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Chemistry: A Molecular Approach, 2nd Ed. Nivaldo Tro Chapter 2 Atoms and Elements Roy Kennedy Massachusetts Bay Community College Wellesley Hills, MA Tro: Chemistry: A Molecular Approach, 2/e 1 Copyright 2011 Pearson Education, Inc. Scanning Tunneling Microscope • Gerd Bennig and Heinrich Rohrer found that as you pass a sharp metal tip over a flat metal surface, the amount of current that flows varies with distance between the tip and the surface • Measuring this “tunneling” current allowed them to scan the surface on an atomic scale – essentially taking pictures of atoms on the surface Tro: Chemistry: A Molecular Approach, 2/e 2 Copyright 2011 Pearson Education, Inc. Operation of a STM Tro: Chemistry: A Molecular Approach, 2/e 3 Copyright 2011 Pearson Education, Inc. Scanning Tunneling Microscope • Later scientists found that not only can you see the atoms on the surface, but the instrument allows you to move individual atoms across the surface Tro: Chemistry: A Molecular Approach, 2/e 4 Copyright 2011 Pearson Education, Inc. Early Philosophy of Matter • Some early philosophers believed that matter had an ultimate, tiny, indivisible particle Leucippus and Democritus • Other philosophers believed that matter was infinitely divisible Plato and Aristotle • Because there was no experimental way of proving who was correct, the best debater was the person assumed correct, i.e., Aristotle Tro: Chemistry: A Molecular Approach, 2/e 5 Copyright 2011 Pearson Education, Inc. Scientific Revolution • In the late 17th century, the scientific approach • to understanding nature became established For the next 150+ years, observations about nature were made that could not easily be explained by the infinitely divisible matter concept Tro: Chemistry: A Molecular Approach, 2/e 6 Copyright 2011 Pearson Education, Inc. Law of Conservation of Mass • In a chemical reaction, • matter is neither created nor destroyed Total mass of the materials you have before the reaction must equal the total mass of the materials you have at the end total mass of reactants = total mass of products Tro: Chemistry: A Molecular Approach, 2/e 7 Antoine Lavoisier 1743-1794 Copyright 2011 Pearson Education, Inc. Reaction of Sodium with Chlorine to Make Sodium Chloride • The mass of sodium and chlorine used is determined by • the number of atoms that combine Because only whole atoms combine and atoms are not changed or destroyed in the process, the mass of sodium chloride made must equal the total mass of sodium and chlorine atoms that combine together 7.7 g Na Tro: Chemistry: A Molecular Approach, 2/e + 11.9 g Cl2 8 19.6 g NaCl Copyright 2011 Pearson Education, Inc. Law of Definite Proportions Joseph Proust 1754-1826 • All samples of a given compound, regardless of their source or how they were prepared, have the same proportions of their constituent elements Tro: Chemistry: A Molecular Approach, 2/e 9 Copyright 2011 Pearson Education, Inc. Proportions in Sodium Chloride A 100.0 g sample of sodium chloride contains 39.3 g of sodium and 60.7 g of chlorine A 200.0 g sample of sodium chloride contains 78.6 g of sodium and 121.4 g of chlorine A 58.44 g sample of sodium chloride contains 22.99 g of sodium and 35.44 g of chlorine Tro: Chemistry: A Molecular Approach, 2/e 10 Copyright 2011 Pearson Education, Inc. Example 2.1: Show that two samples of carbon dioxide obey the Law of Definite Proportions Given: Find: Conceptual Plan: Relationships: Sample 1: 25.6 g O and 9.60 g C Sample 2: 21.6 g O and 8.10 g C proportion O:C g O 1, g C1 g O 2, g C2 O:C in each sample all samples of a compound have the same proportion of elements by mass Solution: Check: both samples have the same O:C ratio, so the result is consistent with the Law of Definite Proportions Tro: Chemistry: A Molecular Approach, 2/e 11 Copyright 2011 Pearson Education, Inc. Practice – If a 10.0 g sample of calcite contains 4.0 g of calcium, how much calcite contains 0.24 g of calcium? Tro: Chemistry: A Molecular Approach, 2/e 12 Copyright 2011 Pearson Education, Inc. Practice – How much calcite contains 0.24 g of calcium? Given: Find: Conceptual Plan: Sample 1: 4.0 g Ca and 10.0 g calcite Sample 2: 0.24 g Ca mass calcite in Sample 2, g g Ca1, g calcite1 g Ca2 g calcite2 because Relationships: Solution: Sig Figs & Round: Tro: Chemistry: A Molecular Approach, 2/e 0.6 g calcite = 0.60 g calcite 13 Copyright 2011 Pearson Education, Inc. Law of Multiple Proportions John Dalton 1766-1844 • When two elements (call them A and B) form two different compounds, the masses of B that combine with 1 g of A can be expressed as a ratio of small, whole numbers Tro: Chemistry: A Molecular Approach, 2/e 14 Copyright 2011 Pearson Education, Inc. Oxides of Carbon • Carbon combines with oxygen to form two different compounds, carbon monoxide and carbon dioxide • Carbon