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ATOMIC THEORY – THE STRUCTURE OF MATTER 1661 ROBERT BOYLE Identified matter as composed of either elements or compounds ELEMENT – Any substance that cannot be decomposed to simpler substances by chemical means COMPOUND – Any substance that can be decomposed to simpler substances by chemical means 1A-1 (of 14) Water ↓electricity Hydrogen and Oxygen ↓electricity ↓electricity Hydrogen and Oxygen 1A-2 (of 14) Compound Elements 1789 ANTOINE LAVOISIER Showed that mass was conserved during chemical changes Charcoal and Rust 1.0 g and 9.0 g ↓∆ Iron and Fixed Air 1A-3 (of 14) 6.3 g and 3.7 g 1797 JOSEPH PROUST Showed that compounds always have the same proportion of elements by mass Mercuric Calc ↓∆ Mercury and Oxygen 5.00 g 8.92 g 4.63 g and 0.37 g 8.26 g and 0.66 g Mercuric Calc is always 92.6% Mercury and 7.4% Oxygen LAW OF DEFINITE PROPORTION – A specific compound always contains exactly the same proportion of elements by mass 1A-4 (of 14) 1803 JOHN DALTON Proposed that elements are composed of tiny particles called ATOMS to explain the Law of Definite Proportion Mass of Carbon Mass of Hydrogen Ethyne 12 grams 1 gram Ethene 12 grams 2 grams Ethane 12 grams 3 grams The elements carbon and hydrogen are composed of atoms 1) Suppose carbon atoms and hydrogen atoms weigh the same Ethyne is C12H1 1A-5 (of 14) Ethene is C12H2 Ethane is C12H3 or C6H1 or C4H1 1803 JOHN DALTON Proposed that elements are composed of tiny particles called ATOMS to explain the Law of Definite Proportion Mass of Carbon Mass of Hydrogen Ethyne 12 grams 1 gram Ethene 12 grams 2 grams Ethane 12 grams 3 grams The elements carbon and hydrogen are composed of atoms 2) Suppose carbon atoms weigh 6 times more than hydrogen atoms Ethyne is C2H1 Ethene is C2H2 or C1H1 1A-6 (of 14) Ethane is C2H3 1803 JOHN DALTON Proposed that elements are composed of tiny particles called ATOMS to explain the Law of Definite Proportion Mass of Carbon Mass of Hydrogen Ethyne 12 grams 1 gram Ethene 12 grams 2 grams Ethane 12 grams 3 grams The elements carbon and hydrogen are composed of atoms 3) Suppose carbon atoms weigh 12 times more than hydrogen atoms 1A-7 (of 14) In 1808 Dalton published his theory of atoms: 1 – Elements are made of indivisible atoms 2 – All atoms of a given element are the same 3 – Compounds are formed when atoms of different elements combine with each other (assuming the Rule of Simplicity) 4 – Atoms of one element cannot change into atoms of another element Dalton prepared the first table of atomic masses Element Hydrogen Carbon Oxygen Magnesium 1A-8 (of 14) Relative Atomic Mass 1 12 16 24 1897 J.J. THOMSON Discovered the first subatomic particle CROOKE’S TUBE 1A-9 (of 14) Experiments showed that cathode rays are 1 - Composed of particles 2 - Negatively charged Thomson called these particles ELECTRONS Electrons were found in atoms of all elements 1A-10 (of 14) 1910 ERNEST RUTHERFORD Determined the structure of the atom Shot alpha particles at a thin gold foil Alpha particles are small, positively charged particles Alpha particles should pass right through the atoms of gold However, several were deflected at severe angles 1A-11 (of 14) Rutherford Model of the Atom The atom has an extremely dense, positively charged nucleus, surrounded by