Presentation on theme: "Principles of Chemistry, Chapt. 2: Atomic Structure and The Elements I.The Structure of Atoms protons, neutrons, and electrons II.Atomic Structure and."— Presentation transcript:
Principles of Chemistry, Chapt. 2: Atomic Structure and The Elements I.The Structure of Atoms protons, neutrons, and electrons II.Atomic Structure and Properties—the Elements atomic mass, atomic number, isotopes III.The Mole Concept: 6.02 x IV.The Periodic Table 1 Homework: Chapt. 2 Problems 26, 29, 37, 43, 75
~ meters = 1 angstrom (Å) _ m Smeared out electron charge cloud Protons and neutrons 2 Most of the mass is here Most of the Chemistry is here Atomic Theory in a single Slide
3 STM Image: Oxygen atoms at the surface of Al 2 O 3 /Ni 3 Al(111) S. Addepalli, et al. Surf. Sci. 442 (1999) 3464 Electronic charge cloud surrounding the nucleus
4 What’s inside the nucleus: ParticleMass (amu) Charge Proton (p + )1.007 amu +1 Neutron (n 0 )1.009 amu 0 What’s outside the nucleus: Electron (e-) amu-1 Note: mass ratio of electron/proton (M p+ /M e- ) = 1836 For any atom: # of electrons = # of protons: Why?
Atomic Theory: Late 19 th Century Atomic theory—everything is made of atoms—generally accepted (thanks to Ludwig Boltzman, and others). Mendeleev/periodic table—accepted, but the basis for periodic behavior not understood What are atoms made of? How are they held together? 5
6 Electrical behavior: “+” attracts “-” but like charges repel Atoms must contain smaller sub-units. Radioactive material Beam of , , and Electrically charged plates β-particles (“–”) Gamma ray (γ) No charge, no deflection α-particle (“+” ) Heavier, deflected less than β – + Alpha particle 2 n 0 + 2p + Beta particel electron (e-) Gamma photon
Electric and magnetic fields deflect the beam. Gives mass/charge of e - = −5.60 x g/C Coulomb (C) = SI unit of charge Thomson (1897) discovered the e - : 7 “Cathode rays” Travel from cathode (-) to anode (+). Negative charge (e − ). Emitted by cathode metal atoms. fluorescent screen – high voltage + cathode ray
+ _ + _ Essence of the Thompson Experiement (and old fashioned TV’s) Electric field exerts Force + plate repels +charged particles - Plate repels – charged particles F = Eq = ma d = displacement = ½ at 2 = Eq/m (t = L/V x ) Therefore, the greater the displacement, the lower the mass of the particle d x y Phosphor screen 8
Charge on each droplet was: n (−1.60 x C) with n = 1, 2, 3,… n (e - charge) Modern value = − x C. = −1 “atomic units”. These experiments give: 10 Modern value = x g = (-1.60 x C)(-5.60 x g/C) = 8.96 x g m e = charge x mass charge
Atoms gain a positive charge when e - are lost. 11 Implies a positive fundamental particle. Hydrogen ions had the lowest mass. Hydrogen nuclei assumed to have “unit mass” protonsCalled protons. Modern science: m p = x g m p ≈ 1800 x m e. Charge = -1 x (e - charge). = x C = +1 atomic units
How were p + and e - arranged? 12 Thompson: Ball of uniform positive charge, with small negative dots (e - ) stuck in it. The “plum-pudding” model Rutherford (former Thompson graduate student) fired α-particles at thin metal foils. Expected them to pass through with minor deflections.
