcium Mass (g) of calcium = 75.0 ​​g calcium carbonate​​ × ______________________    ​​     ​ 20.0 ​g calcium carbonate​ ​= 30.0 g calcium ISTUDY  2.2 The Observations That Led to an Atomic View of Matter    49 SAMPLE PROBLEM 2.2 Calculating the Mass of an Element in a Compound Problem Pitchblende is the most important compound of uranium. Mass analysis of Road Map an 84.2-g sample of pitchblende shows that it contains 71.4 g of uranium, with oxygen the only other element. How many grams of uranium and of oxygen are in 102 kg of Mass (kg) of pitchblende pitchblende? Plan We have to find the masses (in g) of uranium and of oxygen in a known mass multiply by mass ratio of uranium (102 kg) of pitchblende, given the mass of uranium (71.4 g) in a different mass of to pitchblende from analysis pitchblende (84.2 g) and knowing that oxygen is the only other element present. The mass ratio of uranium to pitchblende in grams is the same as it is in kilograms. Mass (kg) of uranium Therefore, using Equation 2.1, we multiply the mass (in kg) of the pitchblende sample by the ratio of uranium to pitchblende (in kg) from the mass analysis. This 1 kg = 1000 g gives the mass (in kg) of uranium, and we convert kilograms to grams. To find the mass of oxygen, the only other element in the pitchblende, we subtract the calculated mass of uranium (in kg) from the given mass of pitchblende (102 kg) and convert Mass (g) of uranium kilograms to grams. Road Map Solution Finding the mass (kg) of uranium in 102 kg of pitchblende: mass (kg)Mass of uranium in pitchblende Mass (kg) of uranium = mass (kg) of pitchblende × _______________________________     ​​      (kg) of pitchblende  ​​ Mass (kg) of pitchblende mass (kg) of pitchblende 71.4 kg uranium multiply by mass ratio of uranium subtract mass (kg) of uranium = 102 ​​kg pitchblende​​ × _________________    ​​      ​​= 86.5 to pitchblende fromkg uranium analysis 84.2 ​kg pitchblende​ Converting the mass of uranium from kg to g: Mass (kg) of uranium Mass (kg) of oxygen 1000 g Mass (g) of uranium = 86.5 ​​kg​​ uranium × ______ ​​   ​​= 8.65×104 g uranium 1 ​kg​ 1 kg = 1000 g 1 kg = 1000 g Finding the mass (in kg) of oxygen in 102 kg of pitchblende: Mass (kg) of oxygen = mass (kg) of pitchblende − massMass (kg) (g) of of uranium uranium Mass (g) of oxygen = 102 kg − 86.5 kg = 15.5 kg oxygen Converting the mass of oxygen from kg to g: 1000 g Mass (g) of oxygen = 15.5 ​​kg​​ oxygen × ​​ ______ ​​= 1.55×104 g oxygen 1 ​kg​ Check The analysis showed that most of the mass of pitchblende is due to uranium, so the large mass of uranium makes sense. Rounding off to check the math gives ​​ 70 ​​ = 82 kg uranium ∼100 kg pitchblende × ___ 85 FOLLOW-UP PROBLEMS 2.2A The mineral “fool’s gold” does not contain any gold; instead, it is a compound composed only of the elements iron and sulfur. A 110.0-g sample of fool’s gold contains 51.2 g of iron. What mass of sulfur is in a sample of fool’s gold that Student Hot Spot contains 86.2 g of iron? 2.2B Silver bromide is the light-sensitive compound coated onto black-and-white film. Student data indicate that you may struggle with using mass fraction to calculate the mass of an A 26.8-g sample contains 15.4 g of silver, with bromine as the only other element. How element in a compound. Access the eBook to many grams of each element are on a roll of film that contains 3.57 g of silver bromide? view an additional Learning Resource video on this topic. SOME SIMILAR PROBLEMS 2.30–2.33 Multiple Proportions It’s quite common for the same two elements to form more than one compound—sulfur and fluorine do this, as do phosphorus and chlorine and many other pairs of elements. The third mass law we consider applies in these cases: ∙ Law of multiple proportions: if elements A and B react to form two compounds, the different masses of B that combine with a fixed mass of A can be expressed as a ratio of small whole numbers. ISTUDY  50   Chapter 2 The Components of Matter Consider two compounds of carbon and oxygen; let’s call them carbon oxides I and II. These compounds have very different properties: the density of carbon oxide I is 1.25 g/L, whereas that of II is 1.98 g/L; I is poisonous and flammable, but II is not. Mass analysis shows that Carbon oxide I is 57.1 mass % oxygen and 42.9 mass % carbon