Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
III Atomic Theory • Democritus – All matter consists of very small indivisible particles “atomes” [Greek for indivisible] • Dalton’s Atomic Theory (beginning of the modern era of chemistry – Postulates • Elements composed of small particles called atoms. All atoms of a given element are identical (size, mass, chemical properties) and can not be changed into another element • Compounds made up of more than one element. Ratio of the number of atoms of the element is an integer or simple fraction • Chem reaction involves only separation, combination, or rearrangement of atoms. It does not result in creation or destruction • Dalton’s New Theory supported the mass laws • Mass Laws – Law of Conversation of Mass • Matter can not be created or destroyed – Proust’s Law of Definite Proportion [Constant Composition] • Different samples of the same compound always have the same elements in the same ratio – Law of Multiple Proportions • If two elements can combine to form more than one compound, the masses of one element with a fixed mass of the second are in a ratio of small whole numbers It is found that 14 grams of nitrogen, can react with 16 grams of oxygen, 32 grams of oxygen and 40 grams of oxygen. This is evidence of of the law of? A) Constant composition B) Conservation of mass C) Multiple proportions It is found that 14 grams of nitrogen, can react with 16 grams of oxygen, 32 grams of oxygen and 40 grams of oxygen. This is evidence of of the law of? – Law of Multiple Proportions • If two elements can combine to form more than one compound, the masses of one element with a fixed mass of the second are in a ratio of small whole numbers • Discovery of Particles Smaller than the Atom • Crookes Cathode Rays anode cathode Gas filled tube – When current applied, the tube begins to glow. Flow goes from cathode to anode. Called CATHODE RAYS – Same beam occurs no matter what the gas. Therefore, cathode rays are part of every atom. – Beams deflected in a magnetic field, so, they have negative charge – Term electron coined by Stanley in 1891. Cathode rays are beams of electrons • Mass to charge ratio (1897) Thomson – Thomson used a Crookes tube and placed it in an adjustable magnetic field. He was able to determine the mass to charge (m/e) ratio of the cathode rays. By comparing the ratio to the smallest mass to charge ratio in solution discovered that the mass of the cathode ray had to be 1/1000 the mass of hydrogen atom. Therefore, contrary to Dalton’s hypothesis, there were particles smaller than the atom. – Found (m/e) = - 5.686 x 10-12 kg/C • C = Coulomb, unit of charge • Millikan Oil Drop experiment (1909) – Measured the speed of charged drops of oil with and without an electric field. – Was able to calculate charge on the particles. – Found, all charges were an integral multiple of the same number. – Determined, this number had to be the charge of an electron. – e = - 1.602 x 10-19 C If you use Thomson’s m/e ratio and Millikan’s charge on the electron (e) you can calculate the mass of the electron. (m/e)(e) = (-5.686 x 10-12kg/C)(- 1.602 x 10-9 C) = 9.109 x 10 -31kg/electron Current accepted value = 9.1093897 x 10-31 kg/electron • X - Rays – Roentgen, working with the Crookes tube noticed that other material in the room fluoresced when the tube was turned off. – Decided that the radiation must be coming from Crookes tube but not regular light. It was some unknown radiation – Called X - ray. • Protons anode cathode – Goldstein used cathode ray tube with a perforated cathode – Noticed, cathode rays went to the anode, but + ions formed also – Mass of the + ions varied depending on the gas in the tube. – Lightest particle produced when hydrogen gas was used. – All other masses were multiples of the mass form when H was used. Therefore, the hydrogen ion had to be a fundamental particle in all matter – Called it the proton • Neutrons – Mass of the + ions formed varied depending on the gas. – But, total mass could not be accounted for only by the mass of the protons. – Therefore, there must be other particles with about the same mass as the proton but neutral • Found by Chadwick (1932), called neutrons. • A) B) C) D) E) His experiment determined the charge on an electron. Crookes Thomson Roentgen Stanley Millikan • • His experiment determined the charge on an electron. Millikan Oil Drop experiment (1909) – Measured the speed of charged drops of oil with and without an electric field. – Was able to calculate charge on the particles. • Models of the Atom developed after the discovery of electrons • Thomson Model (~ 1900) – “plum pudding” model. – Cloud of positive charge with nuggets of negative charge embedded in it so that it is neutral • Rutherford Model (1910) – Devised an experiment to test Thomson model – Shot a beam of particles (He2+; He nucleus) at a thin metal foil – If Thomson is correct, particles should pass through Metal foil particles expected – Concluded: 1) mass of atom and positive charge is in the center of the atom called the nucleus 2) electrons moved at random about the nucleus Atomic Structure • Fundamental Particles of Atom – Proton – Neutron – Electron p+ no e- nucleons • Atom is neutral because the number of p+ = number of e-. • Nuclide – Representation of any specific atomic species A Z X X = element A = atomic mass (p + n) Z = atomic number (p) # neutrons(N) = A - Z # electron in a neutral species = # protons = Z 28 14 Si p+ = ? no = ? e=? 