mostly empty space that contains the negatively charged electrons 1A-12 (of 14) Rutherford Model of the Atom 1A-13 (of 14) Diameter of an Atom: 1 x 10-8 Diameter of a Nucleus: 1 x 10-13 cm cm Rutherford found he could change one element into another by bombarding it with hydrogen nuclei He proposed that hydrogen nuclei were positive subatomic particles called PROTONS Therefore, all atoms are composed of two subatomic particles: positive protons and negative electrons 1A-14 (of 14) LIGHT AND ELECTROMAGNETIC RADIATION Is Light Composed of Particles? Is Light Composed of Waves? 2 Slit Experiment Light produces this diffraction pattern 1B-1 (of 13) AMPLITUDE (A) – The maximum disturbance in the medium during one wave cycle WAVELENGTH (λ) – The distance between two sequential crests FREQUENCY (ν) – The number of complete cycles per unit time Wavelength (λ) and frequency (ν) are inversely related λν = speed 1B-2 (of 13) ELECTROMAGNETIC RADIATION – Forms of energy that consist of oscillating electric and magnetic fields that travel through space at 2.9979 x 108 m/s Light is a form of electromagnetic radiation Different types of EM radiation differ in their wavelengths and frequencies 1B-3 (of 13) For all wave phenomena: λν = speed For EM radiation: λν = c 1B-4 (of 13) (c = speed of light) Find the wavelength of EM radiation with a frequency of 2.4 x 1014 s-1 λν = c λ = c __ = v 2.9979 x 108 m/s ______________________ = 1.2 x 10-6 m 2.4 x 1014 s-1 Convert the wavelength into nm 1 nm = 10-9 m 1.2 x 10-6 m x 1 nm _________ = 1.2 x 103 nm 10-9 m What type of EM radiation is this? Infrared 1B-5 (of 13) 1900 MAX PLANCK Proposed that the energy of EM radiation could only be gained or lost in small, finite amounts, or QUANTA QUANTUM THEORY – Any theory that predicts that values for a property are restricted to multiples of some small, elementary unit 1B-6 (of 13) PHOTON – A particle of EM radiation, acting as a packet of energy, carrying Planck’s quanta of energy The photon energy (E) depends on the EM radiation frequency (ν), and they are directly related E = hν 1B-7 (of 13) (h = Planck’s Constant = 6.626 x 10-34 Js) 1B-8 (of 13) Find the energy of red light photons if the light has a frequency of 3.0 x 1014 s-1 E = hv = (6.626 x 10-34 Js)(3.0 x 1014 s-1) 1B-9 (of 13) = 2.0 x 10-19 J Calculate the energy of green light photons with a wavelength of 5.00 x 10-7 m E = hν λν = c v = c __ = λ 2.9979 x 108 m/s ______________________ = 5.996 x 1014 s-1 5.00 x 10-7 m Always carry at least one guard digit in the middle of calculations E = hv = (6.626 x 10-34 Js)(5.996 x 1014 s-1) = 3.973 x 10-19 J = 3.97 x 10-19 J 1B-10 (of 13) Calculate the energy of one mole of the green light photons 1 mole of photons = 6.022 x 1023 photons 3.973 x 10-19 J x 6.022 x 1023 photons __________________ ___________________________ photon = 2.39 x 105 J/mol mole Calculate the energy of one mole of the green light photons in kJ 1 kJ = 103 J 2.39 x 105 J x _______________ mole 1B-11 (of 13) 1 kJ _______ 103 J = 239 kJ/mol 1905 ALBERT EINSTEIN Used Planck’s quantum theory to explain the photoelectric effect 1B-12 (of 13) 1905 ALBERT EINSTEIN Used Planck’s quantum theory to explain the photoelectric effect 1B-13 (of 13) SCIENTIFIC PROBLEMS OF 1910 (1) The Rutherford Model of the Atom Why don’t