Carbon oxide II is 72.7 mass % oxygen and 27.3 mass % carbon To demonstrate the phenomenon of multiple proportions, we use the mass percents of oxygen and of carbon to find their masses in a given mass, say 100 g, of each com- pound. Then we divide the mass of oxygen by the mass of carbon in each compound to obtain the mass of oxygen that combines with a fixed mass of carbon: Carbon Oxide I Carbon Oxide II g oxygen/100 g compound 57.1 72.7 g carbon/100 g compound 42.9 27.3 57.1 72.7 g oxygen/g carbon = 1.33 = 2.66 42.9 27.3 If we then divide the grams of oxygen per gram of carbon in II by that in I, we obtain a ratio of small whole numbers: 2.66 g oxygen/g carbon in II 2 __________________________ ​​      ​​ = ​​ __      ​​ 1.33 g oxygen/g carbon in I 1 The law of multiple proportions tells us that, in two compounds of the same ele- ments, the mass fraction of one element relative to that of the other element changes in increments based on ratios of small whole numbers. In this case, the ratio is 2/1— for a given mass of carbon, compound II contains 2 times as much oxygen as I, not 1.583 times, 1.716 times, or any other intermediate amount. The simplest arrangement consistent with the mass data for carbon oxides I and II is that one atom of oxygen combines with one atom of carbon in compound I (carbon monoxide) and that two atoms of oxygen combine with one atom of carbon in compound II (carbon dioxide): C O O C O Carbon oxide I Carbon oxide II (carbon monoxide) (carbon dioxide) Dalton’s Atomic Theory With over 200 years of hindsight, it may be easy to see how the mass laws could be explained by the idea that matter exists in indestructible units, each with a particular mass and set of properties, but it was a major breakthrough in 1808 when John Dalton (1766–1844) presented his atomic theory of matter. Postulates of the Atomic Theory Dalton expressed his theory in a series of pos- tulates, presented here in modern terms: 1. All matter consists of atoms—tiny, indivisible units of an element that cannot be created or destroyed. 2. Atoms of one element cannot be converted into atoms of another element. In chemical reactions, the atoms of the original substances recombine to form different substances. 3. Atoms of an element are identical in mass and other properties and are different from atoms of any other element. 4. Compounds result from the chemical combination of a specific ratio of atoms of different elements. How the Theory Explains the Mass Laws Let’s see how Dalton’s postulates explain the mass laws: ∙ Mass conservation. Atoms cannot be created or destroyed (postulate 1) or converted into other types of atoms (postulate 2). Therefore, a chemical reaction cannot pos- sibly result in a mass change because atoms are just combined differently. ISTUDY  2.2 The Observations That Led to an Atomic View of Matter    51 ∙ Definite composition. A compound is a combination of a specific ratio of different atoms (postulate 4), each of which has a particular mass (postulate 3). Thus, each element in a compound must constitute a fixed fraction of the total mass. ∙ Multiple proportions. Atoms of an element have the same mass (postulate 3) and are indivisible (postulate 1). The masses of element B that combine with a fixed mass of element A must give a small, whole-number ratio because different num- bers of B atoms combine with each A atom in different compounds, as we saw in the previous example with carbon monoxide and carbon dioxide. Let’s work through a sample problem that reviews the mass laws. SAMPLE PROBLEM 2.3 Visualizing the Mass Laws Problem The scenes below represent an atomic-scale view of a chemical reaction: Which of the mass laws—mass conservation, definite composition, and/or multiple proportions—is (are) illustrated? Plan From the depictions, we note the numbers, colors, and combinations of atoms (spheres) to see which mass laws pertain. If the numbers of each atom are the same before and after the reaction, the total mass did not change (mass conservation). If a compound forms that always has the same atom ratio, the elements are present in fixed parts by mass (definite composition). If the same elements form different compounds and the ratio of the atoms of one element that combine with one atom of the other element is a small whole number, the ratio of their masses is a small whole number as well (multiple proportions). Solution There are seven purple and nine green atoms in each circle, so mass is conserved. The compound formed has one purple and two green atoms, so it has definite composition. Only one compound forms, so the law of multiple proportions does not pertain. FOLLOW-UP PROBLEMS 2.3A The following scenes represent a chemical change. Which of the mass laws is (are) illustrated? 