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The Atom: From Philosophical Idea to Scientific Theory

This unit discusses the history and development of the atomic theory, from Democritus' concept of the atom to Dalton's atomic theory and Rutherford's discovery of the atomic nucleus.

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The Atom: From Philosophical Idea to Scientific Theory

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  1. SAVE PAPER AND INK!!! When you print out the notes on PowerPoint, print "Handouts" instead of "Slides" in the print setup. Also, get rid of background shading by selecting “Pure Black & White” instead of “Grayscale.” Unit 1 The Atom, Atomic Theory & the Nucleus Chapters 3 & 21

  2. Atoms: The Building Blocks of Matter Chapter 3

  3. Chapter 3 – Section 1: The Atom: From Philosophical Idea to Scientific Theory The Particle Theory of Matter • In 400 B.C. Democritus, a Greek philosopher, first proposed the idea of a basic particle of matter that could not be divided any further. • He called this particle the atom, based on the Greek word atomosmeaning indivisible. • This early theory was not backed up by experimental evidence and was ignored by the scientific community for nearly 2000 years.

  4. Chapter 3 – Section 1: The Atom: From Philosophical Idea to Scientific Theory Foundations of Atomic Theory • By the late 1700s, experiments with chemical reactions led to the discovery of 3 basic laws: • The Law of Conservation of Mass – Mass is neither created nor destroyed during ordinary chemical reactions or physical changes.

  5. Chapter 3 – Section 1:The Atom: From Philosophical Idea to Scientific Theory Foundations of Atomic Theory The Law of Conservation of Mass

  6. Chapter 3 – Section 1: The Atom: From Philosophical Idea to Scientific Theory Foundations of Atomic Theory • The Law of Definite Proportions – A chemical compound contains the same elements in exactly the same proportions by mass regardless of the size of the sample or source of the compound. Visual Concept

  7. Chapter 3 – Section 1: The Atom: From Philosophical Idea to Scientific Theory Foundations of Atomic Theory • The Law of Multiple Proportions – If different compounds are composed of the same two elements, then the ratio of the masses of the elements is always a ratio of small whole numbers.

  8. Chapter 3 – Section 1: The Atom: From Philosophical Idea to Scientific Theory Dalton’s Atomic Theory • In 1808, John Dalton proposed an explanation for the three laws. His atomic theory states: • All matter is composed of atoms. • Atoms of the same element are identical; atoms of different elements are different. • Atoms cannot be subdivided, created, or destroyed. • Atoms of different elements combine in simple whole-number ratios to form chemical compounds. • In chemical reactions, atoms are combined, separated, or rearranged.

  9. Chapter 3 – Section 1: The Atom: From Philosophical Idea to Scientific Theory Corrections to Dalton’s Theory • Dalton turned Democritus’s idea into a scientific theory that could be tested by experiment. • But not all aspects of Dalton’s theory have proven to be correct. We now know that: • Atoms are divisible into even smaller particles. • A given element can have atoms with different masses.

  10. Chapter 3 – Section 2:The Structure of the Atom Dalton’s Atomic Model • An atom is the smallest particle of an element that has all the properties of that element. • Atoms are too small to see…even through the most powerfulmicroscope!! • Dalton thought atoms were solid balls of matter and were indivisible.

  11. Chapter 3 – Section 2:The Structure of the Atom Discovery of the Electron • In 1897, Joseph John (JJ) Thomson showed that cathode rays are composed of identical negatively charged particles, which were named electrons. • The electron was the first subatomic particle to be discovered.

  12. Chapter 3 – Section 2: The Structure of the Atom Thomson’s Cathode Ray Tube Experiment Visual Concept

  13. Chapter 3 – Section 2: The Structure of the Atom Charge and Mass of the Electron • In 1909, Robert Millikan measuredthe charge on the electron during his oil drop experiment. • Using the charge-to-mass ratio, scientists were able to figure out the mass of the electron: about 1/2000 the mass of a hydrogen atom.

  14. Chapter 3 – Section 2:The Structure of the Atom Millikan’s Oil Drop Experiment Visual Concept

  15. Chapter 3 – Section 2: The Structure of the Atom Thomson’s Plum Pudding Model • After the work of Thomson andMillikan, the accepted model of the atom was called the plum pudding model. • The atom was viewed as a ball of positively- charged material with tiny negatively-charged electrons spread evenly throughout.

  16. Chapter 3 – Section 2: The Structure of the Atom Discovery of the Atomic Nucleus • More detail of the atom’s structure was provided in 1911 by Ernest Rutherford and his associates Hans Geiger and Ernest Marsden. • The results of their gold foil experiment led to the discovery of a very densely packed bundle of matter with a positive electric charge. • Rutherford called this positive bundle of matter the nucleus.

