1. Atomic Theory of Matter We now take for granted the idea that all matter is comprised of atoms. But how did the Atomic Theory of Matter develop, and how was it proved? The notion that matter was made up of atoms had to be postulated and proved.

2. Experimental Evidence Experimental evidence for the atomic nature of matter was realized in the 18th century. Since then scientists have proved conclusively the atomic nature of matter In the mid ‘80’s a tool was developed which for the first time allowed scientists to actually “see” individual atoms and molecules.

3. Surface of graphite as imaged by a scanning tunneling microscope

4. http://www.columbia.edu/cu/chemistry/groups/flynn/

5. Law of Conservation of Mass 1775 - Lavoisier “Father of Modern Chemistry” In every chemical operation an equal amount of matter exists before and after the operation. Mass is conserved, the total mass after the chemical operation must be the same as that before.

6. Problem Potassium chlorate (KClO 3 ) decomposes to potassium chloride (KCl) and oxygen (O 2 ) when heated. In one experiment 100.0 g of KClO 3 generated 36.9g of O 2 and 57.3 g of KCl. What mass of KClO 3 remained unreacted? Mass of KClO 3 before reaction = mass of KCl + mass of O 2 + mass of unreacted KClO 3 100.0 g of KClO 3 = 57.3 g KCl + 36.9g O 2 + g unreacted KClO 3 g unreacted KClO 3 = 100.0 g - 57.3 g - 36.9 g = 5.8 g

7. Law of Definite Proportions Joseph Proust In a given chemical compound, the proportions by mass of the elements that compose it are fixed, regardless of the source of the compound. The ratio of elements in a compound is fixed regardless of the source of the compound. Water is made up of 11.1% by mass of hydrogen and 88.9% oxygen.

8. Problem In a set of experiments very pure tin (Sn) was combined with bromine (Br) forming tin tetrabromide (SnBr 4 ). Using the data below, confirm the law of definite proportions by calculating the % of tin in each sample of SnBr 4. Grams of Sn reactedGrams of SnBr 4 formed 2.844510.4914 3.012511.1086 4.523616.6752 Need to determine (mass of Sn reacted) (mass of Br reacted) Mass of Br reacted = Mass of SnBr 4 formed - mass of Sn reacted

9. Grams of Sn reactedGrams of Br reacted 2.844510.4914 - 2.8445 = 7.6469 3.012511.1086 - 3.0125 = 8.0961 4.523616.6752 - 4.5236 = 12.1516 (mass of Sn reacted) (mass of Br reacted) 0.3721 0.3723

10. Dalton’s Atomic Theory All matter consists of indivisible atoms Atoms of one kind of element are identical in mass and properties; atoms of different kinds of elements are different Compounds are made up of definite numbers of atoms of the component elements The weight of a compound equals the sum of the weights of the component elements

11. Gay Lussacs - Law of Combining Volumes For chemical reactions involving gases, combinations occur in simple proportions by volume. Moreover, the ratio of the volume of each product gas to the volume of either reacting gas is a ratio of small integers. For example, if the pressure and temperature are kept constant, two volumes of H 2 gas reacts with one volume of O 2 gas, producing two volumes of water vapor. The two reactants and the product form a 2:1:2 ratio. Based on experimental observation: 2 volumes of hydrogen + 1 volume of oxygen  2 volumes of water vapor How do atoms combine to form compounds?

12. Avogadro’s Hypothesis Equal volumes of different gases (at the same temperature and pressure) contain equal numbers of particles 2 volumes of hydrogen + 1 volume of oxygen  2 volumes of water vapor can be expressed as 2H 2 + O 2  2H 2 O While at this time there was no direct evidence to show that hydrogen and oxygen gas were H 2 and O 2, 50 years later this was proven to be the case.

14. Are Atoms really indivisible? Dalton reached his conclusions about atoms on the basis of evidence gained on a macroscopic level. As scientists developed more instrumentation capable of probing phenomena at a microscopic level, more about atoms was understood. Example: The color of the emitted light characterizes the element.

15. Electrons J.J. Thompson showed that the cathode rays were in fact particles of NEGATIVE charge, the rays could be deflected by a magnetic field. The term ELECTRON was coined for the negative particles. Thompson also calculated the charge of each particle

16. Millikan’s Oil Drop Experiment Robert Millikan’s oil drop experiment calculated the charge/mass ratio of the electron, and combining Thompson’s results the mass of the electron was calculated to be 9.10 x 10 -28 g. (actual mass of the electron 9.10939 x 10 -28 g) There must be a positive species which counters the electron charge.

17. Radioactivity Henri Becquerel in 1896 discovered high-energy radiation was spontaneously emitted from uranium. Later Marie Curie and her husband Pierre further investigated this spontaneous emission of radiation which was termed radioactivity.

18. Further studies of radioactivity by Rutherford showed that the radiation consisted of three types of radiation , ,  radiation.  and  radiation are bent by an electric field, but in opposite directions, and  radiation is unaffected. Rutherford was able to show that  particles have charge of +2 and that  –1 and that  particles combine with electrons to form Helium atoms.  rays are high energy electromagnetic radiation

19. The Structure of the Atom J.J. Thompson, realized that electrons were sub-atomic particles, and presented his theory of the model of the atom. The “PLUM-PUDDING” model

20. Rutherford’s conducted further experiments which contradicted Thompson’s model. To explain his results Rutherford postulated Most of the mass of the atom and all its positive charge was located in a concentrated core, called the nucleus. Most of the total volume of the atom is empty in which electrons move around the positive core.

