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Transcript
The Atom Chapter 4 http://www.youtube.com/watch?v =Uy0m7jnyv6U I. II. III. IV. History of the Atomic Theory A. Democritus B. Aristotle C. Lavoisier D. Proust E. Dalton F. Modern Atomic Theory History of Atomic Structure A. Thomson B. Milikan C. Rutherford D. Bohr E. Chadwick F. Quantum Atom Subatomic Particles A. Atomic Number B. Mass Number and Isotopes C. Electrons and Ions D. Nuclear and Hyphenation Notation E. Average Atomic Mass Weighing and Counting Atoms A. Mole Atoms B. Mole Mass C. Mass Atoms I. History of the Atomic Theory Remember: a scientific theory explains behaviors and the ‘nature’ of things Theories can be revised when new discoveries are made The theory describing the composition of matter has been revised many times I. History of the Atomic Theory Democritus (460-370 BC) 1.Matter is made up of “atoms” that are solid, indivisible and indestructible 2.Atoms constantly move in space 3.Different atoms have different size and shape 4.Changes in matter result from changes in the grouping of atoms 5. Properties of matter result from size, shape and movement A. I. History of the Atomic Theory B. Aristotle (384-322 BC ) & Others 1. Four kinds of matter a. Fire – Earth – Water – Air 2. One kind of matter can transform into another 3. Rejected idea of the “atom” (idea then ignored for almost 2000 years 4. This theory was more popular and it was easier to accept Aristotle’s Theory of Matter I. History of the Atomic Theory C. Antoine Lavoisier (1770s) 1. Experiment: 2 Sn + O2 2 SnO tin oxygen tin (II) oxide mass before reaction = mass after reaction 2. Law of Conservation of Mass a. Matter cannot be created or destroyed (in a chemical or physical change) I. History of the Atomic Theory D. Joseph Proust (1779) 1. Develops Law of Definite Composition- all samples of a specific substance contain the same mass ratio of the same elements a. ex: all samples of CO2 contains 27.3% carbon and 72.7% oxygen b. therefore ‘elements’ are combining in a whole number ratio – WHY???? I. History of the Atomic Theory - Dalton E. John Dalton (1803) 1. Develops Law of Multiple Proportions a. describes the ratio of elements by mass in two different compounds composed of the same elements 2. Example: carbon monoxide carbon dioxide 1 part oxygen : 2 parts oxygen *when compared to the same amount of carbon in each compound I. History of the Atomic Theory- Dalton 3. Dalton collects data and develops his atomic theory in 1803 4. Dalton’s Background a. b. c. d. Dalton became a school teacher at the age of 12 (he left school at age 11) loved meteorology - pioneer in this field studied works of Democritus, Boyle and Proust Wrote New System of Chemical Philosophy in 1808 5. Dalton’s Atomic Theory 1. Matter is made of small particles-atoms 2. Atoms of a given element are identical in size, mass, but differ from those of other elements*. 3. Atoms cannot be subdivided or destroyed*. ( supports law of conservation of mass) 4.Atoms combine in small whole number ratios to form compounds. (def comp,Mult prop) 5. Atoms combine, separate, or rearrange in chemical reactions. * Modified in Modern Atomic Theory JUST A THEORY……. But it lead to the Modern Atomic theory F. Modern Atomic Theory 1. All matter is made up of small particles called atoms. 2. Atoms of the same element have the same chemical properties while atoms of different elements have different properties 3. Not all atoms of an element have the same mass, but they all have a definite average mass which is characteristic. (isotopes) F. Modern Atomic Theory 4. 