monoxide contains 1.33 g of oxygen for every 1.00 g of carbon • Carbon dioxide contains 2.67 g of oxygen for every 1.00 g of carbon • Because there are twice as many oxygen atoms per carbon atom in carbon dioxide of in carbon monoxide, the oxygen mass ratio should be 2 Tro: Chemistry: A Molecular Approach, 2/e 15 Copyright 2011 Pearson Education, Inc. Example 2.2: Show that two oxides of nitrogen are consistent with the Law of Multiple Proportions Given: nitrogen dioxide: 2.28 g O per 1 g N dinitrogen monoxide: 0.570 g O per 1 g N Find: O in nitrogen dioxide:O in dinitrogen monoxide Conceptual g O in nit. dioxide O1:O2 Plan: g O in dinit. monox. Relationships: samples of different compounds that have the same elements show proportions by mass that are small whole number ratios Solution: Check: because the compounds have O:O ratio that is a small whole number, so the results are consistent with the Law of Multiple Proportions. Tro: Chemistry: A Molecular Approach, 2/e 16 Copyright 2011 Pearson Education, Inc. Practice – Hematite contains 2.327 g of Fe for every 1.00 g of oxygen. Wüsite contains 3.490 g of Fe per gram of oxygen. Show these results are consistent with the Law of Multiple Proportions. Tro: Chemistry: A Molecular Approach, 2/e 17 Copyright 2011 Pearson Education, Inc. Practice ─ Show that two oxides of iron are consistent with the Law of Multiple Proportions Given: hematite: 2.327 g Fe per 1 g O wüsite: 3.490 g Fe per 1 g O Find: Fe in wüsite:Fe in hematite Conceptual g Fe in wüsite Plan: g Fe in hematite Relationships: Fe1:Fe2 samples of different compounds that have the same elements show proportions by mass that are small whole number ratios Solution: 1.5 = 3:2 Check: because the compounds have Fe:Fe ratio that is a small whole number, the results are consistent with the Law of Multiple Proportions Tro: Chemistry: A Molecular Approach, 2/e 18 Copyright 2011 Pearson Education, Inc. Dalton’s Atomic Theory • 1. 2. 3. 4. Dalton proposed a theory of matter based on it having ultimate, indivisible particles to explain these laws Each element is composed of tiny, indestructible particles called atoms All atoms of a given element have the same mass and other properties that distinguish them from atoms of other elements Atoms combine in simple, whole-number ratios to form molecules of compounds In a chemical reaction, atoms of one element cannot change into atoms of another element they simply rearrange the way they are attached Tro: Chemistry: A Molecular Approach, 2/e 19 Copyright 2011 Pearson Education, Inc. Practice – Decide if each statement is correct according to Dalton’s model of the atom • Copper atoms can combine with zinc atoms to make gold atoms • Water is composed of many identical molecules that have one oxygen atom and two hydrogen atoms Tro: Chemistry: A Molecular Approach, 2/e 20 Copyright 2011 Pearson Education, Inc. Practice – Decide if each statement is correct according to Dalton’s model of the atom • Copper atoms can combine with zinc atoms to make gold atoms – incorrect; according to Dalton, atoms of one element cannot turn into atoms of another element by a chemical reaction. He knew this because if atoms could change it would change the total mass and violate the Law of Conservation of Mass. • Water is composed of many identical molecules that have one oxygen atom and two hydrogen atoms – correct; according to Dalton, atoms combine together in compounds in small whole-number ratios, so that you could describe a compound by describing the number of atoms of each element in a molecule. He used this idea to explain why compounds obey the Law of Definite Proportions. Tro: Chemistry: A Molecular Approach, 2/e 21 Copyright 2011 Pearson Education, Inc. Practice – Decide if each statement is correct according to Dalton’s Model of the Atom • Some carbon atoms weigh more than other carbon atoms • Because the mass ratio of Fe:O in wüsite is 1.5 times larger than the Fe:O ratio in hematite, there must be 1.5 Fe atoms in a unit of wüsite and 1 Fe atom in a unit of hematite Tro: Chemistry: A Molecular Approach, 2/e 22 Copyright 2011 Pearson Education, Inc. Practice – Decide if each statement is correct according to Dalton’s model of the atom • Some carbon atoms weigh more than other carbon atoms • – incorrect; according to Dalton, all atoms of an element are identical. Because the mass ratio of Fe:O in wüsite is 1.5 times larger than the Fe:O ratio in hematite, there must be 1.5 Fe atoms in a unit of wüsite and 1 Fe atom in a unit of hematite – incorrect; according to Dalton, atoms must combine in small whole-number ratios. If you could combine fractions of atoms, that would mean the atom is breakable and Dalton’s first premise would be incorrect. You can get the Fe:Fe mass ratio to be 1.5 if the formula for wüsite is FeO and the formula for hematite is Fe2O3. Tro: Chemistry: A Molecular