2.3B Which sample(s) best display(s) the fact that compounds of bromine (orange) and fluorine (yellow) exhibit the law of multiple proportions? Explain. A B C SOME SIMILAR PROBLEMS 2.22 and 2.23 ISTUDY  52   Chapter 2 The Components of Matter › Summary of Section 2.2 › The law of mass conservation states that the total mass remains constant during a chemical reaction. › The law of definite composition states that any sample of a given compound has the same elements present in the same parts by mass. › The law of multiple proportions states that, in different compounds of the same elements, the masses of one element that combine with a fixed mass of the other can be expressed as a ratio of small whole numbers. › Dalton’s atomic theory explained the mass laws by proposing that all matter consists of indivisible, unchangeable atoms of fixed, unique mass. › Mass is conserved during a reaction because the atoms retain their identities but are combined differently. › Each compound has a fixed mass fraction of each of its elements because it is composed of a fixed number of each type of atom. › Different compounds of the same elements exhibit multiple proportions because they each consist of whole atoms. 2.3 THE OBSERVATIONS THAT LED TO THE NUCLEAR ATOM MODEL One of the postulates of Dalton’s atomic theory was that atoms were indivisible and thus it did not predict the existence of subatomic charged particles, which were observed in experiments at the turn of the 20th century that led to the discovery of electrons and the atomic nucleus. Let’s examine some of the experiments that resolved questions about Dalton’s model and led to our current model of atomic structure. Discovery of the Electron and Its Properties For many years, scientists knew that matter and electric charge are related. When amber is rubbed with fur, or glass with silk, positive and negative charges form—the same charges that make your hair crackle and cling to your comb on a dry day (see photo). Scientists also knew that an electric current can decompose certain compounds into their elements. But they did not know what a current is made of. Cathode Rays To discover the nature of an electric current, some investigators tried passing it through nearly evacuated glass tubes fitted with metal electrodes. When the Effect of electric charges. electric power source was turned on, a “ray” could be seen striking the phosphor- Source: Bob Coyle/McGraw Hill coated end of the tube, which emitted a glowing spot of light. The rays were called cathode rays because they originated at the negative electrode (cathode) and moved to the positive electrode (anode). Figure 2.7 shows some properties of cathode rays based on these observations. The main conclusion was that cathode rays consist of negatively charged particles found in all matter. The rays become visible as their particles collide with the few remaining gas molecules in the evacuated tube. Cathode ray particles were later named electrons. There are many familiar cases of the effects of charged particles colliding with gas particles or hitting a phosphor-coated screen: ∙ In a “neon” sign, electrons collide with the gas particles in the tube, causing them to give off light. ∙ The aurora borealis, or northern lights (see photo), occurs when Earth’s magnetic Aurora borealis display. field bends streams of charged particles coming from the Sun, which then collide Source: Roman Krochuk/Shutterstock with gas particles in the atmosphere to give off light. ISTUDY  2.3 The Observations That Led to the Nuclear Atom Model    53 Phosphor-coated end of tube 1 Cathode ray – N OBSERVATION CONCLUSION Anode 2 + 1 Ray bends in Consists of S magnetic field. charged 3 particles – + Evacuated tube 2 Ray bends toward Consists of Cathode positive plate negative in electric field. particles Positive