5. Atoms of different elements combine to form compounds and each element in the compound loses its characteristic properties. Atoms cannot be subdivided by chemical or physical changes – only by nuclear changes I. History of the Atomic Theory 1803 1897 1909 1913 1935 Today solid particle electron proton e- orbit nucleus neutron Quantum Atom theory Dalton Thomson Rutherford Bohr Chadwick Schrodinger and others II. History of the Atomic Structure A. J.J. Thomson (1856-1940) J.J. Thomson (1887) 1. Experiments with cathode ray tubes a. atoms have (-) charged particles which are smaller than atoms b. determined charge/mass ratio of the “electron” Voltage source - + Vacuum tube Metal Disks Voltage source - + Voltage source - + Voltage source - + Voltage source + Passing an electric current makes a beam appear to move from the negative to the positive end Voltage source + Passing an electric current makes a beam appear to move from the negative to the positive end Voltage source + Passing an electric current makes a beam appear to move from the negative to the positive end Voltage source + Passing an electric current makes a beam appear to move from the negative to the positive end Voltage source By adding an electric field Voltage source + By adding an electric field Voltage source + By adding an electric field Voltage source + By adding an electric field Voltage source + By adding an electric field Voltage source + By adding an electric field Voltage source + By adding an electric field he found that the moving pieces were negative Demonstration of the cathode ray experiment. 2. Thomson’s Model The Pudding Model a. electrons present b. atom is like plum pudding - bunch of positive stuff (pudding), with the electrons suspended (plums) II. History of the Atomic Structure B. Robert Milikan (1868-1953) 1. Oil Drop Experiment (1909) a. Discovered mass and actual charge of electron (-1) b. Mass is 1/1840 the mass of a hydrogen atom 1) e – has a mass of 9.11 x 10-28 g Oil Drop II. History of the Atomic Structure – Summary thus far So, at this point we know: - Atoms are divisible into smaller particles – Electrons are negatively charged – The mass of an electron is very small HOWEVER – Atoms should have a (+) portion to balance the negative part - Electrons are so small that some other particles must account for mass II. History of the Atomic Structure Ernest Rutherford (1871-1937) C. Ernest Rutherford (1909) 1. Discovered the proton p+ 2. Received Nobel Prize in Chemistry 3. Gold Foil Experiment (Expectations) a. Shot alpha particles at atoms of gold b. expected them to pass straight through Lead block Uranium Florescent Screen Gold Foil He thought this would happen: According to Thomson Model He thought the mass of the positive charge was evenly distributed in the atom Here is what he observed: 4. Gold Foil Experiment Results a. Most positive alpha particles pass right through b. However, a few were deflected c. Rutherford reasoned that the positive alpha particle was deflected or repelled by a concentration of positive charge The positive region accounts for deflection 5. Gold Foil Experiment Conclusions a. the atom is mostly empty space b. the atom has a small, dense positive center surrounded by electrons Rutherford Model of the Atom II. History of the Atomic Structure At this point in 1909, we know: – p+ = 1.67 x 10-24 g – e- = 9.11 x 10-28 g – The charges are balance! But, – How are the electrons arranged? – There is still mass that is unaccounted for II. History of the Atomic Structure D. Niels Bohr (1913) 1. Electrons orbit nucleus in predictable paths II. E. History of the Atomic Structure E. Chadwick (1891 – 1974) Chadwick (1935) 1. Discovers neutron in nucleus 2. Neutron is neutral - does not have a charge n0 3. Mass is 1.67 x 10-24 g a. slightly greater than the mass of a proton II. History of the Atomic Structure F. The Quantum Atom Theory 1. The atom is mostly empty space 2. Two regions: a. Nucleus- protons and neutrons b. Electron cloud- region where you have a 90% chance of finding an electron II. History of the Atomic Structure Charges balanced Mass accounted for However – what about the behavior of the electrons? III. Subatomic Particles A. Comparing Particles Relative Actual mass (g) Name Symbol Charge mass Electron e- -1 Proton p+ +1 1amu 1.67 x 10-24 Neutron n0 0 1amu 0 9.11 x 10-28 1.67 x 10-24 III. Subatomic Particles B. Atomic Number and Mass Number 1. Atomic number 1. the number of protons in the nucleus of an atom a. identifies the element b. no two elements have the same atomic number 2. Ex. C is 6, N is 7 and O is 8 carbon nitrogen oxygen III. Subatomic Particles B. Atomic Number and Mass Number 2. Mass number a. the number of protons plus neutrons in the nucleus of an atom b. mass number is very close to