Approach, 2/e 23 Copyright 2011 Pearson Education, Inc. Some Notes on Charge • • Two kinds of charge called + and – Opposite charges attract • Like charges repel • + attracted to – + repels + – repels – To be neutral, something must have no charge or equal amounts of opposite charges Tro: Chemistry: A Molecular Approach, 2/e 24 Copyright 2011 Pearson Education, Inc. Cathode Ray Tube • • Glass tube containing metal electrodes from which almost all the air has been evacuated When connected to a high voltage power supply, a glowing area is seen emanating from the cathode Tro: Chemistry: A Molecular Approach, 2/e 25 Copyright 2011 Pearson Education, Inc. J.J. Thomson • Believed that the cathode ray was composed of • tiny particles with an electrical charge Designed an experiment to demonstrate that there were particles by measuring the amount of force it takes to deflect their path a given amount like measuring the amount of force it takes to make a car turn Tro: Chemistry: A Molecular Approach, 2/e 26 Copyright 2011 Pearson Education, Inc. Thomson’s Experiment Investigate the effect of placing an electric field around tube 1. charged matter is attracted to an electric field 2. light’s path is not deflected by an electric field +++++++++++ Cathode Anode (+) (-) ------------- Tro: Chemistry: A Molecular Approach, 2/e Power Supply 27 + Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 28 Copyright 2011 Pearson Education, Inc. Thomson’s Results • The cathode rays are made of tiny particles • These particles have a negative charge because the beam always deflected toward the + plate • The amount of deflection was related to two • • factors, the charge and mass of the particles Every material tested contained these same particles The charge:mass ratio of these particles was −1.76 x 108 C/g the charge/mass of the hydrogen ion is +9.58 x 104 C/g Tro: Chemistry: A Molecular Approach, 2/e 29 Copyright 2011 Pearson Education, Inc. Thomson’s Conclusions • If the particle has the same amount of charge as a hydrogen ion, then it must have a mass almost 2000x smaller than hydrogen atoms! later experiments by Millikan showed that the particle did have the same amount of charge as the hydrogen ion • The only way for this to be true is if these particles were pieces of atoms apparently, the atom is not unbreakable Tro: Chemistry: A Molecular Approach, 2/e 30 Copyright 2011 Pearson Education, Inc. Millikan’s Oil Drop Experiment Tro: Chemistry: A Molecular Approach, 2/e 31 Copyright 2011 Pearson Education, Inc. Thomson’s Conclusions, cont’d • Thomson believed that these particles were therefore the ultimate building blocks of matter “We have in the cathode rays matter in a new state, a state in which the subdivision of matter is carried very much further . . . a state in which all matter . . . is of one and the same kind; this matter being the substance from which all the chemical elements are built up.” • These cathode ray particles became known as electrons Tro: Chemistry: A Molecular Approach, 2/e 32 Copyright 2011 Pearson Education, Inc. Electrons • Electrons are tiny, negatively charged • • • particles found in all atoms Cathode rays are made of streams of electrons The electron has a charge of −1.60 x 1019 C The electron has a mass of 9.1 x 10−28 g Tro: Chemistry: A Molecular Approach, 2/e 33 Copyright 2011 Pearson Education, Inc. A New Theory of the Atom • Because the atom is no longer indivisible, Thomson must propose a new model of the atom to replace the first statement in Dalton’s Atomic Theory rest of Dalton’s theory still valid at this point • Thomson proposes that instead of being a hard, marble-like unbreakable sphere, the way Dalton described it, the atom actually had an inner structure Tro: Chemistry: A Molecular Approach, 2/e 34 Copyright 2011 Pearson Education, Inc. Thomson’s Plum Pudding Atom • The structure of the atom contains • many negatively charged electrons These electrons are held in the atom by their attraction for a positively charged electric field within the atom there had to be a source of positive charge because the atom is neutral Thomson assumed there were no positively charged pieces because none showed up in the cathode ray experiment Tro: Chemistry: A Molecular Approach, 2/e 35 Copyright 2011 Pearson Education, Inc. Predictions of the Plum Pudding Atom • The mass of the atom is due to the mass of the electrons within it electrons are the only particles in Plum Pudding atoms, therefore the only source of mass • The atom is mostly empty space should not have a bunch of negatively charged particles near each other as they would repel Tro: Chemistry: A Molecular Approach, 2/e 36 Copyright 2011 Pearson Education, Inc. Radioactivity • In the late 1800s, Henri Becquerel and • • Marie Curie discovered that certain elements would constantly emit small, energetic particles and rays These energetic