Rutherford Scattering Experiment 13
Different Models of the Atom: different scattering results α particles “Plum pudding model” + and – charges evenly distrubted low, uniform density of matter No back scattering Rutherford’s explanation of results: Small regions of very high density + charge in the dense regions - Charges in region around it From wikipedia 14
Some Large Deflections were osbserved α particles Rutherford “It was about as credible as if you had fired a 15-inch shell at a piece of paper and it came back and hit you.” 15
≈10,000 times smaller diameter than the entire atom. e - occupy the remaining space. α particles nucleus Most of the mass and all “+” charge is concentrated in a small core, the nucleus. 16
17 Nucleus diameter~ Å = meters Mass ~ Kg Charge cloud Diameter ~ 1 Å Mass ~ kg
18 Most Chemistry involves rearrangement of outermost electrons, not nuclei Example: H 1p +, 1 e- H + H H 2 +
7 Å Epitaxial Al2O3(111) film on Ni 3 Al(111) (Kelber group): Grown in UHV Uniform No Charging S. Addepalli, et al. Surf. Sci. 442 (1999) 3464 STM Start with ordered films growth studies Proceed to amorphous films on Si(100) Surface terminated by hexagonal array of O anions 20
Atomic mass > mass of all p + and e - in an atom. Rutherford proposed a neutral particle. 21 m n ≈ m p (0.1% larger). m n = x g. Present in all atoms (except ‘normal’ H). neutrons Chadwick (1932) fired -particles at Be atoms. Neutral particles, neutrons, were ejected:
~ meters = 1 angstrom (Å) _ m Smeared out electron charge cloud Protons and neutrons 22
Nucleus Contains p + and n 0 Most of the atomic mass. Small (~10,000x smaller diameter than the atom). Positive (each p + has +1 charge). Small light particles surrounding the nucleus. Occupy most of the volume. Charge = -1. Atoms are neutral. Number of e − = Number of p + Electrons 23
24 A neutron can decay into a proton and electron: n 0 p + + e - This can cause decay of a radioactive element, e.g., 14 6 C # of p + + n 0 Atomic No. (# e- = # p + Elemental symbol (carbon) Carbon with 6 protons and 8 neutrons is unstable (radioactive) Carbon with 6 protons and 6 neutrons is stable (non-radioactive 14 6 C 12 6 C radioctivestable
25 An atom of 14 C can undergo decay to N as a neutron turns into a proton + an emitted electron 14 6 C 7 N + e- 1 p + 1 n 0 + an electron (emitted)
Same element - same number of p + 26 Atomic number Atomic number (Z) = number of p + 1 amu = x g ParticleMass Mass Charge (g) (amu) (atomic units) e − x −1 p x n x Atomic mass unit (amu) = (mass of C atom) that contains 6 p + and 6 n
Isotopes same element Atoms of the same element with different A. equal numbers of p + different numbers of n 0 27 deuterium (D) tritium (T) Hydrogen isotopes:H1 p +, 0 n H1 p +, 1 n H1 p +, 2 n 0
ISOTOPES: SAME Element, Different numbers of neutrons C 12 C 14 Carbon: atomic no. = 6 6 protons in the nucleus+ 6 electrons Atomic mass = 12 amu (12 gr/mole) Therefore, 6 protons + 6 neutrons Atomic mass = 14 amu Therefore, 6 protons+ 8 neutrons 28
Isotopes Display the same chemical reactivities (which depend mainly on the outer arrangement of the electrons) 12 C + O 2 CO 2 14 C +O 2 CO 2 Isotopes display different nuclear properties C 12 stable C 14 Radioactive: spontaneously emits electrons. Half-life ~ 5730 years 29
Isotopes and Moles (more on this later) and isotope abundance: 1 mole = 6.02 x of anything! 1 mole of atoms = 6.02 x atoms Molar Mass (in grams) = average atomic mass (in amu) 1 mole of H atoms = gr. Why not gr?? most atoms are 1 H, but some are 2 H (deuterium) 30
Average atomic mass of H = amu 100 atoms have a mass of amu # of 2 H atoms = n # of 1 H atoms = 100 –n (assume these are the only two isotopes that matter) Mass of 100 atoms = n x (100-n) x = amu 2n n = n = 0.8 So, out of every 100 atoms, have H atoms Out of every 1000 atoms, have 8 2 H atoms Natural abundance of “heavy hydrogen (deuterium) is then 0.8% 31
Most elements occur as a mixture of isotopes. 32 Magnesium is a mixture of: 24 Mg 25 Mg 26 Mg number of p number of n mass/ amu
percent abundance For most elements, the percent abundance of its isotopes are constant (everywhere on earth). The periodic table lists an average atomic weight. 33 Example Boron occurs as a mixture of 2 isotopes, 10 B and 11 B. The abundance of 10 B is 19.91%. Calculate the atomic weight of boron.
34 Atomic weight for B = amu = amu Atomic mass = Σ (fractional abundance)(isotope mass) ( amu) = amu 11 B ( amu) 10 B = amu % abundance of 11 B = 100% % = 80.09% Boron occurs as a mixture of 2 isotopes, 10 B and 11 B. The abundance of 10 B is 19.91%. Calculate the atomic weight of boron.
B Boron 5 Atomic number (Z) Symbol Name Atomic weight Periodic table: 35
A counting unit – a familiar counting unit is a “dozen”: 36 1 dozen eggs= 12 eggs 1 dozen donuts= 12 donuts 1 dozen apples= 12 apples 1 mole (mol) = Number of atoms in 12 g of 12 C Latin for “heap” or “pile” 1 mol = x “units” Avogadro’s numberAvogadro’s number
A green pea has a ¼-inch diameter. 48 peas/foot. (48) 3 / ft 3 ≈ 1 x 10 5 peas/ft 3. V of 1 mol ≈ (6.0 x peas)/(1x 10 5 peas/ft 3 ) ≈ 6.0 x ft 3 37 height = V / area, 1 mol would cover the U.S. to: U.S. surface area = 3.0 x 10 6 mi 2 = 8.4 x ft x ft x ft 2 =7.1 x 10 4 ft = 14 miles !
1 mole of an atom = atomic weight in grams Xe atom has mass = amu 1 mol of Xe atoms has mass = g and There are x atoms in 1 mol of He and 1 mol of Xe – but they have different masses. 1 He atom has mass = amu 1 mol of He has mass = g … 1 dozen eggs is much heavier than 1 dozen peas!
