1. SURVEY OF CHEMISTRY I CHEM 1151 CHAPTER 2 DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university

2. CHAPTER 2 ATOMS AND MOLECULES

3. THE ATOMIC THEORY OF MATTER Law of Constant Composition - The relative numbers and kinds of atoms are constant in a given compound - All samples of a given chemical compound have the same elemental composition Example - Water (H 2 O) always contains 1 g of H for every 8 g of O Law of Conservation of Mass (Matter) - The total mass of materials after a chemical reaction is equal to the total mass before the chemical reaction

4. Law of Multiple Proportions - When two or more elements combine to form a compound, their masses in that compound are in a fixed and definite ratio - Elements combine in a ratio of small whole numbers - If two elements form more than one compound, the ratios of the masses of the second element combined with a fixed mass of the first element will be in ratios of small whole numbers THE ATOMIC THEORY OF MATTER

5. Law of Multiple Proportions - C and O can combine to form CO and CO 2 CO 1.33 g O combine with 1.00 g C CO 2 2.66 g O combine with 1.00 g C - Ratio of O is 2.66 g : 1.33 g = 2 : 1 THE ATOMIC THEORY OF MATTER

6. Dalton’s Atomic Theory 1. All matter (every element) is made up of very small particles called atoms - Atoms are indivisible and indestructible 2. All atoms of a given element are identical in mass and properties - Atoms of a given element are different from atoms of all other elements THE ATOMIC THEORY OF MATTER

7. Dalton’s Atomic Theory 3. Compounds are formed from a combination of two or more different kinds of atoms - A given compound always has the same relative number and kind of atoms 4. A chemical reaction is a rearrangement of atoms - Atoms are neither created nor destroyed in a chemical reaction THE ATOMIC THEORY OF MATTER

8. Modern atomic theory is more involved but based on Dalton’s theory - Atoms can be destroyed by nuclear reactions but not by chemical reactions - There are different kinds of atoms within an element (isotopes - different masses, same properties) THE ATOMIC THEORY OF MATTER

9. THE ATOMIC STRUCTURE Atom - Is the smallest particle of an element that retains the chemical identity of the element - Is t he basic building block of ordinary matter - Made up of smaller particles (the building blocks of an atom) called subatomic particles Three Types of Subatomic Particles Electron: possesses a negative (-) electrical charge Proton: possesses a positive (+) electrical charge Neutron: has no charge (it is neutral)

10. THE ATOMIC STRUCTURE Electronic Charge equals 1.602177 x 10 -19 C (C = coulombs) - Charges are usually expressed as multiples of the electronic charge Charge of an electron = -1.602177 x 10 -19 C = -1 Charge of a proton = +1.602177 x 10 -19 C = +1 Atoms have no net electrical charge since they have equal number of electrons and protons

11. THE ATOMIC STRUCTURE - Protons and neutrons have very large masses (about 2000 x) as compared to electrons - Atoms generally have extremely small masses - Atomic Mass Unit (u) is used to express such small masses 1 u = 1.66054 x 10 -24 g or 1 g = 6.02212 x 10 23 u Charge Negative (-1) Positive (+1) Neutral (0) Particle Electron Proton Neutron Mass (g) 9.109 x 10 -28 1.673 x 10 -24 1.675 x 10 -24 Mass (u) 5.486 x 10 -4 1.0073 1.0087 Relative Mass 1 1837 1839

12. THE ATOMIC STRUCTURE - The center of an atom is small, dense, and positively charged called the nucleus - The nucleus contains all protons and neutrons and are referred to as necleons - The nucleus is, therefore, positively charged and contributes about 99.9% of the mass of an atom

13. THE ATOMIC STRUCTURE - The electrons move rapidly around the nucleus - Outer region called the extranuclear region - Account for most of the volume of an atom Electron Cloud - Volume occupied by electrons - Negatively charged

14. ATOMIC NUMBER (Z) - The number of protons in the nucleus of an atom - determines the identity of the element - Since atoms have no net electrical charge number of protons = number of electrons Z = number of protons = number of electrons

15. MASS NUMBER (A) - The sum of the number of protons and the number of neutrons in the nucleus of an atom -The total number of subatomic particles in the nucleus of an atom - The number of nucleons of an atom A = number of protons + number of neutrons number of neutrons = mass number - atomic number = A - Z

16. ATOMIC AND MASS NUMBERS MASS NUMBER ATOMIC NUMBER CHEMICAL SYMBOL SYMBOL A Z C 12 6 O 16 8 Ca 40 20 Mass number is the superscript to the left Atomic number is the subscript to the left