particles could penetrate matter Ernest Rutherford discovered that there were three different kinds of emissions Marie Curie 1867-1934 alpha, a, rays made of particles with a mass 4x H atom and + charge beta, b, rays made of particles with a mass ~1/2000th H atom and – charge gamma, g, rays that are energy rays, not particles Tro: Chemistry: A Molecular Approach, 2/e 37 Copyright 2011 Pearson Education, Inc. Rutherford’s Experiment • How can you prove something is empty space? • Put something through it! use large target atoms use very thin sheets of target so it will not absorb “bullet” use very small particle as bullet with very high energy but not so small that electrons will affect it • Bullet = alpha particles, target atoms = gold foil a particles have a mass of 4 amu & charge of +2 c.u. gold has a mass of 197 amu & is very malleable Tro: Chemistry: A Molecular Approach, 2/e 38 Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 39 Copyright 2011 Pearson Education, Inc. Rutherford’s Results • Over 98% of the a particles went straight • • through About 2% of the a particles went through but were deflected by large angles About 0.005% of the a particles bounced off the gold foil “...as if you fired a 15” cannon shell at a piece of tissue paper and it came back and hit you.” Tro: Chemistry: A Molecular Approach, 2/e 40 Copyright 2011 Pearson Education, Inc. Rutherford’s Conclusions • Atom mostly empty space because almost all the particles went straight through • Atom contains a dense particle that is small in volume compared to the atom but large in mass because of the few particles that bounced back • This dense particle is positively charged because of the large deflections of some of the particles Tro: Chemistry: A Molecular Approach, 2/e 41 Copyright 2011 Pearson Education, Inc. Plum Pudding Atom • • • • • • • • • • A few of the a particles do not go through • • • • • • • • • • • • If atom was like a plum pudding, all the a particles should go straight through Nuclear Atom . . . Tro: Chemistry: A Molecular Approach, 2/e Almost all a particles go straight through Some a particles go through, but are deflected due to +:+ repulsion from the nucleus 42 Copyright 2011 Pearson Education, Inc. Rutherford’s Interpretation – the Nuclear Model 1. The atom contains a tiny dense center called the nucleus the amount of space taken by the nucleus is only about 1/10 trillionth the volume of the atom 2. The nucleus has essentially the entire mass of the atom the electrons weigh so little they give practically no mass to the atom 3. The nucleus is positively charged the amount of positive charge balances the negative charge of the electrons 4. The electrons are dispersed in the empty space of the atom surrounding the nucleus Tro: Chemistry: A Molecular Approach, 2/e 43 Copyright 2011 Pearson Education, Inc. Structure of the Nucleus • Rutherford proposed that the nucleus had a particle that had the same amount of charge as an electron but opposite sign – these particles are called protons based on measurements of the nuclear charge of the elements • protons are subatomic particles found in the nucleus with a charge = +1.60 x 1019 C and a mass = 1.67262 x 10−24 g • Because protons and electrons have the same amount of charge, for the atom to be neutral there must be equal numbers of protons and electrons Tro: Chemistry: A Molecular Approach, 2/e 44 Copyright 2011 Pearson Education, Inc. Relative Mass and Charge • It is sometimes easier to compare things to each other • • rather than to an outside standard When you do this, the scale of comparison is called a relative scale We generally talk about the size of charge on atoms by comparing it to the amount of charge on an electron, which we call −1 charge units proton has a charge of +1 cu protons and electrons have equal amounts of charge, but opposite signs • We generally talk about the mass of atoms by comparing it to 1/12th the mass of a carbon atom with 6 protons and 6 neutrons, which we call 1 atomic mass unit protons have a mass of 1 amu electrons have a mass of 0.00055 amu, which is generally too small to be relevant Tro: Chemistry: A Molecular Approach, 2/e 45 Copyright 2011 Pearson Education, Inc. Some Problems • How could beryllium have four protons stuck together in the nucleus? shouldn’t they repel each other? • If a beryllium atom has four protons, then it should weigh 4 amu; but it actually weighs 9.01 amu! Where is the extra mass coming from? each proton weighs 1 amu remember, the electron’s mass is only about 0.00055 amu and Be has only four electrons – it can’t account for the extra 5 amu of mass Tro: Chemistry: A Molecular Approach, 2/e 46 Copyright 2011 Pearson Education, Inc. There Must Be Something Else! • To answer these questions, Rutherford and • Chadwick proposed that there was another particle in the nucleus – it is called a neutron Neutrons