Example How many moles of copper are in a g sample? Cu-atom mass = g/mol (periodic table) Conversion factor: 1 mol Cu g = 1 n Cu = g x 1 mol Cu g = mol Cu n = number of moles 39
Calculate the number of atoms in a g sample of boron. 40 n B = (1.000 g) 1 mol B g = mol B B atoms = ( mol B)(6.022 atoms/mol) = B atoms
Dimensional Analysis and Problem Solving Special Homework Problem: Due Tues. Recitation Density = mass/volume Problem: Assume that a hydrogen atom has a spherical diameter of 1 angstrom Assume that the nucleus (1 proton) has a diameter of angstrom 1.Calculate the densities of the nucleus, and of the electron charge cloud in kg/m 3 2.Calculate the ratio of the two densities: R = d nucleus /d electron cloud Mass of proton = 1.67 x kg Mass of electron = 9 x kg 41
Summarizes Atomic numbers. Atomic weights. Physical state (solid/liquid/gas). Type (metal/non-metal/metalloid). Periodicity Elements with similar properties are arranged in vertical groups. 42
In the USA, “A” denotes a main group element… …”B” indicates a transition element. International system uses 1 … 18. The Periodic Table 43
Main group metal Transition metal Metalloid Nonmetal The Periodic Table 44
A period is a horizontal row Period number 45
A group is a vertical column Group 7A Halogens Group 8A Noble gases Group 2A Alkaline earth metals Group 1A Alkali metals (not H) 46
Alkali metals (group 1A; 1) Alkaline earth metals (group 2A; 2) 47 Grey … silvery white colored. Highly reactive. Never found as native metals. Form alkaline solutions.
Transition Elements (groups 1B – 8B) Also called transition metals. Middle of table, periods 4 – 7. Includes the lanthanides & actinides. 48 Lanthanides and Actinides Listed separately at the bottom. Chemically very similar. Relatively rare on earth. (old name: rare earth elements)
Groups 3A to 6A Most abundant elements in the Earth’s crust and atmosphere. Most important elements for living organisms. 49 Halogens (group 7A; 17) Very reactive non metals. Form salts with metals. Colored elements. Noble gases (8A; 18) Very low reactivity. Colorless, odorless gases.
Atoms are very small. 1 tsp of water contains 3x as many atoms as there are tsp of water in the Atlantic Ocean! 50 Impractical to use pounds and inches... Need a universal unit system The metric system. The SI system (Systeme International) - derived from the metric system.
A decimal system. Prefixes multiply or divide a unit by multiples of ten. 51 Prefix Factor Example megaM megaton = 1 x 10 6 tons kilok kilometer (km) = 1 x 10 3 meter (m) decid deciliter (dL) = 1 x liter (L) centic centimeter (cm) = 1 x m millim milligram (mg) = 1 x gram (g) microμ micrometer (μm) = 1 x m nanon nanogram (ng) = 1 x g picop picometer (pm) = 1 x m femtof femtogram (fg) = 1 x g
1 pm = 1 x m ;1 cm= 1 x m 52 How many copper atoms lie across the diameter of a penny? A penny has a diameter of 1.90 cm, and a copper atom has a diameter of 256 pm. x 1 x m 1 cm = 7.42 x 10 7 Cu atoms 1 pm 1 x m x 1.90 cm = 1.90 x pm Number of atoms across the diameter: 1.90 x pm x 1 Cu atom 256 pm
Length Length1 kilometer= mile 1 inch= 2.54 cm (exactly) 1 angstrom (Å)= 1 x m Volume Volume1 liter (L) = 1000 cm 3 = 1000 mL = quarts 1 gallon= 4 quarts = 8 pints Mass Mass1 amu= x g 1 pound= g = 16 ounces 1 ton (metric)= 1000 kg 1 ton (US)= 2000 pounds 53
Example: What is the volume of a 2 gallon container in Liters? 1 gallon x 4 quarts/gallon = 4 quarts 4 quarts x 1Liter/1.057 quarts = Liters (L) 54
165 mg dL A patient’s blood cholesterol level measured 165 mg/dL. Express this value in g/L 1 mg = 1 x g ; 1 dL = 1 x L x 1 x10 -3 g 1 mg = 1.65 g/L x 1 dL 1 x10 -1 L 55
All measurements involve some uncertainty. one Reported numbers include one uncertain digit. 56 Consider a reported mass of g Last digit (“2”) is uncertain Close to 2, but may be 4, 1, 0 … significant figuresFive significant figures in this number.
Read numbers from left to right. starting Count all digits, starting with the 1 st non-zero digit. Allexcept All digits are significant except zeros used to position a decimal point (“placeholders”). 5 sig. figs. ( x ) 57 placeholders significant
Round to 3 significant figures. 1 st non-significant digit Examine the 1 st non-significant digit. If it: 59 > 5, round up. < 5, round down. = 5, check the 2 nd non-significant digit. round up if absent or odd; round down if even. last retained digit 1 st non- significant digit Rounds up to nd non- significant digit