the negative electrons spiral into the positive nucleus? (2) Hydrogen’s Emission Spectrum Passing electricity through hydrogen gas causes light to be emitted Using a prism, this light can be broken down into hydrogen’s EMISSION SPECTRUM 1C-1 (of 12) 1913 NIELS BOHR Proposed that electrons do not spiral into the positive nucleus because they just don’t The electrons orbit the nucleus at only specific distances Verizon Amphitheatre Bohr proposed a quantum theory - the energy of an electron is quantized because the electron can only be specific distances from the nucleus 1C-2 (of 12) Bohr proposed that (1) Electrons orbit the nucleus in circular orbits at specific distances called BOHR ORBITS The GROUND STATE orbit is the one closest to the nucleus The Bohr Model of the Atom 1C-3 (of 12) Bohr proposed that (2) Electrons must absorb energy to jump to higher orbits Higher orbits are called EXCITED STATES The Bohr Model of the Atom 1C-4 (of 12) Bohr proposed that (3) Electrons must release energy to drop to lower orbits A photon of specific energy is released between each pair of Bohr Orbits The Bohr Model of the Atom 1C-5 (of 12) Hydrogen Electronic Energy Diagram The visible emission spectrum of H corresponds to excited e-s dropping to the 2nd Bohr Orbit n = 3 to n = 2 transition: small E, small ν, large λ, so red (~700 nm) n = 5 to n = 2 transition: large E, large ν, small λ, so violet (~400 nm) 1C-6 (of 12) Hydrogen Electronic Energy Diagram Would excited e-s dropping to the 1st Bohr Orbit give off infrared or ultraviolet photons? n = x to n = 1 transition: 1C-7 (of 12) large E, large ν, small λ, so UV (<400 nm) 1924 LOUIS DE BROGLIE Proposed that all particles of matter have a wave-particle duality The deBroglie wavelength of a particle is: λ = m = mass, v = velocity h _____ mv Large particles have small wavelengths Small particles have large wavelengths deBroglie calculated an e-s wavelength and used it to explain Bohr Orbits 1C-8 (of 12) Bohr Orbits are integer multiples of the e-s wavelength 1C-9 (of 12) 1924 WERNER HEISENBERG Because matter possesses wave-particle duality, the position (x) and momentum (p) of a “particle” cannot be simultaneously known accurately The Heisenberg Uncertainty Principle is: Δx = position uncertainty Δp = momentum uncertainty The Bohr model calculates both the exact position and momentum of an electron in a hydrogen atom 1C-10 (of 12) (Δx)(Δp) ≥ h _____ 4π 1926 ERWIN SCHRÖDINGER Proposed a model of the hydrogen atom in which the electron is mathematically treated as a wave, not a particle ORBITAL - A highly probable volume of space in an atom where an electron can be found Orbit Orbital Spectroscopic data now shows that a maximum of 2 electrons can exist in an orbital, but they must be spinning in opposite directions 1C-11 (of 12) 1926 ERWIN SCHRÖDINGER Proposed a model of the hydrogen atom in which the electron is mathematically treated as a wave, not a particle Schrödinger determined all possible orbitals for an electron in a hydrogen atom from the SCHRÖDINGER EQUATION HΨ = EΨ where H is a mathematical operator called the HAMILTONIAN Ψ is the e-s WAVEFUNCTION for a specific orbital E is the e-s energy in the specific orbital Graphing Ψ2 gives the probability distribution of the electron in that orbital 1C-12 (of 