are subatomic particles with a mass = 1.67493 x 10−24 g and no charge, and are found in the nucleus 1 amu slightly heavier than a proton no charge Tro: Chemistry: A Molecular Approach, 2/e 47 Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 48 Copyright 2011 Pearson Education, Inc. Why Does Matter Appear Continuous If the Atom Is Mostly Empty Space? • The emptiness of the atom is on • such a small scale that the variations in density cannot be seen Most of our macroscopic observations involve particles colliding that are so much larger than this scale that the particles appear solid instead of mostly empty Tro: Chemistry: A Molecular Approach, 2/e 49 Copyright 2011 Pearson Education, Inc. Elements • Each element has a unique number of protons • in its nucleus The number of protons in the nucleus of an atom is called the atomic number the elements are arranged on the Periodic Table in order of their atomic numbers • Each element has a unique name and symbol symbol either one or two letters one capital letter or one capital letter and one lowercase letter Tro: Chemistry: A Molecular Approach, 2/e 50 Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 51 Copyright 2011 Pearson Education, Inc. The Periodic Table of the Elements The atomic number tells you how many protons are in the nucleus and how many electrons are in the atom Some symbols are one capital letter, like S, come from the element ‘s C, name, and I. for Others are two letters, and the the second like C carbon. Others come from Latin is lowercase, like Br and name of the element, likeSr Au for gold (aurum) and Cu for copper (cuprium) Tro: Chemistry: A Molecular Approach, 2/e 52 Copyright 2011 Pearson Education, Inc. Structure of the Nucleus • Soddy discovered that the same element could have atoms with different masses, which he called isotopes there are two isotopes of chlorine found in nature, one that has a mass of about 35 amu and another that weighs about 37 amu • The observed mass is a weighted average of the weights of all the naturally occurring atoms the percentage of an element that is one isotope is called the isotope’s natural abundance the atomic mass of chlorine is 35.45 amu Tro: Chemistry: A Molecular Approach, 2/e 53 Copyright 2011 Pearson Education, Inc. Isotopes • All isotopes of an element are chemically identical undergo the exact same chemical reactions • All isotopes of an element have the same number • • • of protons Isotopes of an element have different masses Isotopes of an element have different numbers of neutrons Isotopes are identified by their mass numbers, which is the sum of all the protons and neutrons in the nucleus Tro: Chemistry: A Molecular Approach, 2/e 54 Copyright 2011 Pearson Education, Inc. Isotopes • Atomic number Number of protons Z • Mass Number Protons + neutrons whole number A • Abundance = relative amount found in a sample Tro: Chemistry: A Molecular Approach, 2/e 55 Copyright 2011 Pearson Education, Inc. Neon Symbol Number of Number of A, Mass Protons Neutrons Number Percent Natural Abundance Ne-20 or 20 10 Ne 10 10 20 90.48% 21Ne Ne-21 or 10 10 11 21 0.27% 22 Ne Ne-22 or 10 10 12 22 9.25% Tro: Chemistry: A Molecular Approach, 2/e 56 Copyright 2011 Pearson Education, Inc. Example 2.3b: How many protons, electrons, 52 Cr and neutrons are in an atom of 24 ? Given: Find: Conceptual Plan: 52 Cr 24 therefore A = 52, Z = 24 # p+, # e−, # n0 atomic number symbol symbol atomic & mass numbers # e− # n0 in neutral atom, # p+ = # emass number = # p+ + # n0 Relationships: Solution: # p+ Z = 24 = # p+ # e− = # p+ = 24 Check: Tro: Chemistry: A Molecular Approach, 2/e A = Z + # n0 52 = 24 + # n0 28 = # n0 for most stable isotopes, n0 ≥ p+ 57 Copyright 2011 Pearson Education, Inc. Practice – Complete the table 27 13 Al Tro: Chemistry: A Molecular Approach, 2/e 58 Copyright 2011 Pearson Education, Inc. Practice – Complete the table 13 6C 96 42Mo 27 13 Al 133 55 Cs Tro: Chemistry: A Molecular Approach, 2/e 59 Copyright 2011 Pearson Education, Inc. Reacting Atoms • When elements undergo chemical reactions, the reacting elements do not turn into other elements Statement 4 of Dalton’s Atomic Theory • This requires that all the atoms present when you • • start the reaction will still be there after the reaction Because the number of protons determines the kind of element, the number of protons in the atom does not change in a chemical reaction However, many reactions involve transferring electrons from one atom to another Tro: Chemistry: A Molecular Approach, 2/e 60 Copyright 2011 Pearson Education, Inc. Charged Atoms • When atoms gain or lose electrons, they • • • acquire a charge Charged atoms or groups of atoms are called ions When atoms gain electrons, they become negatively charged ions, called anions When atoms lose electrons, they become positively charged ions, called cations Tro: Chemistry: A Molecular