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.

3. Conservation of Mass and the Law of Definite Proportions 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

4. 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

5. 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):

6. 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.

7. 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. His first experiment was to build a cathode ray tube with a metal cylinder on the end. This cylinder had two slits in it, leading to electrometers, which could measure small electric charges. He found that by applying a magnetic field across the tube, there was no activity recorded by the electrometers and so the charge had been bent away by the magnet. This proved that the negative charge and the ray were inseparable and intertwined.

8. he constructed a slightly different cathode ray tube, with a fluorescent coating at one end and a near perfect vacuum. Halfway down the tube were two electric plates, producing a positive anode and a negative cathode, which he hoped would deflect the rays. the rays were deflected by the electric charge, proving beyond doubt that the rays were made up of charged particles carrying a negative charge Atomic Structure: Electrons

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

10. Atomic Structure: Electrons Millikan’s oil drop experiment Millikan's experiment involved measuring the force on oil droplets in a glass chamber sandwiched between two electrodes, one above and one below. With the electrical field calculated, he could measure the droplet's charge, the charge on a single electron being (1.592×10 −19 C).

11. Atomic Structure: Protons and Neutrons The results of this experiment gave Rutherford the means to arrive at two conclusions: one, an atom was much more than just empty space and scattered electrons (J.J. Thomson model argued), and two, an atom must have a positively charged center that contains most of its mass (which Rutherford termed as the nucleus).

12. 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.

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. Examples 1.How many protons, electrons and neutrons are present in an atom of 2.Write isotopic symbols in both forms for Selenium isotope with 40 neutrons 3.An atom has 32 electrons and 38 neutrons. What is its mass number and what is the element?

18. Atomic Masses and the Mole The mass of 1 atom of carbon-12 is defined to be 12 amu. Atomic Mass: weighed according to the natural abundance of each isotope. Sometimes called average atomic mass or relative atomic mass.

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 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.

22. 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

23. 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

24. A Mole of Atoms A mole is a unit of measurement used in chemistry to express amounts of a chemical substance, the same number of particles as there are carbon atoms in 12.0 g of carbon. 3

25. 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 2

26. 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 Molar Mass: The mass in grams of one mole of any element. It is numerically equivalent to its atomic mass. E.g 1 H atom = 1.01 amu 1 mol H = 1.01 g

27. 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 5

28. Avogadro’s Number and the Mole Copyright © 2008 Pearson Prentice Hall, Inc. Ch apt er 3/2828 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

29. Molar Mass from Periodic Table Molar mass is the atomic mass expressed in grams. 29 1 mole of Ag 1 mole of C 1 mole of S = 107.9 g = 12.01 g = 32.07 g

30. Examples Give the molar mass for each A. 1 mole of Li atoms =________ B. 1 mole of Co atoms =________ 30

31. 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 6

32. 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 10