17. An atom has an atomic number of 56 and a mass number of 138. What are the numbers of protons, electrons, and neutrons present in the atom? What is the number of subatomic particles present in the nucleus of the atom? Number of protons = atomic number = 56 Number of electrons = atomic number = 56 Number of neutrons = mass number – atomic number = 138-56 = 82 Number of subatomic particles in the nucleus = mass number = 138 ATOMIC AND MASS NUMBERS

18. CHEMICAL PROPERTIES OF ATOMS - The number of protons (the atomic number) characterizes an atom - Electrons determine the chemical properties of an atom - Atoms with the same atomic number have the same chemical properties - Atoms with the same atomic number are atoms of the same element

19. Chapter 1 definition of An Element - Is a pure substance that cannot be reduced to a simpler substance by normal chemical means Chapter 2 definition of An Element - Is a pure substance in which all atoms present have the same atomic number CHEMICAL PROPERTIES OF ATOMS

20. ISOTOPES - Atoms of an element with the same atomic number but different mass numbers - Atoms of an element with the same number of protons and the same number of electrons but different numbers of neutrons - Isotopes of an element have the same chemical properties but slightly different physical properties - The atomic number is usually omitted since it is the same for isotopes of a given element

21. ISOTOPES 12 6 C 666 11 1314 H 123 11 1 Si 282930 14 Most abundant is carbon-12 Most abundant is silicon-28 Most abundant is hydrogen-1 CCC HH Si

22. AVERAGE ATOMIC MASS - Determined by using the masses of an element’s various isotopes and their respective natural abundances Units 1 u = 1.66054 x 10 -24 g or 1 g = 6.02214 x 10 23 u u (amu): atomic mass unit u is defined by assigning a mass of exactly 12 u to an atom of carbin-12 (reference point)

23. For an element with n isotopes which have atomic masses in u (m 1, m 2, m 3,….., m n ) and natural abundances expressed as fractions (x 1, x 2, x 3,……,x n ) Average Atomic Mass = m 1 x 1 + m 2 x 2 + m 3 x 3 +….+ m n x n The natural abundance is usually expressed as a percentage Divide by 100 to convert to the decimal form (fractional abundance) AVERAGE ATOMIC MASS

24. The mass spectrometer is an instrument used to measure the masses and relative (natural) abundances of the isotopes present in a sample of an element Homework Describe the operation and uses of the mass spectrometer One page maximum and double spaced AVERAGE ATOMIC MASS

25. Naturally occurring copper is 69.09% 63 Cu, which has a relative mass of 62.93 u, and 30.91% 65 Cu, which has a relative mass of 64.93 u. Calculate the average atomic mass of copper. 63 Cu natural abundance = 69.09% fractional abundance = 69.09/100 = 0.6909 65 Cu natural abundance = 30.91% fractional abundance = 30.91/100 = 0.3091 Average Atomic Mass = (62.93)(0.6909) + (64.93)(0.3091) = 63.5478 = 63.55 u AVERAGE ATOMIC MASS

26. FORMULA MASS - The sum of atomic masses of all the atoms present in the chemical formula of a substance - Relative mass based on the carbon-12 relative-mass scale -It is advisable to use two decimal places for atomic masses

27. Calculate the formula mass of H 2 SO 4 H: 2 x 1.01 u = 2.02 u S: 1 x 32.06 u = 32.06 u O: 4 x 16.00 u = 64.00 u Formula mass = (2.02 + 32.06 + 64.00) u = 98.08 u FORMULA MASS Calculate the formula mass of H 2 O H: 2 x 1.01 u = 2.02 u O: 1 x 16.00 u = 16.00 u Formula mass = (2.02 + 16.00) u = 18.02 u

28. Calculate the formula mass of Fe 2 (SO 4 ) 3 Fe: 2 x 55.85 u = 111.70 u S: 3 x 32.07 u = 96.21 u O: 12 x 16.00 u = 192.00 u Formula mass = (111.70 + 96.21 + 192.00) u = 399.91 u FORMULA MASS Calculate the formula mass of CaCO 3 Ca: 1 x 40.08 u = 40.08 u C: 1 x 12.01 u = 12.01 u O: 3 x 16.00 u = 48.00 u Formula mass = (40.08 + 12.01 + 48.00) u = 100.09 u

29. THE MOLE The amount of substance of a system, which contains as many elementary entities as there are atoms in 12 grams of carbon-12 - abbreviated mol 1 mole (mol) = 6.02214179 x 10 23 entities - known as the Avogadro’s number (after Amedeo Avogadro) - usually rounded to 6.022 x 10 23