Approach, 2/e 61 Copyright 2011 Pearson Education, Inc. Ions and Compounds • Ions behave much differently than the neutral atoms e.g., the metal sodium, made of neutral Na atoms, is highly reactive and quite unstable; however, the sodium cations, Na+, found in table salt are very nonreactive and stable • Because materials such as table salt are neutral, there must be equal amounts of charge from cations and anions in them Tro: Chemistry: A Molecular Approach, 2/e 62 Copyright 2011 Pearson Education, Inc. Atomic Structures of Ions • Nonmetals form anions • For each negative charge, the ion has one more electron than the neutral atom F = 9 p+ and 9 e−, F− = 9 p+ and 10 e− P = 15 p+ and 15 e−, P3− = 15 p+ and 18 e− • Anions are named by changing the ending of the name to -ide fluorine oxygen Tro: Chemistry: A Molecular Approach, 2/e F + 1e− F− O + 2e− O2− 63 fluoride ion oxide ion Copyright 2011 Pearson Education, Inc. Atomic Structures of Ions • Metals form cations • For each positive charge, the ion has one less electron than the neutral atom Na atom = 11 p+ and 11 e−, Na+ ion = 11 p+ and 10 e− Ca atom = 20 p+ and 20 e−, Ca2+ ion = 20 p+ and 18 e− • Cations are named the same as the metal sodium calcium Na Na+ + 1e− sodium ion Ca Ca2+ + 2e− calcium ion Tro: Chemistry: A Molecular Approach, 2/e 64 Copyright 2011 Pearson Education, Inc. Practice – Complete the table Al3 Tro: Chemistry: A Molecular Approach, 2/e 65 Copyright 2011 Pearson Education, Inc. Practice – Complete the table S 2 Mg2 3 Al Br Tro: Chemistry: A Molecular Approach, 2/e 66 Copyright 2011 Pearson Education, Inc. Mendeleev • Ordered elements by atomic mass • Saw a repeating pattern of properties • Periodic Law – when the elements are arranged • • • in order of increasing atomic mass, certain sets of properties recur periodically Put elements with similar properties in the same column Used pattern to predict properties of undiscovered elements Where atomic mass order did not fit other properties, he re-ordered by other properties Te & I Tro: Chemistry: A Molecular Approach, 2/e 67 Copyright 2011 Pearson Education, Inc. Periodic Pattern Tro: Chemistry: A Molecular Approach, 2/e 68 Copyright 2011 Pearson Education, Inc. Periodic Patterns NM H2O a/b H 1.0 H2 M Li 6.9 M Li2O b M LiH 9.0 Be Na2O M b Na 23.0 NaH 24.3 M K2O b K 39.1 KH BeO NM a/b B BeH2 10.8 MgO b M Mg M Al MgH2 27.0 CaO b B2O3 NM a C CO2 NM a N N2O5 NM a O2 NM O F BH3 12.0 CH4 14.0 NH3 16.0 H2O 19.0 HF Al2O3 M/NM a/b SiO2 NM a P4O10 NM a SO3 NM a Cl2O7 a AlH3 28.1 SiH4 31.0 Si P PH3 32.1 S H2S 35.5 Cl HCl Ca 40.1 CaH2 M = metal, NM = nonmetal, M/NM = metalloid a = acidic oxide, b = basic oxide, a/b = amphoteric oxide Tro: Chemistry: A Molecular Approach, 2/e 69 Copyright 2011 Pearson Education, Inc. Most About A fewofelements ¾the of remaining the elements are classified elements are classified asare metalloids. classified as metals. as nonmetals. They have Their solidsaTheir have reflective solids somesurface, characteristics have a non-reflective conductofheat metals and surface, electricity and some dobetter of not nonmetals. conduct than other heatelements, and electricity and are well, and malleable are brittle. and ductile Tro: Chemistry: A Molecular Approach, 2/e 70 Copyright 2011 Pearson Education, Inc. Metals • Solids at room temperature, except Hg • Reflective surface shiny • Conduct heat • Conduct electricity • Malleable can be shaped • Ductile can be drawn or pulled into wires • Lose electrons and form cations in reactions • About 75% of the elements are metals • Lower left on the table Tro: Chemistry: A Molecular Approach, 2/e 71 Copyright 2011 Pearson Education, Inc. Sulfur, S(s) Nonmetals • • • • • • Found in all three states Poor conductors of heat Poor conductors of electricity Solids are brittle Gain electrons in reactions to become anions Upper right on the table Bromine, Br2(l) Chlorine, Cl2(g) except H Tro: Chemistry: A Molecular Approach, 2/e 72 Copyright 2011 Pearson Education, Inc. Metalloids • Show some • properties of metals and some of nonmetals Also known as semiconductors Tro: Chemistry: A Molecular Approach, 2/e 73 Properties of Silicon shiny conducts electricity does not conduct heat well brittle Copyright 2011 Pearson Education, Inc. The Modern Periodic Table • Elements with similar chemical and • physical properties are in the same column Columns are called Groups or Families designated by a number and letter at top • Rows are called Periods • Each period shows the pattern of properties repeated in the next period Tro: Chemistry: A Molecular Approach, 2/e 74 Copyright 2011 Pearson Education, Inc. The Modern Periodic Table • Main group = representative elements = “A” • groups Transition elements = “B” groups all metals • Bottom rows = inner transition elements = rare earth elements metals really belong in Period 6 & 7 Tro: Chemistry: A Molecular