the mass of an atom in amu (atomic mass units) c. two atoms with the same atomic number but different mass number are called isotopes 1) (mass #) – (atomic #) = #n 0 III. Subatomic Particles C. Ions 1. Electrons and Ions a. For neutral atoms, #e- = #p+ b. If there are more electrons, a negative ion forms (anion) c. If there are less electrons, a positive ion forms (cation) For now, we will work only with neutral atoms ions Subatomic Particles C. Formation of Ions Examples of Ions Atom loses electrons Atom gain electrons and form cations and forms anions Cations (+ ions) Anions (- ions) K+ BrCa2+ O2Al3+ N3- C. Formation of Ions From Atoms Na loses an electron and forms a cation Na – e- --> Na+ Cl gains an electron and forms an anion Cl + e- --> Cl- III. Subatomic Particles D. Changing Number of Particles 1. You can never change the number of protons and have the same element 2. If you change the number of neutrons in an atom, you get an isotope 3. If you change the number of electrons in an atom, you get an ion III. Subatomic Particles E. Nuclear Notation 1. Nuclear Notation is one method for depicting isotopes of an element 2. contains the symbol of the element, the mass number, and the atomic number Mass number Atomic number X III. Subatomic Particles E. Nuclear Notation 23 Na 11 How many protons? How many neutrons? How many electrons? III. Subatomic Particles F. Hyphen Notation 1. Element symbol or name – mass # 2. EXAMPLES a. Fluorine-19 b. C-14 c. U-238 IV. Mass of Atoms A. Atomic Mass 1. Mass of an atom a. too small to measure in grams b. use relative mass (amu) 1) atomic mass unit 2) 1 amu is defined as 1/12 the mass of one C-12 atom IV. Mass of Atoms B. Average Atomic Mass 1. weighted average mass of all known isotopes a. weighted means that the frequency of an isotope is considered b. mass of each isotope is multiplied by its percent occurrence in nature – then masses of all isotopes is added to get the average atomic mass IV. Mass of Atoms C. The Mole and Molar Mass 1. measures the amount of substance a. 1 mole = 6.02x1023 (Avogdro’s #) of particles (atoms, molecules, ions, electrons) b. standard – 1mole is the number of atoms in 12g of C-12 isotope 2. Molar mass – mass in grams of one mole (mol) of any substance a. numerically equal to atomic mass in amu b. unit is grams/mol Average Atomic Mass What is average atomic mass? Average atomic mass is the weighted average of the atomic masses of the naturally occurring isotopes of an element Calculating a Weighted Average Example A box contains two size of marbles. If 25.0% have masses of 2.00 g and 75.0% have masses of 3.00 g what is the weighted average? (.250) (2.00) + (.750) (3.00) = .500 + 2.25 = 2.75g Calculating Average Atomic Mass EXAMPLE Boron has two isotopes: B-10 (mass 10.013 amu) 19.8% abundance B-11 (mass 11.009 amu) 80.2% abundance Calculate the average atomic mass. (.198) (10.013) + (.802) ( 11.009) = 1.98 amu + 8.83 amu = 10.81 amu Calculating Average Atomic Mass Calculate the average atomic mass of Mg. Isotope 1 - 23.985 amu (78.99%) Isotope 2 - 24.986 amu (10.00%) Isotope 3 – 25.982 amu (11.01%) (23.985)(.7899)+(24.986)(.1000)+(25.982)(.1101) 18.95 amu + 2.498 amu + 2.861 amu = 24.31 amu Average Atomic Mass Helium has two naturally occurring isotopes, He-3 and He-4. The atomic mass of helium is 4.003 amu. Which isotope is more abundant in nature? He-4 is more abundant in nature because the atomic mass is closer to the mass of He-4 than to the mass of He-3. Calculating Average Atomic Mass Process 1. Multiply %occurrence x mass of isotope 2. Add products for each isotope isotope occurrence isotope occurrence Ex. X- 40 (30.0% ) X-30 (70.0%) (40 x .300) + (30 x .700) = 12 + 21 = 33 amu Isotopic Pennies – number of pre and post 1982 a. Let X be the number of pre-1982 pennies b. Let 10-X be the number of post-1982 pennies c. (X)(3.1g) + (10-X)(2.5g) = mass of 10 pennies pre-82 post-82 EXAMPLE (Mass of a sample of pennies is 31.0g) (X)(3.1g) + [(10-X)(2.5g)] = 31.0 g 3.1X + 25 - 2.5X = 31.0g .6X + 25 = 31.0g .6X = 6.0g X = 6.0g/.6 X = 10 pre-82 pennies 10-X = 0 post-82 pennies Isotopic Penny Lab- Average Atomic Mass Calculate percent of pre-82 and post-82 pennies # of