  17. Chapter 3 – Section 2: The Structure of the Atom The Gold Foil Experiment Visual Concept

  18. Chapter 3 – Section 2: The Structure of the Atom Rutherford’s Atomic Model • After Rutherford’s gold foil experiment, the accepted model of the atom looked like this: • A small, positively-charged nucleus with negative electrons surrounding it at some distance away. Most of the atom is empty space.

  19. Chapter 3 – Section 2: The Structure of the Atom Subatomic Particles • The nucleus is made up of at least one positively charged particle called a protonand usually one or more neutral particles called neutrons. • Protons, neutrons, and electrons are often referred to as subatomic particles.

  20. Chapter 3 – Section 3:Counting Atoms Atomic Number • The atomic number (Z) of an element is the number of protons of each atom of that element. • Atoms of the same element all have the same number of protons.

  21. Chapter 3 – Section 3:Counting Atoms Mass Number • The mass numberis the total number of protons and neutrons in the nucleus of an atom. • Atoms of the same element can havedifferentmass numbers.

  22. Chapter 3 – Section 3:Counting Atoms Isotopes • Isotopes are atoms of the same element that have different masses. • Isotopes have the same number of protons and electrons but different numbers of neutrons. • Most of theelements consist of mixtures ofisotopes.

  23. Chapter 3 – Section 3:Counting Atoms Designating Isotopes • Hyphen notation: The mass number is written with a hyphen after the name of the element. uranium-235 • Nuclear symbol:The superscript indicates the mass number and the subscript indicates the atomic number. Mass number Atomic number

  24. Chapter 3 – Section 3:Counting Atoms Calculating Neutrons • The number of neutrons is found by subtracting the atomic number from the mass number. mass number − atomic number = number of neutrons • Nuclideis a general term for a specific isotope of an element.

  25. Chapter 3 – Section 3:Counting Atoms Calculating Subatomic ParticlesSample Problem How many protons, electrons, and neutrons are there in an atom of chlorine-37? Solution: Number of protons Number of electrons Number of neutrons = atomic number (on periodic table) 17 = number of protons 17 = mass number - protons 20

  26. Significant Figures • the non-place-holding digits in a reported measurement are called significant figures. • All non-zero numbers are significant, but some zeros in a written number are only there to help you locate the decimal point. • How do you remember which zeros are significant and which are not?

  27. Significant Figures • The Atlantic – Pacific Rule: Pacific (Present): If decimal point is present, start with the first non-zero number on the left. Atlantic (Absent): If decimal point is absent, start with the first non-zero number on the right.

  28. Significant Figures • Examples: How many significant figures? 99,000 2 sig figs 99,000. 5 sig figs 0.0099 2 sig figs 0.0990 3 sig figs

  29. Calculating with Significant Figures • When you use your measurements in calculations, your answer may only be as exact as your least exact measurement. • For addition and subtraction, round to the fewest decimal places. Example: (3 decimals) (1 decimal) (unrounded) (rounded) 50.259 + 17.4 = 67.659 67.7 • For multiplication and division, round to the fewest significant figures. Example: (3 sigfigs) (1 sigfig) (unrounded) (rounded) 0.135 x 20 = 2.7 3

  30. Chapter 3 – Section 3:Counting Atoms The Atomic Mass Unit • The standard used by scientists to compare units of atomic mass is the carbon-12 atom. • One atomic mass unit, or 1 amu, is exactly 1/12 the mass of a carbon-12 atom. • The atomic mass of any atom is determined by comparing it with the mass of the carbon-12 atom.

  31. Chapter 3 – Section 3:Counting Atoms Average Atomic Mass • Average atomic massis the weighted average of the atomic masses of the naturally occurring isotopes of an element. • The average atomic mass of an element depends on both the massand the relative abundance of each of the element’s isotopes.

  32. Percent • To convert a decimal to a percent, multiply by 100 (move the decimal point 2 places to the right.) .205 20.5% .090 9.0% • To convert a percent to a decimal, divide by 100 (move the decimal point 2 places to the left.) 65% 0.65 .03% .0003

  33. Multiplying by a Percent • First, convert the percent to a decimal, then multiply. Example: How many grams of carbon are in a 50.0 gram sample of a substance that is 17.5% carbon? 17.5% 0.175 50.0 g x 0.175 = 8.75 g

  34. Chapter 3 – Section 3:Counting Atoms Calculating a Weighted Average If you have the following grades, what would your marking period average be? • First, change percents to decimals. • Next, multiply each grade by its decimal percent. • Finally, add up all the products. ÷ 100 = 0.45 x = 32.4 x = 7.8 ÷ 100 = 0.10 x = 21.0 ÷ 100 = 0.25 x = 9.8 ÷ 100 = 0.10 x = 9.4 ÷ 100 = 0.10 + 80.4

  35. Chapter 3 – Section 3:Counting Atoms Calculating Average Atomic MassSample Problem 1 Copper consists of 69.15% copper-63, which has an atomic mass of 62.929 601 amu, and 30.85% copper-65, which has an atomic mass of 64.927 794 amu. What is the Average Atomic Mass of Copper? Solution • Change percents to decimals. • Multiply the atomic mass of each isotope by its relative abundance. • Add up all of the products. x = 43.52 x = 20.03 + 63.55