12) Schrödinger determined all possible orbitals for an electron in a hydrogen atom from the SCHRÖDINGER EQUATION HΨ = EΨ where Ψ is the e-s WAVEFUNCTION for the orbital E is the e-s energy in the orbital Graphing Ψ2 gives the probability distribution of the electron in that orbital 1D-1 (of 15) Orbitals in atoms are organized by 1) Energy Levels ENERGY LEVEL - A group of orbitals the same distance from the nucleus 1st Energy Level 2nd Energy Level 3rd Energy Level etc. n=1 n=2 n=3 etc. 1D-2 (of 15) Orbitals in Energy Levels are organized by 2) Sublevels SUBLEVEL – A group of orbitals with the same characteristic shape s orbital spherical p orbital dumbbell d orbital 4-leaf clover f orbital 8-leaf clover s orbitals 1 per EL p orbitals 3 per EL d orbitals 5 per EL f orbitals 7 per EL 1D-3 (of 15) etc. Energy Level Sublevels Number of Orbitals 1 1 (s) 1 (1) 2 2 2 (s, p) 4 (1+3) 8 3 3 (s, p, d) 9 (1+3+5) 18 4 4 (s, p, d, f) 16 (1+3+5+7) 32 n n n2 2n2 1D-4 (of 15) Number of Electrons Angel Stadium 45,050 1D-5 (of 15) FIRST ENERGY LEVEL 1 “s” orbital 2 electrons fit in the 1st EL An orbital is named by stating its energy level and its sublevel 1s orbital 1D-6 (of 15) SECOND ENERGY LEVEL 1 “s” and 3 “p” orbitals 8 electrons fit in the 2nd EL 2s orbital 2p orbital or 2px orbital 2p orbital or 2py orbital 2p orbital or 2pz orbital 1D-7 (of 15) THIRD ENERGY LEVEL 1 “s”, 3 “p”, and 5 “d” orbitals 18 electrons fit in the 3rd EL 3d orbital or 3dxy orbital 3s orbital 3d orbital or 3dxz orbital 3px orbital 3d orbital or 3dyz orbital 3py orbital 3d orbital or 3dx2-y2 orbital 3pz orbital 3d orbital or 3dz2 orbital 1D-8 (of 15) FOURTH ENERGY LEVEL 1 “s”, 3 “p”, 5 “d”, and 7 “f” orbitals 32 electrons fit in the 4th EL 1 “4s”, 3 “4p”, 5 “4d” orbitals 4f orbital 4f orbital 4f orbital 4f orbital 4f orbital 4f orbital 4f orbital 1D-9 (of 15) ELECTRON ARRANGEMENTS Orbital Notation H 1s ↑ ___ He 1s ↑↓ ___ ELECTRON PAIR – 2 electrons of opposite spin in the same orbital Electron Configuration Notation 1s1 1s2 Electron Dot Notation H. He : VALENCE ELECTRONS – The electrons in an atom’s highest occupied energy level Electron Dot Notation only shows valence electrons 1D-10 (of 15) Orbital Notation Li 1s ↑↓ ___ Be ___ B ↑↓ ↑↓ ___ ↑↓ C ___ N ___ O ___ F ___ Ne ___ ↑↓ ↑↓ ↑↓ ↑↓ core 1D-11 (of 15) 2s ↑ ___ Electron Conf. Notation 2p ↑↓ ___ ↑↓ ___ ↑ ___ 1s2 2s1 Li . 1s2 2s2 Be : . ___ ___ 1s2 2s22p1 ↑ ↑ ___ ___ ___ 1s2 2s22p2 ↑↓ ↑ ↑ ↑ ___ ___ ___ 1s2 2s22p3 ↑↓ ↑↓ ↑ ↑ ___ ___ ___ 1s2 2s22p4 ↑↓ ↑↓ ↑↓ ↑ ___ ___ ___ 1s2 2s22p5 ↑↓ ↑↓ ↑↓ ↑↓ ___ ___ ___ 1s2 2s22p6 ___ ___ ___ ___ valence B: . ↑↓ ___ Electron Dot Notation .C : . .N : . .. .O : . .. :F : . .. : Ne : .. Picture of a Neon atom: 1D-12 (of 15) Orbital Notation 1s B ↑↓ ___ 2s ↑↓ ___ 2p ↑ ___ ___ ___ 1s e-s SHIELD the 2s and 2p e-s from the nuclear charge 2s e-s have a greater probability of being inside the shielding, so they are more attracted to the nucleus, making them more stable than 2p e-s In each energy level, s orbitals are most stable, then p, then d, etc. 1D-13 (of 15) Orbital Notation 1s C ↑↓ ___ 2s ↑↓ ___ 2p ↑ ↑ ___ ___ ___ HUND’S RULE – When occupying orbitals of equal energy, electrons will remain unpaired as long as possible to minimize repulsion DEGENERATE – Orbitals of equal energy (such as the 2px, 2py, and 2pz) 36 4 