1. C H E M I S T R Y Chapter 2 Atoms, Molecules, and Ions

2. Conservation of Mass and the Law of Definite Proportions Law of Conservation of Mass: Mass is neither created nor destroyed in chemical reactions. Hg I 2 (s) + 2KNO 3 (aq)Hg(NO 3 ) 2 (aq) + 2K I(aq) 4.55 g + 2.02 g = 6.57 g 3.25 g + 3.32 g = 6.57 g

3. Conservation of Mass and the Law of Definite Proportions Law of Definite Proportions: Different samples of a pure chemical substance always contain the same proportion of elements by mass. By mass, water is:88.8 % oxygen 11.2 % hydrogen

4. The Law of Multiple Proportions and Dalton’s Atomic Theory Law of Multiple Proportions: Elements can combine in different ways to form different substances, whose mass ratios are small whole-number multiples of each other. Insert Figure 2.2 p37 7 grams nitrogen per 8 grams oxygen 7 grams nitrogen per 16 grams oxygen nitrogen monoxide (NO): nitrogen dioxide (NO2):

5. The Law of Multiple Proportions and Dalton’s Atomic Theory Elements are made up of tiny particles called atoms. Each element is characterized by the mass of its atoms. Atoms of the same element have the same mass, but atoms of different elements have different masses. The chemical combination of elements to make different chemical compounds occurs when atoms join in small whole-number ratios. Chemical reactions only rearrange how atoms are combined in chemical compounds; the atoms themselves don’t change.

6. Atomic Structure: Electrons Cathode-Ray Tubes: J. J. Thomson (1856-1940) proposed that cathode rays must consist of tiny negatively charged particles. We now call them electrons.

7. Atomic structure: Electrons The strength of deflecting magnetic electric field The size of the negative charge of electron The mass of the electron

8. Atomic Structure: Electrons Millikan’s oil drop experiment Charged to mass ratio

9. Atomic Structure: Protons and Neutrons

10. Rutherford’s Gold Foil Experiment In Rutherford’s gold foil experiment (1871-1937),positively charged particles were aimed at atoms of gold. mostly went straight through the atoms. were deflected only occasionally. Conclusion: There must be a small, dense, positively charged nucleus in the atom that deflects positive particles that come close. Rutherford proposed that the atom must consist mainly of empty space with the mass concentrated in a tiny central core—the nucleus

11. Atomic Structure: Protons and Neutrons The mass of the atom is primarily in the nucleus. The charge of the proton is opposite in sign but equal to that of the electron.

12. Subatomic Particles Atoms contain subatomic particles. Protons have a positive (+) charge. Electrons have a negative (-) charge. Neutrons are neutral. Like charges repel and unlike charges attract.

13. Atomic Numbers Atomic Number (Z): Number of protons in an atom’s nucleus. Equivalent to the number of electrons around an atom’s nucleus Mass Number (A): The sum of the number of protons and the number of neutrons in an atom’s nucleus Isotope: Atoms with identical atomic numbers but different mass numbers

14. Atomic Numbers

15. carbon-13 or C-13 C 13 6 atomic number mass number carbon-12 or C-12 C 12 6 atomic number mass number 6 protons 6 electrons 7 neutrons 6 protons 6 electrons 6 neutrons

16. Isotopic symbols

17. 1.How many protons, electrons and neutrons are present in an atom of 2.Write isotopic symbols in both forms for aluminum isotope with 14 neutrons 3.An atom has 82 electrons and 126 neutrons. What is its mass number and what is the element? Examples

18. Atomic Masses and the Mole Atomic Mass: weighed according to the natural abundance of each isotope. Sometimes called average atomic mass. The calculation for atomic mass requires the percent(%) abundance of each isotope. atomic mass of each isotope of that element. sum of the weighted averages. Atomic mass = mass of isotope(1)x (%) + mass of isotope(2) x (%) 100 100

19. Atomic Masses and the Mole carbon-12:98.89 % natural abundance12 amu carbon-13:1.11 % natural abundance13.0034 amu Why is the atomic mass of the element carbon 12.01 amu? = 12.01 amu mass of carbon = (12 amu)(0.9889) + (13.0034 amu)(0.0111 ) = 11.87 amu + 0.144 amu

20. Atomic Masses and the Mole Atomic mass unit (amu) is the mass in grams of a single atom 1 amu = 1.6605 x 10 -24 g E.g 1 H atom = 1.01 amu

21. Calculating Atomic Mass for Copper Copper has two naturally occurring isotopes: Cu-63 with mass 62.9396 amu and a natural abundance of 69.17% and Cu-65 with mass 64.9278 amu and a natural abundance of 30.83%. Calculate the atomic mass of copper Use atomic mass and percent of each isotope to calculate the contribution of each isotope to the weighted average. Atomic mass Cu-63 x % abundance = Atomic mass Cu-65 x % abundance = Sum is atomic mass of Cu is

22. Example Bromine has two naturally occurring isotopes (Br-79 and Br-81) and has an atomic mass of 79.904 amu. The mass of Br-81 is 80.9163 amu, and its natural abundance is 49.31%. Calculate the mass and natural abundance of Br-79

23. Collection Terms A collection term states a specific number of items. 1 dozen donuts = 12 donuts 1 ream of paper = 500 sheets 1 case = 24 cans Copyright © 2009 by Pearson Education, Inc.

24. A Mole of Atoms A mole is a collection that contains the same number of particles as there are carbon atoms in 12.0 g of carbon. 6.02 x 10 23 atoms of an element (Avogadro’s number).

25. Atomic Masses and the Mole Avogadro’s Number (N A ): One mole of any substance contains 6.022 x 10 23 formula units. 1 mole of anything = 6.02 x10 23

26. Samples of 1 Mole Quantities 1 mole of C atoms= 6.02 x 10 23 C atoms 1 mole of Al atoms= 6.02 x 10 23 Al atoms 1 mole of S atoms= 6.02 x 10 23 S atoms

27. Avogadro’s Number and the Mole Avogadro’s Number (N A ): One mole of any substance contains 6.022 x 10 23 formula units. One mole of any substance is equivalent to its molar mass. 1 mole = 12.01 g C:C: 6.022 x 10 23 molecules = 12.01 g 1 mole = 1.08 g 6.022 x 10 23 molecules = 1.08 g H

28. Molar Mass from Periodic Table Molar mass is the atomic mass expressed in grams. 1 mole of Ag 1 mole of C 1 mole of S = 107.9 g = 12.01 g = 32.07 g Copyright © 2009 by Pearson Education, Inc.

29. Examples Give the molar mass for each A. 1 mole of K atoms =________ B. 1 mole of Sn atoms =________

30. Avogadro’s Number Avogadro’s number, 6.02 x 10 23, can be written as an equality and two conversion factors. Equality: 1 mole = 6.02 x 10 23 particles Conversion Factors: 6.02 x 10 23 particles and 1 mole 1 mole6.02 x 10 23 particles

31. Examples 1. Calculate the number of copper atoms in 2.45 mol of copper 2. The number of moles of S in 1.8 x 10 24 atoms of S 3. The number of atoms in 2.0 g of Al atoms is