plate 3 Ray is identical Particles for any cathode. found in Magnet all matter Figure 2.7 Observations that established the properties of cathode rays. ∙ In older televisions and computer monitors, the cathode ray passes back and forth over the phosphor-coated screen, creating a pattern that we see as a picture. Mass and Charge of the Electron Two classic experiments and their conclusion revealed the mass and charge of the electron: 1. Charge/mass ratio. In 1897, the British physicist J. J. Thomson (1856–1940) used a cathode ray tube to measure the ratio of the electron’s electrical charge to its mass and obtained a value of −1.76×108 coulombs/g (the coulomb, C, is the SI unit of electrical charge). This value indicated that the electron was about 2000 times lighter than hydrogen, the lightest atom! Thomson was shocked because this implied that, contrary to Dalton’s atomic theory, atoms contain even smaller particles. Fellow sci- entists reacted with disbelief to Thomson’s conclusion; some even thought he was joking. 2. Charge. In 1909, the American physicist Robert Millikan (1868–1953) measured the charge of the electron. He did so by observing the movement of tiny oil droplets in an apparatus that contained electrically charged plates and an x-ray source (Figure 2.8). X-rays knocked electrons from gas molecules in the air within the apparatus, and the Fine mist of oil is sprayed into apparatus. Oil droplets fall through hole in positively charged plate. X-rays knock electrons from air molecules, and they stick to droplets. Electrically charged + plates influence droplet’s motion. – Observer times X-ray droplet’s motion and source controls electric field. Figure 2.8 Millikan’s oil-drop experiment for measuring an electron’s charge. The total charge on an oil droplet is some whole-number multiple of the charge of the electron. ISTUDY  54   Chapter 2 The Components of Matter electrons stuck to an oil droplet falling through a hole in a positively charged plate. With the electric field off, Millikan measured the mass of the now negatively charged droplet from its rate of fall. Then, by adjusting the field’s strength, he made the drop- let slow and hang suspended, which allowed him to measure its total charge. After many tries, Millikan found that the total charge of the various droplets was always some whole-number multiple of a minimum charge. If different oil droplets picked up different numbers of electrons, Millikan reasoned that this minimum charge must be the charge of the electron itself. Remarkably, the value that he calculated over a century ago was within 1% of the modern value of the electron’s charge, −1.602×10−19 C. 3. Conclusion: calculating the electron’s mass. The value of the electron’s charge multiplied by its mass/charge ratio gives the electron’s mass, which is extremely small: g ​​  mass  ​​ = −1.602×10−19 ​​ C​​ × ____________ Mass of electron = ​​charge​​ × ______ ​​      ​​ = 9.10×10−28 g ​charge​ −1.76×108 ​C​ Discovery of the Atomic Nucleus The presence of electrons in all matter brought up two major questions about the structure of atoms. Matter is electrically neutral, so atoms must be also. But if atoms contain negatively charged electrons, what positive charges balance them? And if an electron has such a tiny mass, what accounts for an atom’s much larger mass? To address these issues, Thomson proposed his “plum-pudding” model—a spherical atom composed of diffuse, positively charged matter with electrons embedded in it like “raisins in a plum pudding.” In 1910, New Zealand–born physicist Ernest Rutherford (1871–1937) tested this model and obtained a very unexpected result (Figure 2.9): 1. Experimental design. Figure 2.9A shows the experimental setup, in which tiny, dense, positively charged alpha (α) particles emitted from radium are aimed at gold foil. A circular, zinc-sulfide screen registers the deflection (scattering angle) of the α par- ticles emerging from the foil by emitting light flashes when the particles strike it. A Experiment B Hypothesis: All α particles will go straight through “plum-pudding” atoms. Incoming α particles Radioactive sample emits beam of Beam strikes α particles. gold foil. Particles strike zinc-sulfide screen and produce flashes. Almost no deflection Cross section of gold foil: "plum-pudding" atoms C Actual result: A few α particles undergo major deflections by nuclear atoms. Gold foil Incoming Major deflection: α particles seen very rarely No deflection: seen most often Minor deflection: Major Minor seen occasionally deflection deflection Cross section of gold foil: atoms with tiny, massive, positive nucleus Figure 2.9 Rutherford’s α-scattering experiment and discovery of the atomic nucleus. ISTUDY  2.4 The Atomic Theory Today    55 2. Hypothesis and expected results. With Thomson’s model in mind (Figure 2.9B), Rutherford expected only minor, if any, deflections of the α particles because they should act as bullets and go right through the gold atoms. After all, he reasoned, an electron should not be able to deflect an α particle any more than a Ping-Pong ball could deflect a baseball. 3. Actual results. Initial results were consistent with this hypothesis, but then the unex- pected happened (Figure 2.9C). As Rutherford recalled: “I remember two or three days later Geiger [one of his coworkers] coming to me in great excitement and saying, ‘We have been able to get some of the α particles coming backwards... ’ It was quite the most incredible event that has ever happened to me in my life. It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” The data showed very few α particles deflected at all and only 1 in 20,000 deflected by more than 90° (“coming backwards”). 4. Rutherford’s conclusion. Rutherford concluded that these few α particles were being repelled by something small, dense, and positive within the gold atoms. Calculations based on the mass, charge, and velocity of the α particles and the fraction of these large-angle deflections showed that ∙ An atom is mostly space occupied by electrons. ∙ In the center is a tiny region, which Rutherford called the nucleus, that contains all the positive charge and essentially all the mass of the atom. He proposed that positive particles lay within the nucleus and called them protons. Rutherford’s model explained the charged nature of matter, but it could not account for all the atom’s mass. After more than 20 years, in 1932, James Chadwick (1891–1974) discovered the neutron, an uncharged, dense particle that also resides in the nucleus. › Summary of Section 2.3 › Several major discoveries at the turn of the 20th century resolved questions about Dalton’s model and led to our current model of atomic structure. › Cathode rays were shown to consist of negative particles (electrons) that exist in all matter. J. J. Thomson measured their charge/mass ratio and concluded that they are much smaller and lighter than atoms. › Robert Millikan determined the charge of the electron, which he combined with other data to calculate its mass. › Ernest Rutherford proposed that atoms consist of a tiny, massive, positively charged nucleus surrounded by electrons. 2.4 THE ATOMIC THEORY TODAY Dalton’s model of an indivisible particle has given way to our current model of an atom with an elaborate internal architecture of subatomic particles. Structure of the Atom The Tiny, Massive Nucleus An atom is an electrically neutral, spherical entity composed of a positively charged central nucleus surrounded by one or more negatively charged electrons (Figure 2.10). A few analogies can help you grasp the incredible properties of the Let’s examine some of its features: atomic nucleus. A nucleus the size of ∙ The electrons move rapidly within the available volume, held there by the attraction of the ­period at the end of a sentence the nucleus. (To indicate their motion, they are often depicted as a cloudy blur of color, would weigh about 100 tons, as much darkest around a central dot—the nucleus—and becoming paler toward the outer edge.) as 50 cars! An atom the size of the ∙ The nucleus consists of protons and neutrons (the only exception is the simplest Houston Astrodome would have a nucleus of the element hydrogen, which is just a proton). ­nucleus the size of a green pea that ∙ The proton (p+) has a positive charge and the neutron (n0) has no charge; all of would contain virtually all the stadium’s mass. If the nucleus were about 1 cm the positive charge of the nucleus results from its protons. in diameter, the atom would be about ∙ The charge of the proton is the same magnitude as the charge of an