14 14 14 For the nuclide 158 O -2 1. The number of protons is? For the nuclide 158 O -2 1. The number of protons is? 8 2. The number of neutrons is? For the nuclide 158 O -2 1. The number of protons is? 8 2. The number of neutrons is? 7 3. The number of electrons is? For the nuclide 158 O -2 1. The number of protons is? 8 2. The number of neutrons is? 7 3. The number of electrons is? 10 • Isotopes – Same element (same Z), but have different mass, therefore have different number of neutrons 12 C 13 C 14 C – The value of z = 6 is understood • Atomic Mass (Atomic Weight) (AW) – Mass of atom measured relative to the atomic standard. – Current standard is 126 C • Unit of atomic mass is atomic mass unit (u) or (amu) • 1 u defined as 1/12 the mass of 1 atom of 126 C • So 1 atom of 12C has a mass of 12 u exactly! • All other atomic masses based on this specific isotope • On periodic table each element has it own “box” Z X mass X = element Z = atomic number Mass = mass of naturally occurring material • u is an extremely small quantity – Difficult to measure – So: defined an amount that can be measured • Mole – SI unit for amount of a substance – Defined as: the amount of a substance that contains the same number of entities as there are atoms is exactly 12 grams of 12C. – Therefore: 1 mole 12C = 12.00 grams exactly – Defined conversion so it is exact • The quantity of 12C atoms that have a mass of 12.00 grams (i.e. contained in 1 mole of 12C) is 6.022 x 1023 atoms. • Because of the definition of a mole, then, 1 mole of anything contains 6.022 x 1023 particles. Called Avogadro’s Number • Because of the way the mole is defined, Avogadro’s number can actually be used for two different conversions. – The atom to mole conversion • 6.022 x 1023 atoms/mole – A mass conversion • 6.022 x 1023 u/gram • Mass of Elements (Weighted Average) • Most elements exist naturally as more than one isotope. So, “mass” of the element is actually the average of the mass of each isotope. • It is, however, not a direct average, but a weighted average. That is, we have to account for the abundance of each isotope. • Example: lithium exists and 6Li and 7Li. Isotope mass 6Li % abundance 7.42 7Li 92.58 7.01600 6.01512 Average (%1 /100)(mass 1 )(%2 /100)(mass 2 ) Average (0.0742)(6.01512) (0.9258)(7.01600) Average (0.4463219) (6.495412) Average 6.9417339 Average 6.942 • How are isotope mass and abundance determined? • By using mass spectrometer • Lightest mass most strongly deflected, so have greatest arc and are determined first. • Find the number of protons, neutrons and electrons in:(n.b. periodic table on back cover of notes) A) 24Mg+2P = . N = . E = • Find the number of protons, neutrons and electrons in:(n.b. periodic table on back cover of notes) A) 24Mg+2P = 12 . N = 12 . E = 10 • Find the number of protons, neutrons and electrons in:(n.b. periodic table inside back cover of notes) A) 24Mg+2 P = 12 . N = 12 . E = 10 B) 15N -3 P= .N= . E= • Find the number of protons, neutrons and electrons in:(n.b. periodic table inside back cover of notes) A) 24Mg+2 P = 12 . N = 12 . E = 10 B) 15N -3 P = 7 . N = 8 . E =10 • Find the number of protons, neutrons and electrons in:(n.b. periodic table inside back cover of notes) A) 24Mg+2 P = 12 . N = 12 . E = 10 B) 15N -3 C) 65Zn P = 7 . N = 8 . E =10 P= . N= . E= • Find the number of protons, neutrons and electrons in:(n.b. periodic table inside back cover of notes) A) 24Mg+2 P = 12 . N = 12 . E = 10 B) 15N -3 C) 65Zn P = 7 . N = 8 . E =10 P = 30 . N = 35 . E = 30 Boron has two isotopes, 10B and 11B. If a the mass of boron is 10.81, the mass of 10B = 10.0129, and 11B = 11.0093, what is the % abundance of 10B? A) 80.% B) 20.% C) 15.% D) 85.% Boron has two isotopes, 10B and 11B. If a the mass of boron is 10.81, the mass of 10B = 10.0129, and 11B = 11.0093, what is the % abundance of 10B? Average (%1 /100)(mass 1 )(%2 /100)(mass 2 ) Let x = fraction 10B, so 1-x = fraction 11B 10.81(x)(10.0129) (1- x)(11.0093) 10.8110.0129x 11.0093 -11.0093 x 0.9964x = 0.1993 x = 0.20 = 10B 10B = 20.% 1 - x = 0.80 = 11B 11B = 80. % • Determining the mass of a compound – A compound is a collection of elements – Made up of specific atoms in a specific ratio e.g. water = H2O = 2 H + 1 O; methane = CH4 = 1 C + 4 H – So, mass of the compound is the sum of the mass of each element • Mass of the element called Atomic Mass (Atomic Weight) AW • Mass of the compound called Molar Mass (Molecular Weight) M or MW • Examples • Mass of water: H2O Mass = 2(AW H) + AW O = 2(1.01) + (16.00) = 18.02 • Units of mass – If you are talking about an atom or a collection of atoms (molecule), Unit = u/ atom or u/ molecule – If you are talking about a mole of atoms or molecules Unit = grams/ mole What is the mass of 1 mole of Na2CO3, given: Na = 22.99; C = 12.01; O = 16.00 (give numeric answer) What is the mass of 1 mole of Na2CO3, given: Na = 22.99; C = 12.01; O = 16.00 (give numeric answer) 2(22.99) + 12.01 + 3(16.00) = 45.98 + 12.01 + 48.00 = 105.99