7 X 21 58 OCTET – An electron arrangement in which the s and p sublevels are filled in an energy level of an atom 1D-14 (of 15) 3rd Series: Electron Conf. Notation Electron Dot Notation Na 1s2 2s22p63s1 Na . Mg 1s22s22p63s2 Mg : Al 1s2 2s22p63s23p1 . Al : … .. Ar 1s2 2s22p63s23p6 K 1s2 2s22p63s23p64s1 : Ar .. : K. A 3d orbital is closer to the nucleus than a 4s orbital, but the 4s orbital has more probability inside the shielding done by the first two energy levels, so the 4s orbital is more stable 1D-15 (of 15) SUBLEVEL FILLING PATTERN The position of each element tells the sublevel of its final electron 1E-1 (of 16) SUBLEVEL FILLING PATTERN Arsenic 1s22s22p63s23p64s23d104p3 [Ar]4s23d104p3 1E-2 (of 16) SUBLEVEL FILLING PATTERN Iodine 1s22s22p63s23p64s23d104p65s24d105p5 [Kr]5s24d105p5 1E-3 (of 16) SUBLEVEL FILLING PATTERN Platinum [Xe]6s24f145d8 1E-4 (of 16) VALENCE (OUTER SHELL) ELECTRONS .. . . 1E-5 (of 16) As . : . I .. : Pt : 3d Cr 1E-6 (of 16) [Ar] ↑ ___ ↑ ___ ↑ ___ 4s ↑ ___ ___ ↑↓ ___ 3d Cr 1E-7 (of 16) [Ar] ↑ ___ ↑ ___ ↑ ___ 4s ↑ ___ ___ ↑↓ ___ 3d 1E-8 (of 16) ↑ ↑ Cr [Ar] ___ ___ Cu ↑↓ [Ar] ___ ___ ↑ 4s ↑ ___ ___ ↑↓ ___ ↑↓ ___ ↑↓ ↑ ___ ↑ ___ ↑ ___ ↑↓ ___ 3d 1E-9 (of 16) ↑ ↑ Cr [Ar] ___ ___ Cu ↑↓ [Ar] ___ ___ ↑ 4s ↑ ___ ___ ↑↓ ___ ↑↓ ___ ↑↓ ↑ ___ ↑ ___ ↑ ___ ↑↓ ___ 3d 1E-10 (of 16) ↑ ↑ Cr [Ar] ___ ___ Cu ↑↓ [Ar] ___ ___ ↑ 4s ↑ ___ ___ ↑↓ ___ ↑↓ ___ ↑ ↑ ___ ___ ↑↓ ___ ↑↓ ___ ↑ 1932 JAMES CHADWICK Discovered the third subatomic particle Shot alpha particles at beryllium metal, releasing neutral particles with about the same mass as protons Nuclei of most atoms also contain subatomic particles that are not charged, called NEUTRONS 1E-11 (of 16) The Nuclear Model of the Atom Proton Neutron Electron 1E-12 (of 16) Location Charge Relative Masses Nucleus Nucleus Around Nucleus + 0 - 1 1 1/1836 ATOMIC NUMBER (Z) – The number of protons in an atom Each element’s atomic number is found on the Periodic Table All atoms of a given element contain the same number of protons Element Hydrogen Carbon Oxygen Magnesium Atomic Number Protons 1 6 8 12 1 6 8 12 Electrons 1 6 8 12 Because atoms are electrically neutral, their number of negative electrons must equal their number of positive protons 1E-13 (of 16) Hydrogen Hydrogen Hydrogen Atoms of a given element can have different numbers of neutrons ISOTOPES – Atoms of the same element (same number of protons), but with different numbers of neutrons MASS NUMBER (A) – The sum of the protons and neutrons in an atom 1E-14 (of 16) Protons Neutrons 1 0 1 1 1 2 Mass Number 1 2 3 Isotope Name 1H 2H 3H Hydrogen-1 Hydrogen-2 Hydrogen-3 Mass Numbers ARE NOT found on the Periodic Table 1E-15 (of 16) 196 Hg Protons: 80 Neutrons: 196 - 80 = 116 Electrons: 80 1E-16 (of 16) 198 Hg2+ Protons: 80 Neutrons: 198 - 80 = 118 Electrons: 78 STABLE ISOTOPES – Atoms with nuclei that last forever RADIOACTIVE ISOTOPES – Atoms with nuclei that eventually break down to more stable nuclei Isotopes are stable when their nuclei have enough neutrons to minimize proton-proton repulsion (a) For Z < 20 Stable nuclei need 1 neutron per proton (b) For Z > 20 Stable nuclei approach needing 1.5 neutrons per proton 1F-1 (of 13) Stable 200Hg atom: 120 n, 80 p 1.5:1 ratio Stable 16O atom: 8 n, 8 p (1:1 ratio) 1F-2 (of 13) NUCLEAR DECAY Each radioactive