electron (e−) 200 m across, or more than the length but with the opposite sign. An atom is neutral because the number of protons in of two football fields! the nucleus equals the number of electrons surrounding the nucleus. ISTUDY  56   Chapter 2 The Components of Matter Approximately 1×10–10 m Approximately 5×10–15 m Nucleus Proton, p+ Electrons, e– (positive charge) (negative charge) Neutron, n0 (no charge) Atom Nucleus Figure 2.10 General features of the atom. ∙ The mass of a proton or neutron is nearly 2000 times larger than the mass of the electron. ∙ An atom’s diameter (~1×10−10 m) is about 20,000 times the diameter of its nucleus (~5×10−15 m). ∙ The nucleus contributes 99.97% of the atom’s mass but occupies only about 1 quadrillionth of its volume, so it is incredibly dense: about 1014 g/mL! › Some properties of these three subatomic particles are listed in Table 2.3. Atomic Number, Mass Number, and Atomic Symbol We use a standard atom notation to describe an atom of an element; this atom nota- tion has three parts (Figure 2.11A): 1. Atomic number: ∙ The atomic number (Z) of an element equals the number of protons in the nucleus of each of its atoms. ∙ The Z value is written as a left subscript in the notation. ∙ Since the number of protons must equal the number of electrons in an atom, the atomic number also equals the number of electrons in the atom. ∙ All atoms of an element have the same atomic number, and the atomic number of each element is different from that of any other element. All carbon atoms (Z = 6) have 6 protons, all oxygen atoms (Z = 8) have 8 protons, and all uranium atoms (Z = 92) have 92 protons. 2. Mass number: ∙ The mass number (A) is the total number of protons and neutrons in the nucleus of an atom. ∙ The A value is written as a left superscript in the notation. ∙ Each proton and each neutron contributes one unit to the mass number. Thus, a carbon atom with 6 protons and 6 neutrons in its nucleus has a mass number of 12, and a uranium atom with 92 protons and 146 neutrons in its nucleus has a mass number of 238. Table 2.3 Properties of the Three Key Subatomic Particles Charge Mass Name (Symbol) Relative Absolute (C)* Relative (amu)† Absolute (g) Location in Atom + −19 Proton (p ) 1+ +1.60218×10 1.00727 1.67262×10 −24 Nucleus Neutron (n0) 0   0 1.00866 1.67493×10−24 Nucleus − Electron (e ) 1− −1.60218×10−19 0.00054858 9.10939×10−28 Outside nucleus *The coulomb (C) is the SI unit of charge. − †The atomic mass unit (amu) equals 1.66054×10 24 g; it is discussed later in this section. ISTUDY  2.4 The Atomic Theory Today    57 3. Atomic symbol: Mass number ∙ The atomic symbol (or element symbol), designated as X in Figure 2.11A, is a A X (p+ + n0) Atomic one- or two-letter abbreviation for the English, Latin, or Greek name of an element. symbol ∙ Examples are C for carbon, Cl for chlorine, and Na for sodium (Latin natrium). Atomic number Z The number of neutrons (N) is not formally part of the atom notation. But since the A (p+) mass number is the sum of protons and neutrons, the number of neutrons in an atom can be calculated by subtracting the atomic number from the mass number: 6e– Number of neutrons = mass number − atomic number, or N=A−Z (2.2) 6p+ 12 35 6n0 6 C Thus, a chlorine atom, symbolized 17Cl, has A = 35, Z = 17, and N = 35 − 17 = 18. Because each element has its own atomic number, we also know the atomic number An atom of carbon -12 from the symbol. For example, instead of writing 126 C for carbon with mass number 12, we can write 12C (spoken “carbon twelve”), with Z = 6 understood. Another way to 8e– name this atom is carbon-12. 8p+ 16 O 8n0 8 Isotopes An atom of oxygen -16 All atoms of an element have the same atomic number but not the same mass number. Isotopes of an element are atoms that have different numbers of neutrons and therefore 92e– different mass numbers. Most elements occur in nature in a particular isotopic composition, which specifies the proportional abundance of each of its isotopes. For example, all carbon 92p+ atoms (Z = 6) have 6 protons and 6 electrons, but only 98.89% of naturally occurring 235 U 143n0 92 carbon atoms have 6 neutrons (A = 12). A small percentage (1.11%) have 7 neutrons (A = 13), and even fewer (less than 0.01%) have 8 (A = 14). These are carbon’s three natu- rally