  36. Chapter 3 – Section 3:Counting Atoms Calculating Average Atomic MassSample Problem 2 A student believed that she had discovered a new element and named it mythium. Analysis found it contained two isotopes. The composition of the isotopes was 19.9% of atomic mass 10.013 and 80.1% of atomic mass 11.009. What is the average atomic mass, and do you think mythium was a new element? Solution: • Average Atomic Mass:(.199 x 10.013) + (.801 x 11.009) = 10.811 • Because the atomic mass is the same as the atomic mass of boron, mythium was not a new element. Round off to: 10.8

  37. Scientific Notation • A shorthand system of writing very large or very small numbers. • The power of 10 (called the exponent), is the number of times the decimal point is moved. • The number in front of the times sign(called the coefficient) must be greater than 1 and less than 10.

  38. Scientific Notation • Examples: What would each of these numbers be in scientific notation? 3000 3 x 103 32,000 3.2 x 104 0.05 5 x 10-2 0.0058 5.8 x 10-3

  39. Multiplying in Scientific Notation • First, multiply the coefficients. • Then, add the exponents together. • Rewrite your answer in correct scientific notation (coefficient less than 10), and remember to round to the correct number of significant figures. Example: 5.5 x 103 x 3.7 x 104 = 20.35 x 107 = 2.0 x 108

  40. Dividing in Scientific Notation • First, divide the coefficients. • Then, subtract the exponents. • Rewrite your answer in correct scientific notation (coefficient less than 10), and remember to round to the correct number of significant figures. Example: 1.8 x 10-2 ÷ 3.2 x 10-6 = 0.5625 x 104 = 5.6 x 103

  41. Chapter 3 – Section 3:Counting Atoms The Mole • A mole (mol) is the amount of a substance that contains as many particles as there are atoms in exactly 12 g of carbon-12. • It is a counting unit, similar to a dozen. In a dozen, there are 12 things. In a mole, there are 6.02 x 1023things. Visual Concept

  42. Chapter 3 – Section 3:Counting Atoms Avogadro’s Number • Avogadro’s number: 6.022 1415 × 1023— the number of particles in exactly one mole of a pure substance. • Named for nineteenth-century Italian scientist Amedeo Avogadro. Visual Concept

  43. Chapter 3 – Section 3:Counting Atoms Molar Mass • Molar Mass is the mass of one mole of a pure substance. • Molar mass units are g/mol. • The molar mass of an element is the same number as its atomic mass, only the units are different. Try some - What is the molar mass of: • Magnesium? • Carbon? 24.3 g/mol 12.0 g/mol

  44. Chapter 3 – Section 3:Counting Atoms Gram/Mole Conversions • Chemists use molar mass as a conversion factor in chemical calculations. • Remember, molar mass means grams per mole. Example: • What is the mass of 2.5 moles of Helium gas? Use molar mass as a conversion factor This is what we’re given: g He 4.0 x 2.5 mol He = 10. g He 1 mol He

  45. Chapter 3 – Section 3:Counting Atoms Gram/Mole ConversionsSample Problem 1 What is the mass in grams of 3.50 mol of the element copper, Cu? Solution: Conversion factor Given g Cu 63.5 x 3.5 mol Cu = 220 g Cu mol Cu 1

  46. Chapter 3 – Section 3:Counting Atoms Gram/Mole ConversionsSample Problem 2 A chemist produced 11.9 g of aluminum, Al. How many moles of aluminum were produced? Solution: Conversion factor Given 1 mol Al x 11.9 g Al = 0.441 mol Al 27.0 g Al

  47. Chapter 3 – Section 3:Counting Atoms Conversions with Avogadro’s Number • Avogadro’s number can be used as a conversion factor between atoms and moles. • Avogadro’s number units are atoms per mole. Example: • How many moles of silver, Ag, are in 3.01  1023 atoms of silver? Conversion factor Given mol Ag 1 3.01 x 1023 atoms Ag x = 0.500 mol Ag 6.02 x 1023 atoms Ag

  48. Chapter 3 – Section 3:Counting Atoms Conversions with Avogadro’s NumberSample Problem 1 What is the mass in grams of 1.20  108 atoms of copper, Cu? Solution: 2nd Conversion factor 1st Conversion factor Given 1 63.5 mol Cu g Cu x x 1.20 x 108 atoms Cu 6.02 x 1023 1 atoms Cu mol Cu 1.27 x 10-14 g Cu =

  49. Nuclear Chemistry Chapter 21

  50. Chapter 3 – Section 2: The Structure of the Atom Forces in the Atom • Electrons and protons attract because of opposite electrical charges, but protons and protons repel since they have the same charge. • The nucleus is heldtogether by a mysterious force called the strong nuclear force which only exists between nucleons (protons and neutrons) whichare very close together.

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