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CHAPTER 2

CHAPTER 2 . AP CHEM. 2.2 Review. Atomic number is the number of protons An atomic symbol represents the element The mass number is equal to the number of protons +neutrons. Isotopes.

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CHAPTER 2

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  1. CHAPTER 2 AP CHEM

  2. 2.2 Review Atomic number is the number of protons An atomic symbol represents the element The mass number is equal to the number of protons +neutrons

  3. Isotopes Definition: Two or more forms of the same element containing different numbers of neutrons. Isotopes of an element have the same chemical properties but differing atomic mass Examples: Carbon -12 Carbon -13 Chlorine -35 Chlorine - 37

  4. Percent abundance Most elements exist as atoms of different masses - isotopes. Examples: Carbon -12 98.90 % natural abundance Carbon -13 1.10 % abundance Carbon -14 less than 1/trillionth Chlorine-35 75.77 % Chlorine 37 24.23 %

  5. MASS SPECTROMETER 4. Results measured in detector • Low pressure gas sample ionized • Gas bombarded by e-’s, ejecting e-’s from sample creating cations (positively charged ions). • Ions sorted & separated by mass and charge using magnetic field (using Inertia – heavier particles deflected less than lighter and using charges – highly charged particles interact more strongly with magnets and electric fields and are deflected more that particles of equal mass with smaller charges).

  6. ATOMIC WEIGHT Mass or weight??? one amu (atomic mass unit) = exactly 1/12 of the mass of a carbon-12 atom Atomic weight is the weighted average of the masses of an elements’ isotopes

  7. EX2.1 Boron exists in two naturally, occurring isotopes. B-10 ( 10.016 amu) makes up 18.83% of each natural sample of this element. The remaining 81.17% of the sample is B-11 (11.013 amu). What atomic mass would be calculated for this mixture of isotopes? Solve by multiplying isotope mass by % Boron -10 (10.016)x(.1883) = 1.8860 Boron -11 (11.013)x(0.8117) = +8.9393 10.825 amu Keep 5 sig figs – use percent as exact numbers 

  8. Ex2.2 The two natural isotopes of Lithium are 6Li (6.01512 amu) which accounts for 7.42% of the total and 7Li which accounts for the remaining amount. If the mass of lithium is shown as 6.942 on the periodic table, what is the mass of the Li -7 isotope. Lithium -6 (6.01512)(0.0742) = 0.4463219 Lithium – 7 (Lithium 7)(.9258) = 0.9258 x 6.942 amu 0.446322 + 0.9258x = 6.942 0.9258 x = 6.4959 Lithium-7 = x x = 7.016 amu

  9. Most abundant natural isotope Hydrogen is the most abundant in the universe

  10. 2.5 Atoms and the mole Definition: the amount of a substance that contains as many entities ( atoms, molecules, or other particles) as there are atoms in 0.012 kg of carbon-12 atoms. Current accepted number 1 mole = 6.022045 x 1023 particles

  11. ANALOGY Baker’s count in dozens 1 dozen = 12 donuts Chemists count atoms in Moles http://www.youtube.com/watch?v=1R7NiIum2TI Avogadro’s # 6.022 x 1023 = 1 mole 2nd Definition The mass of one mole of atoms of a pure element in grams is numerically equal to the atomic weight of that element in amu

  12. Ex 2.3 One mole of sulfur contains 6.022 x 10 23 atoms of sulfur and has a mass of 32.06 grams. What is the mass of one atom of sulfur? Sample = 1 atom sulfur 1 atom S 1 mole 32.06 grams S = 5.324 x10-23g Sulfur 6.022 x 1023 atoms 1 mole S

  13. Ex 2.4 How many sulfur atoms are present in 1.00 grams of sulfur? Given: 1 mole S = 6.022 x 10 23 atoms S Sulfur atomic weight = 32.06 grams/mole ( P.T.) 1.00 grams S 1 mole S 6.022 x 10 23 atoms = 1.88 x 10 22 atoms Sulfur 32.06 g S 1 mole

  14. Equivalencies and conversion factors

  15. Sample mole problems Ex2.5 How many moles are equivalent to 5.00 g CaCO3? (Add molar masses of calcium, oxygen and carbon present to get the molar mass) 5.00 grams CaCO31 mole CaCO3 = 0.0500 moles CaCO3 100.089 g CaCO3 Ex2.6 A microchemical experiment requires 0.0100 moles of Al(NO3)3. How many grams is this? 0.0100 molesAl(NO3)3 213.0017 gAl(NO3)3 = 2.13 g Al(NO3)3 1 mole Al(NO3)3

  16. More mole problems Ex. 2.7 For exactly 1.000 g of carbon disulfide (CS2), how many molecules are present? How many atoms of sulfur? 1.000 g CS2 1 mole CS2 6.022 x 1023 molecules CS2= 76.131 g CS2 1 mole CS2 7.910 x1021 CS2 To convert to atoms (continue 3 atoms / I molecule CS2) 1.000 g CS2 1 mole CS2 6.022 x 1023 molecules CS2 2 atoms sulfur =1.582 x 1022 76.131 g CS2 1 mole CS2 1 molecule CS2atoms S Ex2.8 A sample of sulfur hexafluoride (SF6) contains 1.69 x 1022 atoms of fluorine. What is the mass of the sample?

  17. More mole problems Ex2.8 A sample of sulfur hexafluoride (SF6) contains 1.69 x 1022 atoms of fluorine. What is the mass of the sample? 1.69 x 1022atoms F 1 molecules SF6 1 mole SF6 146.054 g = 6 atoms F 6.022 x 1023 molecules SF6 1 mol 0.683 g SF6

  18. The Periodic table • History Mendeleev, 1869 - developed the first periodic table by identifying similarities among the elements based on increasing atomic weights (did not know atom was divisible) Periodic Law – The properties of elements are periodic functions of their atomic numbers Moseley , 1913 - organized the P.T. in order of atomic number

  19. The periodic table Groups or families – columns down the P.T. Periods or series – rows across P.T. Metals - on the left side of the stair case , metalliods along the staricase, nonmetals on the right side Main group elements on the left and right side of P.T. Transistion metals – in the middle Lanthanide and actinide – on th every bottom Specific names Group 1 – alkali metals Group 2 – alkaline earth metals Group 17 (7A) – halogens Group 18 (8A) – noble gases

  20. Periodic table http://www.youtube.com/watch?v=SmwlzwGMMwc

  21. Coulomb’s Law – a helpful tool in explaining P.T. trends • An Inverse Square Law • The magnitude of the electrostatic force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them. • If the two charges have the same sign, the electrostatic force between them is repulsive; if they have different sign, the force between them is attractive. • Like charges repel • unlike charges attract

  22. The Equation

  23. Applying Coulomb’s Law to P.T. • The nucleus of the atom (with a positive charge) is attracted to each electron (with a negative charge) and visa versa. • The attraction increases exponentially as the distance between the e- and the nucleus decreases. • Note: electrons repel each other

  24. Homework Chapter 2 HW#1 -21, 23, 25, 27, 29, 47 Composition Atoms, Isotopes HW#2 - 31, 33, 35 Atomic Symbols Wkst HW #3 – 37, 39, 59, 61, 62, 63 HW #4 - Mole Relationships Wkst. HW #5 - 43, 45, 57 PT

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