occurring isotopes—12C, 13C, and 14C. A natural sample of carbon has these three isotopes in these relative proportions. Five other carbon isotopes—9C, 10C, 11C, 15C, and An atom of uranium-235 16 C—have been created in the laboratory. Figure 2.11B depicts the atomic number, mass 92e– number, and symbol for four atoms, two of which are isotopes of the element uranium. A key point is that the chemical properties of an element are primarily determined 92p+ by the number of electrons, so all isotopes of an element have nearly identical chemi- 238 U 146n0 92 cal behavior, even though they have different masses. SAMPLE PROBLEM 2.4 Determining the Numbers of Subatomic Particles in the Isotopes of an Element B An atom of uranium-238 Figure 2.11 Atom notation. A, The Problem Silicon (Si) is a major component of semiconductor chips. It has three naturally 28 29 30 meaning of the ZAX notation. B, Notations occurring isotopes: Si, Si, and Si. Determine the numbers of protons, electrons, and and spherical representations for four neutrons in an atom of each silicon isotope. ­atoms. (The nuclei are not drawn to scale.) Plan The mass number (A; left superscript) of each of the three isotopes is given, which is the sum of protons and neutrons. From the List of Elements in the front of the book, we find the atomic number (Z, number of protons), which equals the number of electrons. We obtain the number of neutrons by subtracting Z from A (Equation 2.2). Solution From the List of Elements, the atomic number of silicon is 14. Therefore, 28 Si has 14p+, 14e−, and 14n0 (28 − 14) 29 Si has 14p+, 14e−, and 15n0 (29 − 14) 30 Si has 14p+, 14e−, and 16n0 (30 − 14) FOLLOW-UP PROBLEMS 2.4A Titanium, the ninth most abundant element, is used structurally in many objects, such as electric turbines, aircraft bodies, and bicycle frames. It has five naturally occurring isotopes: 46Ti, 47Ti, 48Ti, 49Ti, and 50Ti. How many protons, electrons, and neutrons are in an atom of each isotope? 2.4B An atom has a mass number of 88 and has 50 neutrons. (a) What is the element? (b) Another isotope of this element has a mass number of 86. How many neutrons does this isotope have? SOME SIMILAR PROBLEMS 2.46–2.49 ISTUDY  58   Chapter 2 The Components of Matter Atomic Masses of the Elements The mass of an atom is measured relative to the mass of an atomic standard. The modern standard is the carbon-12 atom, whose mass is defined as exactly 12 atomic mass units. Thus, the atomic mass unit (amu) is ​​ __1 12 ​​the mass of a carbon-12 atom. Based on this standard, the H atom has a mass of 1.008 amu; in other words, a 12C 1 atom has almost 12 times the mass of an 1H atom. We will continue to use the term atomic mass unit in the text, although the name of the unit has been changed to the dalton (Da); thus, one 12C atom has a mass of 12 daltons (12 Da, or 12 amu). The atomic mass unit is a relative unit of mass, but it is equivalent to an absolute mass of 1.66054×10−24 g. Finding Atomic Mass from Isotopic Composition The isotopic composition of an element is determined by mass spectrometry, a method for measuring the relative masses and abundances of atomic-scale particles very precisely. In this powerful technique, high-energy electrons collide with a particle—say, an atom of neon-20—knocking away one of the atom’s electrons to produce a particle that has one positive charge, Ne+ (Figure 2.12A). Thus, the particle has a mass/charge ratio (m /e) that equals its mass divided by 1+. Figure 2.12B depicts the process for a sample of neon’s three naturally occurring isotopes. The positively charged particles are attracted toward a series of negatively charged plates with slits in them. Some particles pass through the slits into an evacuated tube exposed to a magnetic field. As the particles enter this region, their paths are bent; the lightest particles (lowest m/e) are deflected most and the heaviest particles (highest m/e) least. At the end of the region, the particles strike a detector, which records their relative positions and abundances (Figure 2.12C). The relative abundance is the percentage (or fraction) of each isotope in a sample of the element. For example, the relative abundance of 20Ne is 90.48% (or 0.9048), which Figure 2.12 The mass spectrometer and its data. High-energy electron 20