Approach, 2/e 75 Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 76 Copyright 2011 Pearson Education, Inc. = Alkali metals = Halogens = Alkali earth metals = Lanthanides = Noble gases = Actinides = Transition metals Tro: Chemistry: A Molecular Approach, 2/e 77 Copyright 2011 Pearson Education, Inc. Important Groups - Hydrogen • Nonmetal • Colorless, diatomic gas very low melting point and density • Reacts with nonmetals to form molecular compounds HCl is acidic gas H2O is a liquid • Reacts with metals to form hydrides metal hydrides react with water to form H2 • HX dissolves in water to form acids Tro: Chemistry: A Molecular Approach, 2/e 78 Copyright 2011 Pearson Education, Inc. Important Groups – Alkali Metals • Group IA = Alkali Metals • Hydrogen usually placed here, • • • • • though it doesn’t really belong Soft, low melting points, low density Flame tests Li = red, Na = yellow, K = violet Very reactive, never find uncombined in nature Tend to form water-soluble compounds, therefore salt is crystallized from seawater then molten salt is electrolyzed colorless solutions React with water to form basic (alkaline) solutions and H2 2 Na + 2 H2O 2 NaOH + H2 releases a lot of heat Tro: Chemistry: A Molecular Approach, 2/e 79 lithium sodium potassium rubidium cesium Copyright 2011 Pearson Education, Inc. Important Groups – Alkali Earth Metals • Group IIA = Alkali earth metals • Harder, higher melting, and denser than alkali metals Mg alloys used as structural materials • Flame tests Ca = red, Sr = red, Ba = green • Reactive, but less than corresponding alkali metal • Form stable, insoluble oxides from which they are • • normally extracted Oxides are basic = alkaline earth Reactivity with water to form H2 Be = none; Mg = steam; Ca, Sr, Ba = cold water Tro: Chemistry: A Molecular Approach, 2/e 80 Copyright 2011 Pearson Education, Inc. Important Groups – Halogens • Group VIIA = halogens • Nonmetals • F2 and Cl2 gases; Br2 liquid; I2 • • • • • solid All diatomic Very reactive Cl2, Br2 react slowly with water Br2 + H2O HBr + HOBr React with metals to form ionic compounds HX all acids HF weak < HCl < HBr < HI Tro: Chemistry: A Molecular Approach, 2/e 81 fluorine chlorine bromine iodine astatine Copyright 2011 Pearson Education, Inc. Important Groups – Noble Gases • Group VIIIA = Noble Gases • All gases at room temperature very low melting and boiling points • Very unreactive, practically inert • Very hard to remove electron from or give electron to Tro: Chemistry: A Molecular Approach, 2/e 82 Copyright 2011 Pearson Education, Inc. Ion Charge and the Periodic Table • The charge on an ion can often be • • • • determined from an element’s position on the Periodic Table Metals always form positively charged cations For many main group metals, the charge = the group number Nonmetals form negatively charged anions For nonmetals, the charge = the group number − 8 Tro: Chemistry: A Molecular Approach, 2/e 83 Copyright 2011 Pearson Education, Inc. Tro: Chemistry: A Molecular Approach, 2/e 84 Copyright 2011 Pearson Education, Inc. Practice – What is the charge on each of the following ions? • • • • • potassium cation sulfide anion calcium cation bromide anion aluminum cation Tro: Chemistry: A Molecular Approach, 2/e K+ S2− Ca2+ Br− Al3+ 85 Copyright 2011 Pearson Education, Inc. Atomic Mass • We previously learned that not all atoms of an element have the same mass isotopes • We generally use the average mass of all an element’s atoms found in a sample in calculations However, the average must take into account the abundance of each isotope in the sample • We call the average mass the atomic mass Tro: Chemistry: A Molecular Approach, 2/e 86 Copyright 2011 Pearson Education, Inc. Mass Spectrometry • Masses and abundances of isotopes are measured with a mass spectrometer • Atoms or molecules are ionized, then accelerated down a tube some molecules are broken into fragments during the ionization process these fragments can be used to help determine the structure of the molecule • Their path is bent by a magnetic field, separating them by mass similar to Thomson’s cathode ray experiment Tro: Chemistry: A Molecular Approach, 2/e 87 Copyright 2011 Pearson Education, Inc. Mass Spectrum • A mass spectrum is a • • graph that gives the relative mass and relative abundance of each particle Relative mass of the particle is plotted in the xaxis Relative abundance of the particle is plotted in the yaxis Tro: Chemistry: A Molecular Approach, 2/e 88 Copyright 2011 Pearson Education, Inc. Mass Spectrometer Tro: Chemistry: A Molecular Approach, 2/e 89 Copyright 2011 Pearson Education, Inc. Example 2.5: If copper is 69.17% Cu-63 with a mass of 62.9396 amu and the rest Cu-65 with a mass of 64.9278 amu, find copper’s atomic mass Given: Find: Conceptual Plan: Relationships: Cu-63 = 69.17%, 62.9396 amu Cu-65 = 100-69.17%, 64.9278 amu atomic mass, amu