pre-82 pennies x 100% 10 # post-82 pennies x 100% 10 Calculate the average atomic mass of coinium (% pre-82)(3.1g) + (% post-82)(2.5) = average atomic mass V. Radioactive Decay A. The Process 1. What is radioactive decay? a. spontaneous nuclear change in which unstable nuclei emit radiation and lose energy 1) radiation – rays and particles emitted by radioactive materials 2. Why do atoms undergo decay? a. produce a nucleus that is more stable V. Radioactive Decay of Elements B. Comparison of alpha, beta, gamma Alpha Form particle Beta particle Gamma ____ electromagnetic radiation Symbol Mass Charge Notation 4 amu +2 no mass -1 no mass none Alpha Particle V. 1. 2. Radioactive Decay C. Types of Decay Beta Decay (neutronproton + electron) a. beta particle (electron) is given off b. atomic number increases by one c. mass number stays the same Alpha Decay a. alpha particle (2p++2n0) is given off b. atomic number decreases by 2 c. mass number decreases by 4 V. Radioactive Decay D. Examples of Decay Beta Decay (n0 p+ + e-) e- released Parent Nuclei Daughter Nuclei Co-60 --------------> Ni-60 + e(z = 27) (z = 28) C-14 ---------------> N-14 + e(z = 6) (z = 7) Beta Decay Beta Decay Beta Decay + beta particle + beta particle + beta particle + beta particle + beta particle Radioactive Decay D. Examples of Decay Alpha Decay (alpha particle (2n0 + 2p+) released) Parent Nuclei Daughter Nuclei Th-232 ----------------> Ra-228 + alpha (z = 90) (z = 88) Ra-226 ---------------> Rn-222 + alpha (z = 88) (z = 86) Alpha Decay Alpha Decay Alpha Decay Alpha Decay + alpha particle + alpha particle + alpha particle + alpha particle + alpha particle Radioactive Decay of Uranium Radioactive Decay of Uranium Alpha Decay + alpha particle + alpha particle + alpha particle + alpha particle + alpha particle Radiation Radiation Radiation The Mole What is a mole in chemistry? What conversion factors are associated with the mole? Types of conversions involving mole equalities What is a Mole? What are mole equalities A mole is equal to 6.02 x 1023 particles Particles can be atoms, molecules or ions 6.02 x 1023 is Avogadro's Number Mole Equalities - 1 mole = molar mass - 1 mole = 6.02 x 1023 particles Mole Conversions [mass-mole-atoms] Type Equality Used 1. MOLES MASS 2. MASS MOLES 1 mole= molar mass (g) 3. MOLES ATOMS 4. ATOMS MOLES 1 mole = 6.02 x 1023 atoms 5. MASS ATOMS 6. ATOMS MASS 1 mole = 6.02 x 1023 atoms 1 mole = molar mass (g) Mole Calculations Using Conversion Factors 1 mole 6.02 x 1023 molar mass 1 mole PARTICLES <----> MOLES <----> MASS 6.02 x 1023 1 mole 1 mole molar mass Solving Mole Problems EXAMPLES 1. 2. 1.00 mole of He = 4.00 g. 2.00 mole of He = _____g 2.00 mol He X 4.00g He = 1 1 mole He He 8.00 g EXAMPLES 3. 1.00 mole He = 6.02 X 1023 atoms 2.00 mole He = ________atoms He 2.00 mole He x 6.02 x 1023 atoms He = 12.04 x 1023 1 1 mole He 1.20 x 1024 atoms He 5. 16.00g He = _____ moles He 16.0 g He x 1 mole He = 1 4.00 g He 4.00 moles He 4. EXAMPLES 6. 3.01 X 1023 atoms He = _____ moles 3.01 x 1023 atoms He x 1 1 mole He = 6.02 x 1023 atoms He .500 mol He 7. 8.00g He =______atoms He 8.00 g He x 1 mole He x 6.02 x 1023 atoms He = 1 4.00g He 1 mole He 12.04 x 1023 atoms He = 1.20 x 1024 atoms He Sample Problems 1. Moles to mass. Find the mass of 3.50 moles of carbon. 2. Mass to Moles How many moles of carbon are contained in 60.0 g of carbon? 3. Moles to Atoms How many atoms of carbon are found in 4.00 moles of carbon? Sample Problems 4. Atoms to Moles How many moles of carbon are represented by 1.806 x 1024 atoms of carbon? 5. Mass to Atoms How many carbon atoms are found in 36.0g of carbon? 6. Atoms to Mass What is the mass of 1.204 x 1024 atoms of carbon? More Sample Problems 2.00 moles of Cu = 60.0 grams of C = 3.00 x 1023 atoms He = 2.50 moles Al = 28.0 grams N = 1.80 x 1023 atoms Mg = atoms of Cu moles of C moles of He grams of Al atoms of N grams of Mg Mole Calculations (Mass of Helium is 4.00 ) 1. 1.00 mole of helium = 4.00g 2. 2.00 mole of helium = 8.00g 3. 1.00 mole of helium = 6.02 x 1023 atoms 4. 2.00 mole of helium = 1.20 x 1024 atoms 5. 16.0g of helium = 4.00 mol 6. 3.0l x 1023 atoms of helium = .500 moles 7. 8.00g of helium = 1.20 x 1024 atoms.