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SURVEY OF CHEMISTRY I CHEM 1151 CHAPTER 2

SURVEY OF CHEMISTRY I CHEM 1151 CHAPTER 2. DR. AUGUSTINE OFORI AGYEMAN Assistant professor of chemistry Department of natural sciences Clayton state university. CHAPTER 2 ATOMS AND MOLECULES. THE ATOMIC THEORY OF MATTER. Law of Constant Composition

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SURVEY OF CHEMISTRY I CHEM 1151 CHAPTER 2

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  1. SURVEY OF CHEMISTRY I CHEM 1151CHAPTER 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 (H2O) 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. THE ATOMIC THEORY OF MATTER • 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

  5. THE ATOMIC THEORY OF MATTER Law of Multiple Proportions - C and O can combine to form CO and CO2 CO 1.33 g O combine with 1.00 g C CO2 2.66 g O combine with 1.00 g C - Ratio of O is 2.66 g : 1.33 g = 2 : 1

  6. THE ATOMIC THEORY OF MATTER 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

  7. THE ATOMIC THEORY OF MATTER 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

  8. THE ATOMIC THEORY OF MATTER 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)

  9. THE ATOMIC STRUCTURE Atom - Is the smallest particle of an element that retains the chemical identity of the element - Is the 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 1023 u Mass (u) 5.486 x 10-4 1.0073 1.0087 Relative Mass 1 1837 1839 Particle Electron Proton Neutron Charge Negative (-1) Positive (+1) Neutral (0) Mass (g) 9.109 x 10-28 1.673 x 10-24 1.675 x 10-24

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

  17. ATOMIC AND MASS NUMBERS 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

  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. CHEMICAL PROPERTIES OF ATOMS 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

  20. ISOTOPES - Atoms of an element with the sameatomic 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 samechemical properties but slightly differentphysical properties - The atomic number is usually omitted since it is the same for isotopes of a given element

  21. ISOTOPES 11 12 13 14 C C C C Most abundant is carbon-12 6 6 6 6 1 2 3 H H H Most abundant is hydrogen-1 1 1 1 28 29 30 Si Si Si Most abundant is silicon-28 14 14 14

  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 1023 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. AVERAGE ATOMIC MASS For an element with n isotopes which have atomic masses in u (m1, m2, m3,….., mn) and natural abundances expressed as fractions (x1, x2, x3,……,xn) Average Atomic Mass = m1x1 + m2x2 + m3x3 +….+ mnxn The natural abundance is usually expressed as a percentage Divide by 100 to convert to the decimal form (fractional abundance)

  24. AVERAGE ATOMIC MASS 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

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

  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. FORMULA MASS Calculate the formula mass of H2O 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 Calculate the formula mass of H2SO4 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

  28. FORMULA MASS Calculate the formula mass of CaCO3 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 Calculate the formula mass of Fe2(SO4)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

  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 1023 entities - known as the Avogadro’s number (after Amedeo Avogadro) - usually rounded to 6.022 x 1023

  30. THE MOLE The number of entities (or objects) can be atoms or molecules 1 mol C = 6.022 x 1023 atoms C 1 mol CO2 = 6.022 x 1023 molecules CO2 2 conversion factors can be derived from each

  31. THE MOLE How many atoms are there in 0.40 mole nitrogen? = 2.4 x 1023 nitrogen atoms How many molecules are there in 1.2 moles water? = 7.2 x 1023 water molecules

  32. THE MOLE How many H atoms are there in 1.2 moles water? = 1.4 x 1024 H atoms

  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 CaCO3 = 100.09 u The mass of 1 mole of CaCO3 = 100.09 g Molar mass of CaCO3 = 100.09 g/mol

  35. MOLAR MASS Calculate the mass of 2.4 moles of NaNO3 Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00) = 85.00 g /mol NaNO3 = 204 g NaNO3 = 2.0 x 102 g NaNO3

  36. MOLAR MASS How many moles are present in 2.4 g NaNO3 Molar mass NaNO3 = 22.99 + 14.01 + 3(16.00) = 85.00 g /mol NaNO3 = 0.028 mol NaNO3 = 2.8 x 10-2 mol NaNO3

  37. CHEMICAL FORMULA Subscripts represent both atomic and molar amounts Consider Na2S2O3: - Two atoms of sodium, two atoms of sulfur, and three atoms of oxygen are present in one molecule of Na2S2O3 - Two moles of sodium, two moles of sulfur, and three moles of oxygen are present in one mole of Na2S2O3

  38. CHEMICAL FORMULA How many moles of sodium atoms, sulfur atoms, and oxygen atoms are present in 1.8 moles of a sample of Na2S2O3? I mol Na2S2O3 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, (NH2)2CO Given mass of urea: - Convert to moles of urea using molar mass - Convert to molecules of urea using Avogadro’s number = 7.5 x 1020 molecules (NH2)2CO

  40. CHEMICAL CALCULATIONS How many grams of carbon are present in a 0.125 g of vitamin C, C6H8O6? Given mass of vitamin C: - Convert to moles of vitamin C using molar mass - Convert to moles of C (1 mole C6H8O6 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 (C2H5OH)

  43. PERCENTAGE COMPOSITION Calculate the percent composition by mass of each element in the following compounds C9H8O4 (NH4)2PtCl4 C2H2F4 C8H10N4O2 Pt(NH3)2Cl2

  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 2 or __.33 x 3 or __ .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 % Cl 24.27 % C 4.07 % 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 CH2Cl

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