21. Model of the Atom Since the times of Rutherford, many more subatomic particles have been discovered. However, for chemists three sub-atomic particles are all that we need to focus on – ELECTRON, PROTON, NEUTRON. Electrons are –1, protons +1 and neutrons are neutral. Atoms have an equal number of electrons and protons they are electrically neutral.

22. Protons and neutrons make up the heavy, positive core, the NUCLEUS which occupies a small volume of the atom.

23. Isotopes Atoms of different elements are distinguished by the difference in the number of protons and the number of electrons. Since atoms are electrically neutral, the number of protons = number of electrons Since the number of protons (and electrons) differ, their MASS differ. Hence atoms of different elements have different masses. So for example, hydrogen has ONE proton and ONE electron Carbon has SIX protons and SIX electrons

24. Atoms of the SAME element can have different number of NEUTRONS. These atoms of the SAME elements but with different number of neutrons are called ISOTOPES. Hence isotopes of the same elements have the same number of protons and electrons, but different number of neutrons and hence different masses. EXAMPLE – Carbon has three isotopes C12, C13, C14. Each of these isotopes differ by the number of neutrons – ALL have SIX protons. C12 has SIX neutron, C13 has SEVEN and C14 has EIGHT.

25. To denote the number of protons and neutrons in an atom the following symbol notation is used 12 6 C where 12 denotes SUM OF PROTONS + NEUTRONS 6 denotes the number of PROTONS So for the isotopes of carbon the complete chemical symbols are: 12 6 C, 13 6 C, 14 6 C The superscript, which is the sum of the number of protons and neutrons, is called the MASS NUMBER (A). The subscript indicates the number of protons and is called the ATOMIC NUMBER (Z).

26. How many protons, neutrons, and electrons are there in 197 is the mass number and refers to the total number of protons and neutrons. 79 is the atomic number and refers to the number of protons. Hence this atom has 79 electrons and 197-79 = 118 neutrons

27. Atomic Units Atoms are very, very light and very, very small Since atomic dimensions are so small, it would be cumbersome to use units we typically use for length (cm, m) or mass (g). Hence, on the atomic scale we define units appropriate for this scale

28. MASS – unit typically used is an ATOMIC MASS UNIT (amu) 1 amu = 1.66054 x 10 – 24 g ParticleChargeMass (g)Mass (amu) Proton+11.6727x10 -24 1.0073 NeutronNeutral1.6750x10 -24 1.0087 Electron-19.109x10 -28 5.486 x 10 -4 LENGTH – ANGSTROM(Å) = 10 -10 m Typical atomic dimensions are 1 to 5 x10 -10 m which corresponds to 1 to 5 Å.

29. Relative Atomic Mass A relative scale has been developed to compare the relative masses of atoms. The ATOMIC MASS of an atom is its relative mass on this scale. Carbon- 12 ( 12 C) has been set as the standard and assigned a RELATIVE MASS of exactly 12. Relative atomic masses have no units since they are the ratio of two masses.

30. Average Relative Atomic Mass Because the abundance of the isotopes of different elements are essentially constant, we can define an AVERAGE RELATIVE ATOMIC mass Average Relative Atomic Mass = average mass of atoms of an element = (Abundance) A (Mass) A + (Abundance) B (Mass) A + … (Table at the back of the text lists relative atomic masses of elements)

31. Problem Naturally occurring chlorine has two isotopes, 35 17 Cl, 37 17 Cl. The 35-Cl isotope has a relative atomic mass of 34.9688 and an abundance of 75.77% and the 37-Cl isotope has a relative atomic mass of 36.9659 and an abundance of 24.23%. Calculate the average atomic mass of Cl. Average Atomic Mass of Cl = (0.7577x34.9659) + (.2423x36.9659) = 35.4527 Average relative atomic mass of C is 12.0107 accounting for 12 C (98.892%, relative atomic mass 12.000000) and 13 C (1.108%, relative atomic mass 13.003354)

32. Relative Molecular Mass The relative molecular mass is the sum of the relative atomic masses of the atoms that make up the molecule. Example, the chemical formula for water is H 2 O Its relative molecular mass = 2 (1.00794) + 15.9994 = 18.0153

33. Avogadro’s Number Avogadro’s number: the number of atoms in exactly 12 g of 12 C. N o = 6.022137 x 10 23 Sodium (Na) has a relative atomic mass of 22.98977 12 Hence a sodium atom is (22.98977) times as heavy as 12 C If N o atoms of 12 C have a mass of 12g then, the mass of N o atoms of sodium must be (22.98977) 12g = 22.98977 g 12

34. The mass, in grams, of N o atoms of ANY element is numerically equal to the relative atomic mass of that element. Same applies to molecules. Since the relative molecular mass of water is 18.0152, the mass of N o water molecules is 18.0152g

35. MOLE: A mole has been defined as a unit containing 6.022137 x 10 23, Avogadro’s number, atoms or molecules, One mole of any atom or molecule contains the same number of atoms or molecules The mass, in grams, of ONE MOLE of atoms or molecules is numerically equal to relative atomic or molecular mass. Hence 1 mole of Na weighs 22.9898 g, 1 mole of H 2 O weighs 18.0153 g The MASS of one mole of atoms or molecules is called its MOLAR MASS and has UNITS of g/mol

36. PROBLEM How many moles of Fe are there in 8.232 g of Fe? How many atoms are there in 8.232 g of Fe? 1 mol H 2 O 0.2000 mol H 2 O x 18.015 g H 2 O = 3.603 g H 2 O 55.85 g Fe Moles of Fe = 8.232 g Fe x 1 mole = 0.1474 mol Fe How many grams of water are there in 0.2000 moles of water?