isotope undergoes nuclear decay at its own unique rate HALF-LIFE (t1/2) – The time required for half of the radioactive isotopes in a sample to decay Half-lives range from 1 x 10-21 seconds for 18Na 5 x 1015 years for 142Ce The shorter the half-life, the more unstable the isotope 1F-3 (of 13) Half-life for 125I = 60 days At 0 days: 16 125I atoms 60 days: 8 125I atoms At 120 days: 4 125I atoms At 180 days: 2 125I atoms At 240 days: 1 125I atom At 1F-4 (of 13) Radioactive 14C has a half-life of 5,730 years How old is an axe with an elk antler sleeve if it has 25% the 14C content of antlers in living elks? Living Elk: 100% 14C content 5,730 yrs: 50% 14C content At 11,460 yrs: 25% 14C content At 11,460 years old 1F-5 (of 13) When a nucleus decays, it becomes more stable by releasing one of several types of “radiation” ALPHA RADIATION – the release of alpha particles (heium-4 nuclei) BETA RADIATION – the release of electrons GAMMA RADITION – the release of high energy EM photons 1F-6 (of 13) Large nuclei can also decay by breaking in half SPONTANEOUS FISSION – When a large nucleus (Z > 80) breaks into two, approximately equal halves Fission occurs in nuclear reactors and nuclear bombs 1F-7 (of 13) 1964 MURRAY GELL-MANN Proposed that protons and neutrons are made of smaller particles called QUARKS Up Quark (+⅔) Down Quark (-⅓) Proton Neutron All stable matter is composed of up quarks, down quarks, and electrons 1F-8 (of 13) THE HISTORY OF ATOMS 0s No matter, only energy exists 10-35 s The Period of Inflation – The universe undergoes a giant expansion (The Big Bang), and energy is converted into quarks and electrons 10-6 s Quarks are pulled together into protons and neutrons 3m Protons and neutrons form simple nuclei 105 y Electrons are captured forming atoms, almost all H and He 108 y Gravity consolidates huge clouds of H and He into galaxies with billions of stars each 1F-9 (of 13) THE HISTORY OF ATOMS The center of stars get heated to very high temperatures because of gravity, causing H to fuse into He FUSION – The combining of small nuclei to produce large nuclei 4 11 H → 4 2 He Fusion requires very high temperatures or pressures, but releases much more energy than fission This is why stars shine 1F-10 (of 13) THE HISTORY OF ATOMS The cores of very massive stars get so hot other fusion occurs After 1H fuses into 4He, the 4He fuses into 12C, 16O, 20Ne, 24Mg, 28Si, … , up to 56Fe Because no H is present, fusion does not form atoms with odd atomic numbers Once 56Fe is formed at the core, fusion stops, the star cools, and collapses in seconds The heat explodes the star into a SUPERNOVA, dispersing the star material into space All other elements are produced by collisions of atoms after supernova 1F-11 (of 13) THE HISTORY OF ATOMS Dispersed atoms from a supernova collect into second generation stars Our sun is a second generation star Planets coalesce around second generation stars, and are rich in heavy elements 1F-12 (of 13) THE HISTORY OF ATOMS In 5 billion years, the H in the core of our sun will be used up, the fusion will spread to the outer layers, and the sun will become a RED GIANT When the H is used up the sun will collapse, starting the fusion of He into C and O As He in the sun’s core is used up, fusion spreads to the outer layers, and the star