30. THE MOLE The number of entities (or objects) can be atoms or molecules 1 mol C = 6.022 x 10 23 atoms C 1 mol CO 2 = 6.022 x 10 23 molecules CO 2 2 conversion factors can be derived from each

31. THE MOLE How many atoms are there in 0.40 mole nitrogen? = 2.4 x 10 23 nitrogen atoms How many molecules are there in 1.2 moles water? = 7.2 x 10 23 water molecules

32. How many H atoms are there in 1.2 moles water? = 1.4 x 10 24 H atoms THE MOLE

33. MOLAR MASS - The mass of a substance in grams that is numerically equal to the formula mass of that substance - Add atomic masses to get the formula mass (in u) = molar mass (in g/mol) - The mass, in grams, of 1 mole of the substance

34. MOLAR MASS Consider the following Sodium (Na) has an atomic mass of 22.99 u This implies that the mass of 1 mole of Na = 22.99 g Molar mass of Na = 22.99 g/mol Formula mass of NaCl = 58.44 u The mass of 1 mole of NaCl = 58.44 g Molar mass of NaCl = 58.88 g/mol Formula mass of CaCO 3 = 100.09 u The mass of 1 mole of CaCO 3 = 100.09 g Molar mass of CaCO 3 = 100.09 g/mol

35. MOLAR MASS Calculate the mass of 2.4 moles of NaNO 3 Molar mass NaNO 3 = 22.99 + 14.01 + 3(16.00) = 85.00 g /mol NaNO 3 = 204 g NaNO 3 = 2.0 x 10 2 g NaNO 3

36. MOLAR MASS How many moles are present in 2.4 g NaNO 3 Molar mass NaNO 3 = 22.99 + 14.01 + 3(16.00) = 85.00 g /mol NaNO 3 = 0.028 mol NaNO 3 = 2.8 x 10 -2 mol NaNO 3

37. CHEMICAL FORMULA Subscripts represent both atomic and molar amounts Consider Na 2 S 2 O 3 : - Two atoms of sodium, two atoms of sulfur, and three atoms of oxygen are present in one molecule of Na 2 S 2 O 3 - Two moles of sodium, two moles of sulfur, and three moles of oxygen are present in one mole of Na 2 S 2 O 3

38. CHEMICAL FORMULA How many moles of sodium atoms, sulfur atoms, and oxygen atoms are present in 1.8 moles of a sample of Na 2 S 2 O 3 ? I mol Na 2 S 2 O 3 contains 2 mol Na, 2 mol S, and 3 mol O

39. CHEMICAL CALCULATIONS Calculate the number of molecules present in 0.075 g of urea, (NH 2 ) 2 CO Given mass of urea: - Convert to moles of urea using molar mass - Convert to molecules of urea using Avogadro’s number = 7.5 x 10 20 molecules (NH 2 ) 2 CO

40. CHEMICAL CALCULATIONS How many grams of carbon are present in a 0.125 g of vitamin C, C 6 H 8 O 6 ? Given mass of vitamin C: - Convert to moles of vitamin C using molar mass - Convert to moles of C (1 mole C 6 H 8 O 6 contains 6 moles C) - Convert moles carbon to g carbon using molar mass = 0.0511 g carbon

41. PERCENTAGE COMPOSITION - Percentage by mass contributed by individual elements in a compound

42. PERCENTAGE COMPOSITION Calculate the percentage of carbon, hydrogen, and oxygen, in ethanol (C 2 H 5 OH)

43. PERCENTAGE COMPOSITION Calculate the percent composition by mass of each element in the following compounds C 9 H 8 O 4 (NH 4 ) 2 PtCl 4 C 2 H 2 F 4 C 8 H 10 N 4 O 2 Pt(NH 3 ) 2 Cl 2

44. EMPIRICAL FORMULA Given mass % elements: - Convert to g elements assuming 100.0 g sample - Convert to mole elements using molar mass - Calculate mole ratio (divide each by the smallest number of moles) - Round each to the nearest integer - Multiply through by a suitable factor if necessary ( __.5 x 2or __.33 x 3or__.25 x 4)

45. EMPIRICAL FORMULA Determine the empirical formula for a compound that gives the following percentages upon analysis (in mass percents): 71.65 % Cl24.27 % C4.07 % H - Assume 100.0 g of sample and convert grams to moles 71.65 g Cl 24.27 g C 4.07 g H

46. EMPIRICAL FORMULA - Calculate mol ratios - Round to nearest integers and write empirical formula Cl: 1, C: 1, H: 2 giving CH 2 Cl