isotope masses, isotope fractions avg. atomic mass Solution: Check: the average is between the two masses, closer to the major isotope Tro: Chemistry: A Molecular Approach, 2/e 90 Copyright 2011 Pearson Education, Inc. Practice ─ Ga-69 with mass 68.9256 amu and abundance of 60.11% and Ga-71 with mass 70.9247 amu and abundance of 39.89%. Calculate the atomic mass of gallium. Tro: Chemistry: A Molecular Approach, 2/e 91 Copyright 2011 Pearson Education, Inc. Practice – Ga-69 with mass 68.9256 amu and abundance of 60.11% and Ga-71 with mass 70.9247 amu and abundance of 39.89%. Calculate the atomic mass of gallium. Given: Ga-69 = 60.11%, 68.9256 amu Ga-71 = 39.89%, 70.9247 amu Find: atomic mass, amu Conceptual isotope masses, avg. atomic mass Plan: isotope fractions Relationships: Solution: Check: the average is between the two masses, closer to the major isotope Tro: Chemistry: A Molecular Approach, 2/e 92 Copyright 2011 Pearson Education, Inc. Counting Atoms by Moles • If we can find the mass of a particular number of atoms, we can use this information to convert the mass of an element sample into the number of atoms in the sample • The number of atoms we will use is 6.022 x 1023, and we call this a mole 1 mole = 6.022 x 1023 things like 1 dozen = 12 things Tro: Chemistry: A Molecular Approach, 2/e 93 Copyright 2011 Pearson Education, Inc. Example 2.6: Calculate the number of atoms in 2.45 mol of copper Given: Find: Conceptual Plan: Relationships: 2.45 mol Cu atoms Cu mol Cu atoms Cu 1 mol = 6.022 x 1023 atoms Solution: Check: because atoms are small, the large number of atoms makes sense Tro: Chemistry: A Molecular Approach, 2/e 94 Copyright 2011 Pearson Education, Inc. Practice — A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Tro: Chemistry: A Molecular Approach, 2/e 95 Copyright 2011 Pearson Education, Inc. Practice — A silver ring contains 1.1 x 1022 silver atoms. How many moles of silver are in the ring? Given: 1.1 x 1022 atoms Ag Find: moles Ag Conceptual Plan: Relationships: atoms Ag mol Ag 1 mol = 6.022 x 1023 atoms Solution: Check: because the number of atoms given is less than Avogadro’s number, the answer makes sense Tro: Chemistry: A Molecular Approach, 2/e 96 Copyright 2011 Pearson Education, Inc. Chemical Packages - Moles • Mole = number of particles equal to the number of atoms in 12 g of C-12 1 atom of C-12 weighs exactly 12 amu 1 mole of C-12 weighs exactly 12 g • The number of particles in 1 mole is called Avogadro’s Number = 6.0221421 x 1023 1 mole of C atoms weighs 12.01 g and has 6.022 x 1023 atoms the average mass of a C atom is 12.01 amu Tro: Chemistry: A Molecular Approach, 2/e 97 Copyright 2011 Pearson Education, Inc. Relationship Between Moles and Mass • • • • The mass of one mole of atoms is called the molar mass The molar mass of an element, in grams, is numerically equal to the element’s atomic mass, in amu The lighter the atom, the less a mole weighs The lighter the atom, the more atoms there are in 1 g Tro: Chemistry: A Molecular Approach, 2/e 98 Copyright 2011 Pearson Education, Inc. Mole and Mass Relationships 1 mole sulfur 32.06 g Tro: Chemistry: A Molecular Approach, 2/e 1 mole carbon 12.01 g 99 Copyright 2011 Pearson Education, Inc. Example 2.7: Calculate the moles of carbon in 0.0265 g of pencil lead Given: 0.0265 g C Find: mol C Conceptual Plan: gC mol C Relationships: 1 mol C = 12.01 g Solution: Check: because the given amount is much less than 1 mol C, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 100 Copyright 2011 Pearson Education, Inc. Practice — Calculate the moles of sulfur in 57.8 g of sulfur Tro: Chemistry: A Molecular Approach, 2/e 101 Copyright 2011 Pearson Education, Inc. Practice — Calculate the moles of sulfur in 57.8 g of sulfur Given: Find: 57.8 g S mol S Conceptual Plan: gS mol S Relationships: 1 mol S = 32.07 g Solution: Check: because the given amount is much less than 1 mol S, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 102 Copyright 2011 Pearson Education, Inc. Example 2.8: How many copper atoms are in a penny weighing 3.10 g? Given: Find: Conceptual Plan: Relationships: 3.10 g Cu atoms Cu g Cu mol Cu atoms Cu 1 mol Cu = 63.55 g, 1 mol = 6.022 x 1023 Solution: Check: because the given amount is much less than 1 mol Cu, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 103 Copyright 2011 Pearson Education, Inc. Practice — How many aluminum atoms are in a can weighing 16.2 g? Tro: Chemistry: A Molecular Approach, 2/e 104 Copyright 2011 Pearson Education, Inc. Practice — How many aluminum atoms are in a can weighing 16.2 g? Given: 16.2 g Al Find: atoms Al Conceptual Plan: g Al mol Al atoms Al Relationships: 1 mol Al = 26.98 g, 1 mol = 6.022 x 1023 Solution: Check: because the given amount is much less than 1 mol Al, the number makes sense Tro: Chemistry: A Molecular Approach, 2/e 105 Copyright 2011 Pearson Education, Inc.