swells, becoming a larger RED GIANT No other fusion can occur so the sun sheds its outer layers, and becomes a WHITE DWARF 1F-13 (of 13) ATOMIC MASS – The mass of an individual atom ATOMIC MASS UNIT (amu or u) – One twelfth the mass of a carbon-12 atom, equal to 1.66 x 10-24 g Atomic masses are measured today with mass spectrometers 1G-1 (of 11) 1H 2H 1.00782 u 2.00140 u 1G-2 (of 11) 99.985 % 0.015 % 12C 13C 12.00000 u 13.00335 u 98.90 % 1.10 % ← Atomic Number ← Elemental Symbol ← The ELEMENT’S Atomic Mass, measured in amu’s ELEMENT’S ATOMIC MASS - Found on the Periodic Table, the average atomic mass of all of the element’s naturally occurring isotopes 1G-3 (of 11) Mass Spectrometer Data Isotope 16O 17O 18O Atomic Mass Percentage 15.99491 u 16.99913 u 17.99916 u 99.7587 % 0.0374 % 0.2039 % Element’s Atomic Mass = (mass 1)(dec % 1) + (mass 2)(dec % 2) + (mass 3)(dec % 3) (15.99491 u)(.997587) + (16.99913 u)(.000374) + (17.99916 u)(.002039) _______________________________ 15.95631 u .006357 u .036700 u ________________ 15.999367 u = 15.9994 u Always carry at least one guard digit in the middle of calculations 1G-4 (of 11) Atomic masses allowed chemists to count atoms by weighing them Gummy Worms Jordan Almonds 8 g each 4 g each 200 g 100 g These 2 samples contain the same number of pieces of candy 1G-5 (of 11) 1 C atom 12 amu 1 Mg atom 24 amu 12 g C 24 g Mg If the mass of the magnesium sample is twice the mass of the carbon sample, then the two samples will contain the same number of atoms The number of atoms contained in 12 g carbon and 24 g magnesium (numerically the element’s atomic mass, but measured in grams) is called a MOLE of atoms For carbon: 12 g C = 1 mol C atoms For magnesium: 24 g Mg = 1 mol Mg atoms 1G-6 (of 11) The Element’s Atomic Mass The Element’s Molar Mass The average atomic mass of the element’s isotopes, measured in amu’s The mass necessary to have 1 mole of atoms of the element, measured in grams 1 average O atom = 15.9994 u 1 mole O atoms = 15.9994 g A Molar Mass may be rounded to four significant figures as long as it doesn’t limit the number of significant figures in a calculation 1G-7 (of 11) Calculate the number of moles of boron atoms in a 75.0 gram sample of boron For boron: 10.81 g B = 1 mol B 75.0 g B x 1 mol B _______________ = 6.94 mol B 10.81 g B Calculate the mass of iron that would contain 6.94 moles of iron atoms. For iron: 55.85 g Fe = 1 mol Fe 6.94 mol Fe x 55.85 g Fe _______________ 1 mol Fe 1G-8 (of 11) = 388 g Fe 1898 MICHAEL FARADAY Using his relationship between mass and charge, determined electrochemically the number of atoms in one mole 1 mole of atoms = 6.022 x 1023 atoms 1G-9 (of 11) For shoes: 1 pair = 2 shoes For eggs: 1 dozen = 12 eggs For atoms: 1 mole = 6.022 x 1023 atoms Calculate the number of atoms contained in 2.50 moles of potassium For any matter: 6.022 x 1023 atoms = 1 mol 2.50 mol K x 6.022 x 1023 atoms K _____________________________ 1 mol K 1G-10 (of 11) = 1.51 x 1024 atoms K Calculate the number of grams of calcium needed to have 5.00 x 1025 calcium atoms For any matter: 6.022 x 1023 atoms = 1 mol For calcium: 40.08 g Ca = 1 mol Ca 5.00 x 1025 atoms Ca x 1 mol Ca _____________________________ 6.022 x 1023 atoms Ca = 3,327 g Ca 1G-11 (of 11) = 3,330 g Ca x 40.08 g Ca _______________ 1 mol Ca
